Contents

How to use these notes v

Supplementary material and references vi

1 A preview of vector calculus 1

1.1 A preview of vector calculus . . . . . . . . . . . . . . . . . . . . 1

2 One-forms and vector fields 4

2.1 One-forms and vector fields . . . . . . . . . . . . . . . . . . . . . 4
2.2 Exact one-forms and conservative vector fields . . . . . . . . . . . . . 7
2.3 Changes of variables . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4 The pullback of a one-form . . . . . . . . . . . . . . . . . . . . . 19

3 Integrating one-forms: line integrals 26

3.1 Integrating a one-form over an interval in R . . . . . . . . . . . . . . 26
3.2 Parametric curves in Rn . . . . . . . . . . . . . . . . . . . . . . 31
3.3 Line integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.4 Fundamental Theorem of line integrals . . . . . . . . . . . . . . . . 47
3.5 Applications of line integrals . . . . . . . . . . . . . . . . . . . . 53
3.6 Poincare’s lemma for one-forms . . . . . . . . . . . . . . . . . . . 57

4 k-forms 66
4.1 Differential forms revisited: an algebraic approach . . . . . . . . . . . . 66
4.2 Multiplying k-forms: the wedge product . . . . . . . . . . . . . . . . 73
4.3 Differentiating k-forms: the exterior derivative . . . . . . . . . . . . . 79
4.4 The exterior derivative and vector calculus . . . . . . . . . . . . . . . 90
4.5 Physical interpretation of grad, curl, div . . . . . . . . . . . . . . . . 99
4.6 Exact and closed k-forms . . . . . . . . . . . . . . . . . . . . . . 109
4.7 The pullback of a k-form . . . . . . . . . . . . . . . . . . . . . . 119
4.8 Hodge star . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

vii
CONTENTS viii

5 Integrating two-forms: surface integrals 144

5.1 Integrating zero-forms and one-forms . . . . . . . . . . . . . . . . . 144
5.2 Orientation of a region in R2 . . . . . . . . . . . . . . . . . . . . 150
5.3 Integrating a two-form over a region in R2 . . . . . . . . . . . . . . . 158
5.4 Parametric surfaces in Rn . . . . . . . . . . . . . . . . . . . . . 172
5.5 Orientation of parametric surfaces in R3 . . . . . . . . . . . . . . . . 182
5.6 Surface integrals . . . . . . . . . . . . . . . . . . . . . . . . . 192
5.7 Green’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . 205
5.8 Stokes’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . 222
5.9 Applications of surface integrals . . . . . . . . . . . . . . . . . . . 236

6 Beyond one- and two-forms 245

6.1 Generalized Stokes’ theorem . . . . . . . . . . . . . . . . . . . . 245
6.2 Divergence theorem in R3 . . . . . . . . . . . . . . . . . . . . . 248
6.3 Divergence theorem in Rn . . . . . . . . . . . . . . . . . . . . . 259
6.4 Applications of the divergence theorem . . . . . . . . . . . . . . . . 265
6.5 Integral theorems: when to use what . . . . . . . . . . . . . . . . . 269

7 Unoriented line and surface integrals 271

7.1 Unoriented line integrals . . . . . . . . . . . . . . . . . . . . . . 271
7.2 Unoriented surface integrals . . . . . . . . . . . . . . . . . . . . . 277
7.3 Applications of unoriented line and surface integrals . . . . . . . . . . . 285

Appendices

A List of results 294

B List of definitions 297

C List of examples 300

D List of exercises 303

Chapter 1

A preview of vector calculus

1.1 A preview of vector calculus

1.1.1 Motivation
Have you ever wondered why, in the definition of definite integrals
⁄ b
f (x) dx,
a

there is always a “dx” inside the integrand? What does it mean, and why is it there? When
you were introduced to definite integrals in Calculus I, they were defined as limits of Riemann
sums: ⁄ n
b ÿ
f (x) dx = lim f (a + i x) x,
a næŒ
i=1

where x is the width of the rectangles in the Riemann sum. It was then argued that in the
limit as n æ Œ, the width of the rectangles goes to zero, and somehow x becomes “dx”,
which is some sort of infinitesimal width of the rectangles. But what does that mean, really?
What is this “dx” thing inside the definite integral?
And it’s not like we can just forget about it, even though many students actually do in
first year. :-) We all know how important dx is: just think of substitution for definite integrals.
Without the dx, substitution would fail miserably. It must be there. Why?
Things become even more interesting with double and triple integrals:
⁄⁄ ⁄⁄⁄
f dA, f dV,
D D

with dA = dxdy and dV = dxdydz in Cartesian coordinates. What are these objects dA and
dV , and why must they be there in the integrand?
What we will do in this course is provide an answer to this question. Our goal is to
define a unified theory of integration for curves, surfaces, and volumes, which will make
the appearance of dx, dA, and dV natural. To achieve this, we must define a new type of
objects that will play the role of integrands: those are called differential forms, or n-forms.
More specifically, one-forms are objects that can be integrated over curves, two-forms over

1
CHAPTER 1. A PREVIEW OF VECTOR CALCULUS 2

surfaces, and three-forms over volumes. In other words, it was all a big lie: what you should
be integrating is not functions, but rather differential forms!
But before we start, we can already identify two key guiding principles for the construction,
using what we already know about integration.

• We want our theory to be “reparametrization-invariant”.

s
Consider a definite integral
s
in
one dimension. Instead of writing ab f (x) dx, we would like to write something like C Ê,
where C stands for a curve, and Ê for the integrand, which will be called a "one-form".
(We will also allow C to be a curve in R2 and R3 ). More precisely, to make s
sense of
this expression, we will need C to be a parametric curve. However, we want C Ê to be
defined intrinsically in terms of the geometry of the curve itself: we do not want the
integral to depend on the choice of parametrization. This is key. This constraint will
be satisfied if the integrand Ê transforms in a specific way under reparametrizations
of the curve; in one dimension this will reproduce the substitution formula for definite
integrals.
s
• We want our theory to be “oriented”. Consider C Ê as above. As mentioned, to make
sense of this expression we will work with a parametric curve C. But once we parametrize
a curve, we introduce a choice of “orientation”: we select one of the two endpoints as the
starting point, and we introduce a “direction of travel along the curve” (the orientation
is given by the direction of the tangent vector, or velocity, for a parametric curve). If
we do a reparametrization of the curve that reverses the orientation, should the integral
remain
sa
invariant?
sb
The answer is no! We see this directly in basic calculus: we know that
b f (x) dx = ≠ a f (x) dx. If we interpret the first integral as being over the interval
[a, b] with direction of travel from a to b, and the second integral as being over the same
interval but with the reverse orientation, then we see that exchanging the direction of
travel over the interval [a, b] changes the sign of the integral.

Putting these two properties together, we see that if C and s

C Õ ares two parametrizations of
the same curve that preserve the s
orientation,
s
we must have C Ê = C Õ Ê, while if they reverse
the orientation, we must have C Ê = ≠ C Õ Ê. This is what it means to say that the theory
should be “oriented” and “reparametrization-invariant”.
This brief exposition focused on curves, but following these two guiding principles we
will develop a unified theory of oriented integrals not only over curves, but over surfaces and
volumes as well, using the machinery of differential forms. Along the way we will discover the
beautiful intricacies of vector calculus, culminating with the very important Stokes’ Theorem.
Let us embark on this journey together!

1.1.2 Vector calculus and differential forms: two sides of the same coin
What we will study in this course is known as vector calculus. There are two main approaches
to vector calculus. On the one hand, there is the “traditional” approach, which involves
concepts such as gradient, curl, div, etc. It relies heavily on the geometry of R3 , and is very
explicit. But at first it seems like a complicated amalgation of strange constructions and
definitions that satisfy all kinds of intricate identities. My recollections of learning vector
calculus is that it involves many formulae that appear to come out of nowhere and that one
needs to learn by heart. Not fun.
CHAPTER 1. A PREVIEW OF VECTOR CALCULUS 3

On the other hand, there is the “modern” approach, pioneered by Cartan, which relies
on the definition of differential forms. This approach is a little more abstract, but is much
more unified and elegant. It brings together all the concepts of vector calculus in a unified
formalism, from which all the identities and formulae come out naturally. It also does not
rely on the geometry of R3 , and is naturally generalized to Rn (even though we will focus on
R3 in this course). I remember this feeling of “ah, now this all makes sense!” when I learned
differential forms later on in my studies.
In this course we will take the perhaps non-traditional approach of introducing vector
calculus directly through the unified formalism of differential forms, guided by the exposition
of the previous subsection. The challenge is to make the concepts accessible to second-year
students, stripping them down from the fancier definitions of differential geometry. But I
truly believe that this is possible, and that it will make the whole theory of vector calculus
much more interesting and unified, and less reliant on brute force memorization.
But, at the same time, it is important for students to be fluent with the traditional
concepts of vector calculus. Indeed, students who will study topics like fluid mechanics,
electromagnetism, applied mathematics, etc. will repeadtedly encounter vector calculus,
usually expressed in the traditional language. Moreover, traditional concepts such as grad, div,
curl, are often useful for explicit calculations. So in this course we will translate all concepts
from differential forms to standard vector calculus every step of the way.
In the end, the goal is for students to be fluent with both approaches: to see the beauty
and elegance of differential forms, and to be able to use the traditional approach for explicit
calculations.
Chapter 2

One-forms and vector fields

In this section we study one-forms, which will become the objects that can be integrated along
curves. The counterpart to one-forms in traditional vector calculus is vector fields.

2.1 One-forms and vector fields

We define the concept of differential one-forms (or more simply one-forms), which will become
the objects that can be integrated along curves in our theory of integration. We also define
vector fields, and show that the two are closely connected: given a one-form, there is an
associated vector field, and vice-versa.

Objectives
You should be able to:

• Define one-forms and vector fields in R, R2 , R3 , and more generally Rn .

• Translate between one-forms and vector fields.

• Visualize a vector field in R2 by plotting vectors on a grid.

2.1.1 One-forms
Let us start by defining one-forms in R, R2 , R3 , and more generally Rn .
Definition 2.1.1 One-forms. A differential one-form (or simply one-form) on an open
interval (or union of open intervals) U ™ R is an expression of the form

Ê = f (x) dx

with f : U æ R a function with continuous derivatives (we say that the function f is “smooth”,
or C Π, on U ).
A one-form on an open subset1 U ™ R2 is an expression of the form

Ê = f (x, y) dx + g(x, y) dy

4
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 5

with f, g : U æ R functions with continuous partial derivatives (again, we say that they are
smooth, or C Π, on U ).
A one-form on an open subset U ™ R3 is an expression of the form

Ê = f (x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz

for smooth functions f, g, h : U æ R.

More generally, a one-form on an open subset U ™ Rn is an expression of the form
n
ÿ
Ê= fi (x1 , . . . , xn ) dxi
i=1

for smooth functions fi : U æ R, i = 1, . . . , n. ⌃

For the time being, you can think of the objects dx, dy and dz (or more generally dxi ) as
placeholders; they will be given an interpretation shortly.
We note here two obvious properties of one-forms, which follow from the definition.

1. If Ê and ÷ are one-forms on U , then the sum Ê + ÷ is also a one-form on U .

2. If Ê is a one-form on U and f a smooth function on U , then the product f Ê is also a

one-form on U .

2.1.2 Vector fields

There is a closely related concept in vector calculus, which is called a vector field.
Definition 2.1.2 Vector fields. A vector field on an open subset U ™ Rn is a function
F : U æ Rn . In other words, it is a rule which assigns to each point in U ™ Rn a vector with
n components. ⌃
For example, we can write a vector field F in U ™ R as2

F(x, y) = (f1 (x, y), f2 (x, y)),

where f1 , f2 are functions f1 , f2 : U æ R. We call these functions the component functions

of the vector field. We say that the vector field is smooth if the component functions are
smooth.
There are many examples of vector fields in physics: think of the velocity vector field of a
moving fluid, or the gravitational force vector field produced by a mass.
Looking at the two definitions above, we see a natural correspondence between one-forms
and vector fields.
Principle 2.1.3 Correspondence between one-forms and vector fields in R3 . Given
a one-form Ê = f dx + g dy + h dz, we can define an associated smooth vector field with
component functions (f, g, h). Conversely, given a smooth vector field F = (f, g, h), we can
define an associated one-form Ê = f dx + g dy + h dz.
1
A subset U ™ R2 is open if for every point P œ U there is an open ball (an open disk in R2 , an open
sphere in R3 ) centered at P that lies entirely in U . This is a generalization of open intervals (and unions of
open integrals) to two and three dimensions.
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 6

Of course a similar correspondence holds in Rn . This is the starting point for the dictionary
between the modern approach (one-forms) and the traditional approach (vector fields) to
vector calculus.
Example 2.1.4 A one-form and its associated vector field. The expression

Ê = x dx + x sin(y) dy

is a one-form on R2 . The associated vector field is the function F : R2 æ R2 given by

F(x, y) = (x, x sin(y)), which assigns the vector (x, x sin(y)) to the point (x, y) œ R2 . ⇤

2.1.3 Exercises
1. Consider the vector field in R3 given by F(x, y, z) = (x, sin y, xyz). Write the correspond-
ing one-form.
Solution. The corresponding one-form is Ê = x dx + sin y dy + xyz dz.
2. Consider the vector field v(x, y) = (≠y, x) in R2 .

(a) Plot the given vector field at points with integer coordinates.

(b) If the vector field represents the velocity of a flowing fluid, what kind of motion is
this vector field describing?

Solution.

(a) We can plot the vector field by plotting arrows for points in R2 with integer
coordinates. For instance, at the point (1, 1), the vector field would be the vector
v(1, 1) = (≠1, 1). At (0, 1), we would get v(0, 1) = (≠1, 0). And so on and so forth.
The result is the following plot, which was obtained via this geogebra app.

Figure 2.1.5 A plot of the vector field v(x, y) = (≠y, x).

CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 7

(b) Looking at the plot, we see that the fluid is rotating around the origin counter-
clockwise. This type of fluid motion is called “vortex”.
3. Consider a vector field v(x, y) = (5, 1). Suppose that it is the velocity field of a moving
fluid. Suppose that you drop an object at the origin at time t = 0. Where will the object
be at time t = 2?
Solution. The velocity field v(x, y) = (5, 1) is constant, which means that all points in
the fluid are moving at the same constant velocity. In other words, each point in the
fluid is going through a linear motion of the form

x(t) = x0 + t(5, 1),

where x0 is some initial position. If we drop an object at the origin at t = 0, its position
vector is then described by x(t) = t(5, 1) since its initial position is x0 = (0, 0). We
conclude that at time t = 2, the object is at position x(2) = 2(5, 1) = (10, 2).
4. Consider the object Ê = ≠ x2 +y
y
2 dx +
x
x2 +y 2
dy. Is this a one-form on R2 ?
Solution. Ê is not a one-form on R2 , because its component functions are not defined
at the origin (0, 0) œ R2 (we would be dividing by zero). However, it is a one-form on
the open subset U = R2 \ {(0, 0)} where we removed the origin,2 since the component
functions are smooth on U .

2.2 Exact one-forms and conservative vector fields

We introduce a class of one-forms that arise as differentials of a function; the associated vector
field is the gradient of the function. But not all one-forms are differentials of functions: we
call such one-forms “exact”, and their corresponding vector fields “conservative”.

Objectives
You should be able to:

• Calculate the differential of a function, and intepret it as a one-form.

• Translate between the differential of a function and its gradient.

• Define exact one-forms.

• Define conservative vector fields and their associated potentials.

• Translate between exact one-forms and conservative vector fields.

• Define closed one-forms using partial derivatives.

• Show that exact one-forms are always closed.

• Rephrase the statement as the "screening test" for conservative vector fields.

• Use the screening test to show that a given vector field cannot be conservative.
2
The notation R2 \ {(0, 0)} means “R2 minus the point (0, 0)”.
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 8

2.2.1 Differential of a function

One-forms may seem strange, but in fact there is a large class of one-forms that can be
obtained directly from functions.
Definition 2.2.1 Differential of a function. Let f be a smooth function on an open
subset U ™ R3 . Its differential (it will become known as “exterior derivative” shortly) is

ˆf ˆf ˆf
df = dx + dy + dz.
ˆx ˆy ˆz

It is a one-form on U . A similar definition of course holds for R, R2 , and more generally Rn .

⌃
Using the correspondence in Principle 2.1.3 between one-forms and vector fields, we can
find the vector field associated to the differential of a function.
Fact 2.2.2 Correspondence between the differential and the gradient of a function.
The vector field associated to the differential df of a function f is the vector field with component
functions 3 4
ˆf ˆf ˆf
, , ,
ˆx ˆy ˆz
which is nothing else but the gradient Òf .
Recall from Calculus III that at a given point (x0 , y0 , z0 ) œ U , the gradient vector field
Òf (x0 , y0 , z0 ) gives the direction of fastest increase of f . In three dimensions, this direction
is orthogonal to the level surfaces. In two dimensions, it is perpendicular to the level curves:
think of a topographical map, the direction of fastest increase is the direction perpendicular
to the contour lines.
Example 2.2.3 The differential and gradient of a function. Consider the function
f (x, y) = x2 ey . Its differential is

ˆf ˆf
df = dx + dy = 2xey dx + x2 ey dy,
ˆx ˆy

which is a one-form on R2 . Its associated vector field is the function F : R2 æ R2 with

1 2 3 4
ˆf ˆf
F(x, y) = 2xey , x2 ey = , = Òf.
ˆx ˆy
⇤
Remark 2.2.4 Using the definition of the differential of a function, we can somewhat make
sense of the placeholders dx, dy, and dz. Consider for instance the function f (x, y, z) = x. Its
differential becomes
ˆf ˆf ˆf
df = dx + dy + dz = dx.
ˆx ˆy ˆz
In other words, the placeholders dx, dy, and dz are simply the differentials of the component
functions x, y, and z on R3 .
Note that we will give a more satisfying algebraic meaning for these placeholders later on
in Subsection 4.1.1.
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 9

2.2.2 Exact one-forms and conservative vector fields

Differentials of functions are one-forms, but not all one-forms are differentials of functions,
just like not all vector fields can be written as the gradient of a function. Such one-forms and
vector fields are special, and hence have their own name.
Definition 2.2.5 Exact one-forms. We say that a one-form Ê on U ™ R3 is exact if
Ê = df for some function f on U . ⌃
We can define a similar concept for vector fields.
Definition 2.2.6 Conservative vector fields. A vector field F on U ™ R3 is conservative
if F = ÒV for some function V on U . We call V a potential for F.1 ⌃
Clearly, by Fact 2.2.2, we see that if a one-form is exact, its associated vector field is
conservative, and vice-versa.
Example 2.2.7 An exact one-form and its associated conservative vector field.
Consider the one-form Ê = cos x dx ≠ sin y dy on R2 . Ê is an exact one-form, since it can be
written as Ê = df for the function f (x, y) = sin x + cos y. In the language of vector fields,
this is the statement that the vector field F : R2 æ R2 associated to Ê, which is given by
F(x, y) = (cos x, ≠ sin y), is conservative, since it can be written as F = Òf for the function
f above. The function f is called the potential of the vector field F. ⇤
Example 2.2.8 The gravitational force field is conservative. The gravitational force
field F : R3 \ {(0, 0, 0)} æ R3 that a mass M at the origin exerts on a mass m at position
(x, y, z) is given by
GM m
F(x, y, z) = ≠ (x, y, z),
r3

where r = x2 + y 2 + z 2 and G is the gravitional constant. The corresponding one-form is

GM m
Ê=≠ (x dx + y dy + z dz) .
r3
It is easy to see that the gravitational force field is conservative, or equivalently that the
one-form Ê is exact, since Ê = dÏ with the function Ï given by
GM m
Ï(x, y, z) = .
r
Ï is the potential function, which from physics you may recognize as minus the gravitational
potential energy. ⇤

2.2.3 Closed one-forms in R2

Since not all one-forms can be written as differentials of functions, i.e. not all one-forms
are exact, a natural question arises: can we determine, looking at a one-form, whether it
is exact or not? Similarly, can we easily determine whether a vector field is conservative?
Unfortunately we will not be able to fully answer this question at the moment, we will come
1
Note that in physics, the potential of a conservative vector field is usually defined as F = ≠ÒV , with an
extra minus sign. The difference is purely conventional; the minus sign is introduced so that when F is a force
field, then V becomes the potential energy physically.
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 10

back to it in Section 3.6. For the time being, we will be able to find a necessary condition for a
one-form to be exact, which in the context of vector calculus is sometimes called a "screening
test" for conservative vector fields.
Let us focus first on one-forms and vector fields on U ™ R2 .
Definition 2.2.9 Closed one-forms in R2 . We say that a one-form Ê = f dx + g dy on
U ™ R2 is closed if
ˆf ˆg
= .
ˆy ˆx
⌃
This definition may seem ad hoc, but it is important because of the following lemma.
Lemma 2.2.10 Exact one-forms in R2 are closed. If a one-form Ê on U ™ R2 is exact,
then it is closed.
Proof. Suppose that Ê is exact: then it can be written as
ˆF ˆF
Ê = f dx + g dy = dF = dx + dy
ˆx ˆy

for some smooth function F on U ™ R2 . It will then be closed if

ˆf ˆ2F
=
ˆy ˆyˆx
is equal to
ˆg ˆ2F
= .
ˆx ˆxˆy
Equality of the two expressions follows from the Clairaut-Schwarz theorem, which states that
partial derivatives commute, as long as they are continuous. But continuity of the partial
derivatives is guaranteed by the fact that all partial derivatives of F exist and are continuous,
since by definition (see Definition 2.1.1) one-forms are assumed to have smooth component
functions. Therefore the one-form is closed. ⌅
Note however that the converse statement is not necessarily true: not all closed one-forms
on U ™ R2 are exact. In fact, the question of when closed one-forms are exact is an important
one; the result is known as Poincare’s lemma. We will come back to this in Section 3.6. But
what Lemma 2.2.10 tells us is that one-forms that are not closed cannot be exact.
There is of course an analogous statement for conservative vector fields. The only minor
difference is that we need to impose a condition on the component functions of vector fields,
since vector fields are not always smooth by definition (see Definition Definition 2.1.2.
Lemma 2.2.11 Screening test for conservative vector fields in R2 . If a vector field
F = (f1 , f2 ) on U ™ R2 is conservative and has component functions that are continuously
differentiable, then it passes the screening test:
ˆf1 ˆf2
= .
ˆy ˆx
Proof. Same as for Lemma 2.2.10, but for the associated vector fields. ⌅
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 11

In the context of vector fields, it is called a screening test, because it is a quick test to
determine whether a vector field has a chance at all to be conservative. In other words, if a
vector field does not pass the screening test, then it is certainly not conservative. However,
if it passes the screening test, at this stage we cannot conclude anything. Just like closed
one-forms are not necessarily exact.
Example 2.2.12 Exact one-forms are closed. Consider the exact one-form on R2 from
Example 2.2.7, Ê = cos x dx ≠ sin y dy. We show that it is closed, according to Definition 2.2.9.
The partial derivatives are easily calculated:
ˆ ˆ
cos x = 0, (≠ sin y) = 0.
ˆy ˆx
Thus Definition 2.2.9 is satisfied, and Ê is closed. ⇤
Example 2.2.13 Closed one-forms are not necessarily exact. If a vector field is
conservative, then it passes the screening test. Correspondingly, if a one-form is exact, then
it is closed. But the converse statement is not necessarily true (we will revisit it later in
Section 3.6). Consider for instance the one-form
y x
Ê=≠ dx + 2 dy.
x2 + y 2 x + y2
Calculating the partial derivatives for Definition 2.2.9, we get that
3 4 3 4
ˆ y y 2 ≠ x2 ˆ x y 2 ≠ x2
≠ 2 = , = .
ˆy x + y2 x2 + y 2 ˆx x + y2
2 x2 + y 2
The two expressions are equal, and thus Ê is closed. However, one can show that Ê is not
exact: there does not exist a function f such that Ê = df (see Exercise 3.4.3.2). ⇤

2.2.4 Closed one-forms in R3

We focused on R2 for simplicity, but similar results hold for one-forms and vector fields in R3 .
Definition 2.2.14 Closed one-forms in R3 . We say that a one-form Ê = f dx+g dy +h dz
on U ™ R3 is closed if
ˆf ˆg ˆf ˆh ˆg ˆh
= , = , = .
ˆy ˆx ˆz ˆx ˆz ˆy
⌃
We then have the lemma:
Lemma 2.2.15 Exact one-forms in R3 are closed. If a one-form Ê on U ™ R3 is exact,
then it is closed.
Proof. Suppose that Ê is exact: then it can be written as
ˆF ˆF ˆF
Ê = f dx + g dy + h dz = dF = dx + dy + dz
ˆx ˆy ˆz
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 12

for some smooth function F on U ™ R3 . It will then be closed if

ˆ2F ˆ2F ˆ2F ˆ2F ˆ2F ˆ2F

= , = , = .
ˆyˆx ˆxˆy ˆzˆx ˆxˆz ˆzˆy ˆyˆz
As before, these equalities follow from the Clairaut-Schwarz theorem, which states that partial
derivatives commute as long as they are continuous. ⌅
The analogous statement for vector fields goes as follows:
Lemma 2.2.16 Screening test for conservative vector fields in R3 . If a vector field
F = (f1 , f2 , f3 ) on U ™ R3 is conservative and has component functions that are continuously
differentiable, then it passes the screening test:
ˆf1 ˆf2 ˆf1 ˆf3 ˆf2 ˆf3
= , = , = .
ˆy ˆx ˆz ˆx ˆz ˆy
Proof. Same as for Lemma 2.2.15, but for the associated vector fields. ⌅
Remark 2.2.17 At this stage the definition of closeness for one-forms and the associated
screening test for vector fields appear to be rather ad hoc. Sure, they are necessary conditions
for a one-form to be exact and a vector field to be conservative, but is that it? No, not really.
In fact, those conditions will come out very naturally when we go beyond one-forms and
introduce the theory of k-forms in general. We will see how we can "differentiate" forms - this
is the notion of exterior derivative. Then the definition of closed one-forms in Definition 2.2.9
will be simply that a one-form Ê is closed if its exterior derivative dÊ vanishes. Furthermore,
if F is the vector field associated to a one-form Ê, then the vector field associated to its
derivative dÊ will be called the "curl" of F, and denoted by Ò ◊ F. The screening test for
vector fields will then be that F is "curl-free", that is Ò ◊ F = 0. All fun stuff, but it will
have to wait for a bit! Coming in Chapter 4.

2.2.5 Exercises
1. Consider the function f (x, y, z) = sin(y) + zx. Find its differential df .
Solution. df is given by:
ˆf ˆf ˆf
df = dx + dy + dz
ˆx ˆy ˆz
=z dx + cos(y) dy + x dz.
2. Consider the one-form Ê = x dx ≠ y dy on R2 . Show that it is exact, and find a function
f such that Ê = df .
Solution. Suppose that there exists a function f (x, y) such that df = ˆf
ˆx dx + ˆf
ˆy dy =
x dx ≠ y dy. Then we must have
ˆf ˆf
= x, = ≠y.
ˆx ˆy

Integrating the first equation (recalling that this is a partial derivative, so the “constant
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 13

of integration” is any function of y alone), we get:

x2
f (x, y) = + g(y)
2
for some function g(y). Substituting into the second equation, we get

ˆf
= g Õ (y) = ≠y, ,
ˆy
which can be integrated to
y2
g(y) = ≠ +C
2
for any constant C. Since we are only interested in finding one f such that df = Ê, we
can choose C = 0. We conclude that the function
x2 y 2
f (x, y) = ≠
2 2
is such that df = Ê, and hence that Ê is an exact one-form.
3. True or False. If F and G are both conservative, then F + G is also conservative.
Solution. True. If F and G are both conservative, then F = Òf and G = Òg for some
functions f and g. But then F + G = Ò(f + g), and hence F + G is also conservative.
4. Consider the vector field F(x, y, z) = (yzexy , xzexy , exy ). Show that it is conservative
and find a potential.
Solution. F is conservative if there exists a function f such that F = Òf . So we need
to solve the equations
ˆf ˆf ˆf
= yzexy , = xzexy , = exy .
ˆx ˆy ˆz
Let us start by integrating the last one. We get:

f (x, y, z) = zexy + g(x, y)

for some function g(x, y). Substituting back into the second one, we get:
ˆf ˆg
= xzexy + = xzexy ,
ˆy ˆy

from which we conclude that ˆy ˆg

= 0, that is, g(x, y) = h(x) for some function h(x).
Finally, substituting back into the first equation, we get:
ˆf dh
= yzexy + = yzexy ,
ˆx dx
from which we get hÕ (x) = 0, that is h(x) = C for some constant C. We choose C = 0,
and we conclude that F is conservative, with potential

f (x, y, z) = zexy .
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 14

5. Show that the one-form Ê = f Õ (x) dx + g Õ (y) dy on R2 is exact, for any smooth functions
f, g : R æ R.
Solution. Consider the function F (x, y) = f (x) + g(y). Its differential is

ˆF ˆF
dF = dx + dy = f Õ (x) dx + g Õ (y) dy.
ˆx ˆy

Therefore Ê is exact, and it is a well defined one-form on R2 as f and g are smooth.

6. Show that the one-form Ê = cos(y) dx ≠ x sin(y) dy is closed. Is it exact?
Solution. Let us write Ê = f dx + g dy for f (x, y) = cos(y) and g(x, y) = ≠x sin(y).
To show that it is closed, we calculate:
ˆf ˆg
= ≠ sin(y), = ≠ sin(y).
ˆy ˆx
As the two partial derivatives are equal, we conclude that Ê is a closed one-form.
Is it exact? We are looking for a function F (x, y) such that dF = Ê, that is,

ˆF ˆF
= cos(y), = ≠x sin(y).
ˆx ˆy

Integrating the first equation, we get F (x, y) = x cos(y) + h(y) for some function h(y).
Substituting in the second equation, we get
ˆF
= ≠x sin(y) + hÕ (y) = ≠x sin(y),
ˆy

from which we conclude that hÕ (y) = 0, that is h(y) = C. We pick C = 0. We conclude

that Ê = dF with F (x, y) = x cos(y), and hence it is exact.
7. Show that the vector field F(x, y, z) = (y + z, x + z, x + y) passes the screening test.
Solution. We simply need to calculate partial derivatives. If we denote the component
functions by (f1 , f2 , f3 ), we get:

ˆf1 ˆf1 ˆf2

= 1, = 1, =1
ˆy ˆz ˆz
ˆf2 ˆf3 ˆf3
= 1, = 1, = 1.
ˆx ˆx ˆy
It follows that F passes the screening test. In fact, it is easy to show that it is conservative,
with a potential given by f (x, y, z) = xy + xz + yz.
8. Determine whether the one-form Ê = x dx + x dy + z dz is exact. If it is, find a function
f such that Ê = df .
Solution. If we write Ê = f dx + g dy + h dz, we see right away that
ˆf
= 0,
ˆy
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 15

while
ˆg
= 1.
ˆx
Thus Ê is not closed, and hence it cannot exact.
9. Consider the vector field F(x, y) = (xy n , 2x2 y 3 ) for some positive integer n. Find the
value of n for which F is conservative on R2 , and find a potential for this value of n.
Solution. For F to be conservative, it must pass the screening test. We write F =
(f1 , f2 ), and calculate:
ˆf1 ˆf2
= nxy n≠1 , = 4xy 3 .
ˆy ˆx
We see that the two partial derivatives are equal for all (x, y) if and only if n = 4, in which
case the vector field becomes F = (xy 4 , 2x2 y 3 ). To show that it is conservative and find
a potential function, we are looking for a function f (x, y) such that Òf = (xy 4 , 2x2 y 3 ).
So we must have
ˆf ˆf
= xy 4 , = 2x2 y 3 .
ˆx ˆy
Integrating the first equation, we get
1
f (x, y) = x2 y 4 + g(y)
2
for some function g(y). Substituting in the second equation, we get

ˆf
= 2x2 y 3 + g Õ (y) = 2x2 y 3 ,
ˆy

from which we get g Õ (y) = 0, that is g(y) = C. We pick C = 0. Thus F is conservative

for n = 4, and a potential function is given by
1
f (x, y) = x2 y 4 .
2

2.3 Changes of variables

Recall from the introduction that our goal is to construct a theory of integration over curves,
surfaces, and volumes. Our guiding light is that we want the theory to be “reparametrization-
invariant” and “oriented”. In the last few subsections, we introduced the concept of one-forms
and their associated vector fields. In the next section, we will see that one-forms become the
objects that will be integrated over curves. Thus, to study reparametrization-invariance, we
need to know how one-forms transform under changes of variables. This is what we study in
this section. For the time being, we focuse only on functions and one-forms on R, and remain
informal and hand-wavy; our goal is to get a feeling for how one-forms transform. Studying
how one-forms transform will encourage us to introduce a little bit of mathematical formalism:
the concept of “pullbacks”. We will generalize further the concept of pullbacks in the next
section.
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 16

Objectives
You should be able to:

• Determine how one-forms transform under changes of variables on R.

• State the transformation property of one-forms in terms of the mathematical notion of

pullback.

2.3.1 How functions and one-forms on R transform under changes of variables

Consider a smooth function f (x) of a single variable x, i.e. f : U æ R for some open subset
U ™ R. Now suppose that we think of x itself as a smooth function of another variable t, that
is, x = x(t). What happens to the function f ? Well, it’s pretty simple: our function becomes
f (x(t)), which defines a new function, which we could call g(t) = f (x(t)).
What about one-forms? How do these transform under changes of variables? To see what
is going on, let us first consider a one-form on R that is exact, i.e. a differential of a function:
Ê = dF . We can write the one-form as
dF
Ê = dF (x) = dx
dx
in terms of a real variable x. What happens if we think of x itself as a smooth function of
another variable t, that is, x = x(t)? We can use what we learned above about functions. We
are interested in dF (x(t)). If we define G(t) = F (x(t)) as above, then our one-form can be
written as dG(t) = dGdt dt. But, using the chain rule of calculus, we know that

dG d dF dx
= (F (x(t)) = .
dt dt dx dt
So by changing variable from x to t, we get a new one-form, let’s call it ÷, defined by
3 4
dG dF dx
÷= dt = dt.
dt dx dt

We now generalize this to all one-forms on R (or open subsets thereof), not just exact
one-forms. Given a one-form Ê = f (x) dx written in terms of a variable x, if we think of
x = x(t) as a function of a new variable t, then by changing variable from x to t we get a new
one-form ÷: 3 4
dx
÷ = f (x(t)) dt.
dt
This defines how one-forms transform under changes of variables. As we will see, this
transformation property is what lies behind the substitution formula for definite integrals.
Remark 2.3.1 The upshot of this brief discussion is that it is easy to remember how one-forms
in R transform under changes of variables. If we write Ê = f (x) dx, and do a change of
variable x = x(t), then all we need to do is rewrite the coefficient function as a function of t
by composition f (x(t)), and then “transform the differential” dx as dx(t) = dxdt dt. This gives
us the new one-form ÷ = f (x(t)) dt dt.
dx
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 17

Example 2.3.2 An example of a change of variables. Consider the one-form Ê =

x2 dx = f (x) dx on R. Let us do the change of variables x = sin t, with dx
dt = cos t. By
changing variables from x to t we a get a new one-form
dx
÷ = f (x(t)) dt = (sin t)2 cos t dt.
dt
⇤

2.3.2 The pullback of functions and one-forms on R

Our brief discussion above can be formalized mathematically in terms of the concept of
“pullback”. Let us start with functions again, and be a little more formal. Given a function
f : U æ R for some open subset U ™ R. We write f (x) in terms of a variable x. Now suppose
that x = „(t), for some smooth function „ : V æ U , where V ™ R is also an open subset of
R. As described above, changing variables from x to t amounts to defining a new function
g(t) = f („(t)). This new function is simply the composition of f and „:

g = f ¶ „ : V æ R.

We call this new function “the pullback of f ”.

Definition 2.3.3 The pullback of a function on R. Let U ™ R and V ™ R be open
subsets, and f : U æ R and „ : V æ U be smooth functions. The pullback of f , which is
denoted by „ú f , is the smooth function

„ú f := f ¶ „ : V æ R.

Explicitly, the pullback can be written as „ú f (t) = f („(t)). ⌃

In other words, the pullback of a function by another function just means that we are
composing functions. It is called “pullback” because if we think of the chain of maps:
„ f
V æ U æ R, while our original function was from U to R, by composition we “pull it back”
to a function from V to R.
We can define a similar concept for one-forms on R, using the discussion above about how
they transform under changes of variables.
Definition 2.3.4 The pullback of a one-form on R. Let U ™ R and V ™ R be open
subsets, Ê = f (x) dx be a one-form on U , and „ : V æ U be a smooth function. The pullback
of Ê, which is denoted by „ú Ê, is the one-form on V defined by
3 4 3 4
d„ d„
„ú Ê = f („(t)) dt = „ú f (t) dt
dt dt
⌃
Note that when calculating the pullback of a one-form, it is very important not to forget
the d„
dt term!

Example 2.3.5 Change of variables as pullback. Going back to Example 2.3.2, we

could rephrase it as follows. We have a one-form Ê = x2 dx = f (x) dx on R, and a function
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 18

„ : R æ R given by „(t) = sin t. The pullback one-form „ú Ê is then given by

d„
„ú Ê = f („(t)) dt = (sin t)2 cos t dt.
dt
This is of course the same thing as implementing the change of variables x æ t in the one-form
Ê. ⇤

2.3.3 Exercises
1. Consider the one-form Ê = ex sin(x) dx on R, and the smooth function „ : R>0 æ R
with „(t) = ln(t) (where R>0 is the set of positive real numbers). Find the pullback
one-form „ú Ê. Where is the one-form „ú Ê defined?
Solution. First, since „ : R>0 æ R, and Ê is a one-form defined on all of R, by
definition of the pullback we see that the pullback one-form „ú Ê is defined only on R>0 ,
i.e. for all positive real numbers. Using the definition of pullback, we find its expression
as:
d„
„ú Ê =e„(t) sin(„(t)) dt
dt
1
=eln(t) sin(ln(t)) dt
t
1
=t sin(ln(t)) dt
t
= sin(ln(t)) dt.
2. Consider the one-form Ê = x1 dx defined on R>0 , and the smooth function „ : R æ R>0
defined by „(t) = et . Find the pullback one-form „ú Ê.
Solution. First, by definition of the pullback we see that „ú Ê is defined on all of R.
We find its expression to be:
1 Õ
„ú Ê = „ (t) dt
„(t)
=e≠t et dt
=dt.

How simple! :-)

This is not really a surprise, since Ê = d ln(x). One property of the pullback of
one-forms is that for exact one-forms, „ú df = d(„ú f ). For f = ln(x), the pullback of the
function is simply „ú f (t) = f („(t)) = ln(et ) = t, and hence „ú df = d„ú f = dt, as we
found.
3. Following up on the previous exercise, show that for an exact one-form Ê = df on U ™ R,
and a smooth function „ : V æ U , the following property of the pullback is satisfied:

„ú (df ) = d(„ú f ).

In other words, the pullback commutes with the exterior derivative of a function.
Solution. From the definition of the pullback of a function, we can write the right-
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 19

hand-side as:
df („(t))
d(„ú f ) = d(f („(t)) = dt.
dt
Using the chain rule, this can be written as
d„
d(„ú f ) =f Õ („(t)) dt
dt
=„ú (f Õ (x) dx)
=„ú (df ),

where we used the definition of the pullback of a one-form. This concludes the proof.

2.4 The pullback of a one-form

All right, time to get serious! :-) In the previous section we introduced the notion of the
pullback of a one-form with respect to a function „ : V æ U with U, V ™ R open subsets. But
this notion of pullback can be generalized, and will become essential to develop our theory of
integration (in fact, perhaps this class should be called “the power of the pullback”!). In this
section we provide a more general definition of the pullback of a one-form.

Objectives
You should be able to:

• Determine the pullback of a one-form in general.

• State and use the three fundamental properties of the pullback of one-forms.
Let Ê be a one-form on U ™ Rn , where n is a positive integer. We now consider a smooth
function „ : V æ U , where V ™ Rm , with again m a positive integer. Note that m and n
don’t have to be the same: we could have, say U ™ R3 , and V ™ R2 . Our goal is to define the
pullback „ú Ê, which should be a one-form on V .
Just to be concrete: we could take, for instance, a one-form Ê on R3 , and a smooth
function „ : R2 æ R3 . The pullback „ú Ê should then be a one-form on R2 .
Note that our notion of pullback should generalize the definition of pullback in Defini-
tion 2.3.4, which should consist in the case with m = n = 1.

2.4.1 The pullback of a function

Let us first define the pullback of a function in this context, generalizing Definition 2.3.3.
Definition 2.4.1 The pullback of a function. Let U ™ Rn and V ™ Rm be open subsets,
where m, n are positive integers. Let f : U æ R and „ : V æ U be smooth functions. The
pullback of f , which is denoted by „ú f , is the smooth function
„ú f = f ¶ „ : V æ R.
Explicitely, if we write t œ V for an m-dimensional vector in V , then
„ú f (t) = f („(t)),
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 20

where „(t) is an n-dimensional vector in U . ⌃

Example 2.4.2 The pullback of a function from R3 to R. To make things more
concrete, let us look at a specific example. Suppose that f is a smooth function on R3 , that
is f : R3 æ R. Let „ : R æ R3 be another smooth function, which takes a point in R and
maps it to a vector in R3 (so it is a vector-valued function). We can write f explicitly as
f = f (x, y, z). As for the vector-valued function „, we write „(t) = (x(t), y(t), z(t)). Then the
pullback „ú f : R æ R is simply the composition:

„ú f (t) = f („(t)) = f (x(t), y(t), z(t)).

For instance, if f (x, y, z) = xy + z, and „(t) = (t, t2 , 1), then

„ú f (t) = f (t, t2 , 1) = t3 + 1.

⇤
Example 2.4.3 The pullback of a function from R3to R2 .
We can do the same thing
but pulling back to R instead of R. Suppose that f is a smooth function on R3 , that is
2

f : R3 æ R. Let „ : R2 æ R3 be another smooth function, which takes a point in R2 and

maps it to a vector in R3 . We can write f explicitly as f = f (x, y, z). As for the vector-valued
function „, we write „(t1 , t2 ) = (x(t1 , t2 ), y(t1 , t2 ), z(t1 , t2 )). Then the pullback „ú f : R æ R
is simply the composition:

„ú f (t1 t2 ) = f („(t1 , t2 )) = f (x(t1 , t2 ), y(t1 , t2 ), z(t1 , t2 )).

For instance, if f (x, y, z) = xy + z, and „(t1 , t2 ) = (t1 , t22 , t1 + t2 ), then

„ú f (t1 , t2 ) = f (t1 , t22 , t1 + t2 ) = t1 t22 + t1 + t2 .

2.4.2 An axiomatic definition of the pullback of a one-form

We will take an axiomatic approach to the definition of the pullback of a one-form. Let us first
recall three important properties of one-forms (from Subsection 2.1.1 and Definition 2.2.5):
1. If Ê and ÷ are one-forms on U , then Ê + ÷ is a one-form on U .

2. If Ê is a one-form on U and f a smooth function on U , then f Ê is a one-form on U .

3. An exact one-form is a one-form Ê that can be written as the differential of a function

f on U : Ê = df .
We now want the pullback to be consistent with these properties. More precisely, we
require that the pullback „ú satisfies the following properties:
1. „ú (Ê + ÷) = „ú Ê + „ú ÷.

2. „ú (f Ê) = („ú f )(„ú Ê).

3. „ú (df ) = d(„ú f ).
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 21

It turns out that this is completely sufficient to fully determine the pullback of any one-form.
Let us see why.
Lemma 2.4.4 The pullback of dx. Let dx be the basic one-form on R3 . Let „ : V æ R3
be a smooth function, where V ™ Rm is an open subset, with m a positive integer. Write
t = (t1 , . . . , tm ) for an m-dimensional vector in V , and „(t) = (x(t), y(t), z(t)). Then
m
ÿ ˆx
„ú (dx) = dti .
i=1
ˆti

As an example, if m = 3, we would get

ˆx ˆx ˆx
„ú (dx) = dt1 + dt2 + dt3 .
ˆt1 ˆt2 ˆt3

The same formula for „ú (dx) remains true if dx is considered to be a one-form on R2 , R, or

more generally Rn , instead of R3 .
We also note that the same result holds for the basic one-forms dy and dz, with x replaced
by y and z respectively.
Proof. This follows from the third axiomatic property that we are imposing on the pullback.
Recall from Remark 2.2.4 that we can think of dx as the differential df of the function
f (x, y, z) = x. By the third property of pullbacks, we want to impose that

„ú (dx) = „ú (df ) = d(„ú f ).

From Definition 2.4.1, we can calculate „ú f . We get „ú f (t) = x(t). We thus obtain
m
ÿ ˆx
„ú (dx) = dx(t) = dti ,
i=1
ˆti

where we use the definition of the differential of the function x(t). ⌅

This result enables us to write down a general formula for the pullback of a one-form. For
clarity, we will only write it down for a one-form on R3 , but it is clear what the similar result
should be for a one-form in R2 or R.
Lemma 2.4.5 The pullback of a one-form. Let Ê = f dx + gdy + hdz be a one-form
on an open subset U ™ R3 . Let „ : V æ U be a smooth function, where V ™ Rm is an open
subset, with m a positive integer. Write t = (t1 , . . . , tm ) for an m-dimensional vector in V ,
and „(t) = (x(t), y(t), z(t)). Then
m
ÿ m
ÿ m
ÿ
ˆx ˆy ˆz
„ Ê = f („(t))
ú
dti + g(„(t)) dti + h(„(t)) dti .
i=1
ˆti i=1
ˆti i=1
ˆti

A similar formula holds if we start with a one-form on Rn instead of R3 .

Proof. To prove this result, we use Lemma 2.4.4 (and the similar result for dy and dz), and
the first and second axiomatic properties that we are imposing on the pullback. Using the
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 22

first and second properties, we can write:

„ú Ê = „ú (f dx + gdy + hdz) = („ú f )„ú (dx) + („ú g)„ú (dy) + („ú h)„ú (dz).

We then use Lemma 2.4.4 to evaluate „ú (dx), „ú (dy) and „ú (dz), and from Definition 2.4.1
we know that „ú f (t) = f („(t)), and similarly for g and h. ⌅
Example 2.4.6 The pullback of a one-form from R3 to R. Suppose that Ê =
f dx + gdy + hdz is a one-form on R3 . Let „ : R æ R3 be a smooth function, which takes a
point in R and maps it to a vector in R3 . We write „(t) = (x(t), y(t), z(t)). Then the pullback
„ú Ê is a one-form on R given by:
3 4
dx dy dz
„ú Ê = f („(t)) + g(„(t) + h(„(t)) dt.
dt dt dt

For instance, if Ê = xdx + xydy + z 2 dz, and „(t) = (x(t), y(t), z(t)) = (t2 , t, 1), then
3 4
dx dy dz
„ú Ê = x(t) + x(t)y(t) + z(t)2 dt
dt dt dt
1 2
= (t2 )(2t) + (t2 )(t)(1) + (1)(0) dt
=3t3 dt.

⇤
Example 2.4.7 The pullback of a one-form from to R3 R2 .
Suppose that Ê =
f dx + gdy + hdz is a one-form on R3 . Let „ : R2 æ R3 be a smooth function, which takes a
point in R2 and maps it to a vector in R3 . We write „(t) = (x(t), y(t), z(t)), with t = (t1 , t2 ).
Then the pullback „ú Ê is a one-form on R2 given by:
3 4 3 4
ˆx ˆx ˆy ˆy
„ú Ê =f („(t)) dt1 + dt2 + g(„(t)) dt1 + dt2
ˆt1 ˆt2 ˆt1 ˆt2
3 4
ˆz ˆz
+ h(„(t)) dt1 + dt2 .
ˆt1 ˆt2

For instance, if Ê = xdx + xydy + z 2 dz, and „(t) = (x(t), y(t), z(t)) = (t1 t2 , t2 , t1 + t2 ),
then
3 4 3 4
ˆx ˆx ˆy ˆy
„ú Ê =x(„(t)) dt1 + dt2 + x(„(t))y(„(t)) dt1 + dt2
ˆt1 ˆt2 ˆt1 ˆt2
3 4
ˆz ˆz
+ z(„(t))2 dt1 + dt2 .
ˆt1 ˆt2
=(t1 t2 )(t2 dt1 + t1 dt2 ) + (t1 t2 )(t2 )(dt2 ) + (t1 + t2 )2 (dt1 + dt2 )
=(t1 t22 + (t1 + t2 )2 )dt1 + (t21 t2 + t1 t22 + (t1 + t2 )2 )dt2 .

⇤
Example 2.4.8 Consistency check: the pullback of a one-form from R to R.
As a consistency check, we show that the pullback of a one-from from R to R reduces to
Definition 2.3.4. Let Ê = f (x)dx on U ™ R, and „ : V æ U with V ™ R. We write „(t) = x(t).
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 23

Then the pullback „ú Ê is the one-form on V given by:

3 4
dx
„ú Ê = f (x(t)) dt,
dt

which indeeds reproduces Definition 2.3.4 with our notation „(t) = x(t). ⇤
We now have a very general definition of the pullback of a one-form. This will turn out to
be very useful to define the integral of a one-form, which is what we now turn to.

2.4.3 Exercises
1. Consider the function f : R2 æ R given by f (x, y) = ex+y + x + y, and the function
„ : R3 æ R2 given by „(u, v, w) = (u + v, v + w). Find the pullback „ú f . What is its
domain?
Solution. First, as f : R2 æ R and „ : R3 æ R2 , we see that the composition
„ú f = f ¶ „ : R3 æ R2 æ R, i.e. the pullback „ú f is a function from R3 to R. So its
domain is R3 .
We calculate its expression by composition:

„ú f (u, v, w) =f (u + v, v + w)
=e(u+v)+(v+w) + (u + v) + (v + w)
=eu+2v+w + u + 2v + w.
2. Consider the one-form Ê = x2 dx on R, and the function „ : R2 æ R given by „(u, v) = u,
which projects on the u-axis. Find the pullback one-form „ú Ê on R2 . Interpret the result.
Solution. Let us write Ê = f dx = x2 dx. By the definition of pullback, we get:
3 4
ˆ„ ˆ„
„ Ê =f („(u, v))
ú
du + dv
ˆu ˆv
=u2 (1 du + 0 dv)
=u2 du.

We see that the pullback one-form looks the same, but written in terms of u instead
of x. However, „ú Ê is defined on R2 , while Ê was defined on R. Since the function „
here simply projects on the u-axis, what the pullback does here is extend the one-form
uniformly in the v-coordinate on the uv-plane; at any two points (u, v1 ) and (u, v2 ), the
one-form will be the same. Conceptually, this is what happens when we pullback using
a “forgetful map”, i.e. a map that somehow “forgets” some information (in this case,
the v-coordinate). The pullback then extends the object uniformly across the forgotten
structure.
3. Consider the one-form Ê = x2 dx + y 2 dy on R2 , and the map : R2 æ R2 with
(r, ◊) = (r cos ◊, r sin ◊), which defines polar coordinates. Find the pullback ú Ê.
Solution. We write Ê = f dx + g dy = x2 dx + y 2 dy, and (r, ◊) = (x(r, ◊), y(r, ◊)).
Then:
3 4 3 4
ˆx ˆx ˆy ˆy
ú
Ê =f ( (r, ◊)) dr + d◊ + g( (r, ◊)) dr + d◊
ˆr ˆ◊ ˆr ˆ◊
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 24

=r2 cos2 ◊ (cos ◊ dr ≠ r sin ◊ d◊) + r2 sin2 ◊ (sin ◊ dr + r cos ◊ d◊)

=r2 (cos3 ◊ + sin3 ◊) dr + r3 (sin2 ◊ cos ◊ ≠ cos2 ◊ sin ◊) d◊.

The notion of pullback allows us to easily calculate how one-forms change under
changes of coordinates, such as going from Cartesian to polar coordinates in this case.
2 dz
4. Consider the one-form Ê = Ôz on U = {(x, y, z) œ R3 | (x, y, z) ”= (0, 0, z)} (this is R3
x2 +y 2
with the z-axis removed), and the function „ : V æ U with „(r, ◊, ’) = r(cos ◊, sin ◊, ’),
and V = {(r, ◊, ’) œ R3 | r ”= 0}. Determine the pullback one-form „ú Ê.
Solution. By definition of the pullback, we get:
3 4
1 ˆ ˆ ˆ
„ Ê = r2 ’ 2
ú
(r’) dr + (r’) d◊ + (r’)d’
r ˆr ˆ◊ ˆ’
=r’ 2 (’ dr + r d’).
5. Let Ê be a one-form on R3 , and Id : R3 æ R3 the identity function defined by Id(x, y, z) =
(x, y, z). Show that Idú Ê = Ê.
Solution. Write Ê = f dx + g dy + h dz. By definition of the pullback, we get:
3 4
ˆ ˆ ˆ
Idú Ê =f (x, y, z) (x) dx + (x) dy + (x) dz
ˆx ˆy ˆz
3 4
ˆ ˆ ˆ
+ g(x, y, z) (y) dx + (y) dy + (y) dz
ˆx ˆy ˆz
3 4
ˆ ˆ ˆ
+ h(x, y, z) (z) dx + (z) dy + (z) dz
ˆx ˆy ˆz
=f (x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz
=Ê,

which completes the proof.

6. Let Ê be a one-form U ™ R, and „ : V æ U and – : W æ V be smooth functions, with
V, W ™ R open subsets. Show that

(„ ¶ –)ú Ê = –ú („ú Ê).

In other words, it doesn’t matter whether we pullback in one or two steps through the
„
chain of maps W æ V æ U .
–

We note here that while the exercise is only asking you to prove it for open subsets
U, V, W ™ R, this property is true in general, not just in R.
Solution. Let us write Ê = f (x) dx, „ = „(t), and – = –(u). On the one hand, we
have:
d
(„ ¶ –)ú Ê = f („(–(u))) („(–(u))) du.
du
On the other hand, we have
d
„ú Ê = f („(t)) („(t)) dt,
dt
CHAPTER 2. ONE-FORMS AND VECTOR FIELDS 25

and 3 4-
d - d
– („ Ê) = f („(–(u))
ú ú
„(t) - –(u) du.
dt t=–(u) du

But 3 4-
d d - d
(„(–(u))) = „(t) - –(u)
du dt t=–(u) du

by the chain rule, and hence („ ¶ –)ú Ê = –ú („ú Ê).

Chapter 3

Integrating one-forms: line integrals

We study how one-forms can be integrated along curves, which leads to the definition of
(oriented) line integrals (also called “work integrals”). An important result in this section
is the Fundamental Theorem of line integrals, which is the natural generalization of the
Fundamental Theorem of Calculus to integrals of one-forms along curves in R2 and R3 .

3.1 Integrating a one-form over an interval in R

Our goal is to define the integral of a one-form along a curve. But as a starting point, we
define how to integrate one-forms over a closed interval in R. We attach particular importance
to our guiding principles of reparametrization-invariance and orientability.

Objectives
You should be able to:

• Define the integral of a one-form over an interval in R.

• Determine how the integral changes when reversing the orientation of the interval.

• Relate the transformation property of one-forms to the substitution formula for definte
integrals.

3.1.1 The integral of a one-form over an interval

Consider a one-form Ê = f (x) dx over an open subset U ™ R that contains the interval [a, b].
We define the integral of Ê over [a, b].
Definition 3.1.1 The integral of a one-form over [a, b]. We define the integral of Ê
over [a, b], with a Æ b, as follows:
⁄ ⁄ b
Ê= f (x) dx,
[a,b] a

where on the right-hand-side we use the standard definition of definite integrals from calculus.
⌃

26
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 27

Well, that was simple! This is just the standard notion of definite integrals from calculus.
All we are doing is introducing some fancy notation for it. Great!
Example 3.1.2 An example of an integral of a one-form over an interval. Consider
the one-form Ê = x3 dx on R. Suppose that you want to integrate it over the interval [0, 1].
Then the integral is ⁄ ⁄ 1
1
Ê= x3 dx = ,
[0,1] 0 4
where we used the Fundamental Theorem of Calculus to evaluate the integral as usual, since
we are back in the realm of the definite integrals that we know and love. ⇤
What is interesting however is to study what our guiding principles of reparametrization-
invariance and orientability become in this simple context. Let us start with orientability.

3.1.2 Integrals of one-forms over intervals are oriented

If we look at Definition 3.1.1, there is something a bit peculiar. On the left-hand-side, we
are integrating over an interval [a, b], so by definition we must have a Æ b. However, on the
right-hand-side, we could exchange the limits of integration, and instead of integrating from a
to b, we could integrate from b to a. But what would that correspond to on the left-hand-side?
What is going on here is that thes definition involves an implicit choice of orientation for
the interval. Indeed, when we write ab f (x) dx, we say that we integrate “from a to b”: this
is a choice of direction, of orientation. We think of the interval [a, b] as being implicitly given
a choice of direction of increasing real numbers, i.e. from a to b.
But we could have decided to consider the interval [a, b], but with the opposite choice of
orientation, i.e. going from b to a, in the direction of decreasing real numbers. That would be
another choice of orientation for the interval. Let us be a little more precise.
Definition 3.1.3 The orientation of an interval. We define the orientation of an
interval in R to be a choice of direction. There are two choices: either in the direction of
increasing real numbers, or decreasing real numbers.
Let a Æ b. By [a, b]+ , we mean the interval [a, b] with the orientation of increasing real
numbers, i.e. “from a to b”. By [a, b]≠ , we denote the same interval but with the orientation
of decreasing real numbers (from b to a).
We define the canonical orientation to be the orientation of increasing real numbers.
When we write [a, b] without specifying the orientation, we always mean the interval with its
canonical orientation. ⌃
With this clarification, we can extend Definition 3.1.1 to intervals with the opposite choice
of orientation.
Definition 3.1.4 The oriented integral of a one-form. We define the integral of Ê
over the oriented interval [a, b]± , with a Æ b, as follows:
⁄ ⁄ b
Ê=± f (x) dx.
[a,b]± a

⌃
When the orientation is canonical, that is [a, b]+ = [a, b], we recover our original definition
Definition 3.1.1. But when the interval is oriented in the direction of decreasing real numbers,
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 28

that is [a, b]≠ , we get:

⁄ ⁄ b ⁄ a
Ê=≠ f (x) dx = f (x) dx,
[a,b]≠ a b
where we used properties of definite integrals to exchange the limits of integration. In other
words, we can think of the integral on the right-hand-side as “going from b to a”, which is
consistent with the orientation of decreasing real numbers as a Æ b.

3.1.3 Orientation-preserving reparametrizations

So we know that integrals of one-forms over intervals are oriented. What about reparametrization-
invariance?
Suppose that Ê = f (x) dx is a one-form over an open subset U ™ R that contains the
interval [a, b]. What happens to the integral if we do a change of variables x = „(t)?
We know how to study this question, as we know how one-forms transform under changes
of variables. The key: the pullback. Let us be a bit more precise.
Suppose that we are given a function „ : [c, d] æ [a, b], and assume that „ can be extended
to a C 1 -function1 on an open set V ™ R that contains [c, d]. We know how the one-form
changes under this change of variables: as in Section 2.3, we can pullback Ê to get a new
one-form „ú Ê which can be integrated over [c, d]. The question is: does the integral change
when we do such a change of variables? In other words, is the integral of „ú Ê over [c, d] equal
to the integral of Ê over [a, b]?
To answer this question we need to be a little more precise. We impose a further requirement
on „: we require that „(c) = a and „(d) = b, that is, it maps the left endpoint of the interval
[c, d] to the left endpoint of the interval [a, b], and similarly for the right endpoints. This
means that „ “preserves the orientation” of the intervals: it maps the smallest real number
to the smallest one, and the largest one to the largest one. In other words, it preserves the
direction of increasing real numbers.
Note that we can think of „ as a “reparametrization”, in the sense that we can think of
the interval [a, b] µ R as a “curve” in R, and „ as a parametrization of the curve.
Lemma 3.1.5 Integrals of one-forms over intervals are invariant under orientation-
preserving reparametrizations. Let Ê be a one-form on U ™ R with [a, b] µ U , and
„ : [c, d] æ [a, b] be a bijective function that can be extended to a C 1 -function on an open set
containing [c, d] and such that „(c) = a and „(d) = b. Then
⁄ ⁄
„ Ê=
ú
Ê.
[c,d] [a,b]

Explicitly, Ê = f (x) dx, and „ú Ê = f („(t)) d„

dt dt, thus the statement becomes
⁄ d ⁄ b
d„
f („(t)) dt = f (x) dx.
c dt a

Proof. The proof is clear from the explicit statement

⁄ d ⁄ b
d„
f („(t)) dt = f (x) dx,
c dt a
1
A C 1 -function is a differentiable function whose derivative is continuous.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 29

since this is nothing else but the substitution formula for definite integrals that you proved in
Calculus I! Indeed, since a = „(c) and b = „(d), we can rewrite this equation as
⁄ d ⁄ „(d)
d„
f („(t)) dt = f (x) dx,
c dt „(c)

which is the substitution formula if we do the change of variables x = „(t). ⌅

s
What this means is that we can think of the integral C Ê, where we think of C = [a, b]
as a “curve” in R, as being defined intrinsically in terms of the geometry of the curve and a
choice of orientation. It does not matter how we parametrize the curve, as long as we preserve
the orientation: the integral is the same.
Moreover, as stated in the proof of the lemma, reparametrization-invariance is nothing
else but the substitution formula for definite integrals. Isn’t that cool? What this means is
that the substitution formula for definite integrals is simply the statement that
integrals of one-forms over intervals are invariant under orientation-preserving
reparametrizations, or equivalently that they are invariant under pullback. Neat! The
fact that we need to “transform the differential dx” when we do a substitution is now clear: it
comes from the transformation property for one-forms under changes of variables studied in
Section 2.3, formulated mathematically in terms of pullback.

3.1.4 Orientation-reversing reparametrizations

In the previous lemma we considered functions „ that preserve the orientation. What happens
if instead we consider a function such that „(c) = b and „(d) = a, i.e. that maps the left
endpoint of the interval [c, d] to the right endpoint of the interval [a, b], and vice-versa? Such
a „ reverses the orientation from increasing real numbers to decreasing real numbers.
Lemma 3.1.6 Integrals of one-forms over intervals pick a sign under orientation-
reversing reparametrizations. Let Ê be a one-form on U ™ R with [a, b] µ U , and
„ : [c, d] æ [a, b] be a bijective function that can be extended to a C 1 -function on an open set
containing [c, d] and such that „(c) = b and „(d) = a (it sends the left endpoint of the interval
to the right endpoint and vice-versa). Then
⁄ ⁄ ⁄
„ú Ê = Ê=≠ Ê.
[c,d] [a,b]≠ [a,b]

Proof. From the substitution formula for definite integrals, we know that
⁄ d ⁄ „(d) ⁄ a ⁄ b
d„
f („(t)) dt = f (x) dx = f (x) dx = ≠ f (x) dx.
c dt „(c) b a
s
We recognize the right-hand-side as ≠ [a,b] Ê, which completes the proof. ⌅
It is thus not true that the integral is invariant under all reparametrizations: it is only
invariant under reparametrizations that perserve the orientation of the interval. Under
reparametrizations that reverse the orientation, the integral changes sign, as expected.
We have thus shown that the integral of a one-form over an interval is both oriented and
reparametrization-invariant, our two guiding principles. In the context of definite integrals,
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 30

orientability reduces to the statement that definite integrals pick a sign when we exchange
limits of integration, and reparametrization-invariance to the substitution formula. Neat!

3.1.5 Exercises
2
1. Consider the one-form Ê = xex dx. Compute the integral of Ê over the interval [1, 2],
with both choices of orientation.
s
Solution. We first calculate [1,2] Ê with the canonical choice of orientation in the
direction of increasing real numbers, i.e. from 1 to 2. We get:
⁄ ⁄ 2 ⁄ 4
x2 1 1
Ê= xe dx = eu du = (e4 ≠ e),
[1,2]+ 1 2 1 2

where we used the substitution u = x2 . As for the other choice of orientation in the
direction of decreasing real numbers, i.e. from 2 to 1, we get:
⁄ ⁄ 1 ⁄ 1
x2 1 1
Ê= xe dx = eu du = ≠ (e4 ≠ e),
[1,2]≠ 2 2 4 2

which is minus the other integral as expected.

2. Let Ê = x12 dx be a one-form on R>0 , and „ : R æ R>0 be given by „(t) = et . Show that
we can write ⁄ ⁄
Ê= e≠t dt.
[1,e3 ] [0,3]

Solution. The map „(t) = can be restricted to the interval [0, 3] µ R. We see that
et
its image is the interval [1, e3 ], and that the map is injective. Moreover, „(0) = 1 and
„(3) = e3 , so it preserves orientation. Thus we know that
⁄ ⁄
„ú Ê = Ê.
[0,3] [1,e3 ]

We calculate the pullback one-form:

1 1
„ú Ê = „Õ (t) dt = 2t et dt = e≠t dt.
(„(t))2 e
Therefore, we conclude that ⁄ ⁄
Ê= e≠t dt.
[1,e3 ] [0,3]
Note that this is just a fancy way of doing a substitution. Indeed, we could write the
original integral as follows:
⁄ ⁄ e3
1
Ê= 2
dx.
[1,e3 ] 1 x
We can do the substitution x = et , and the integral becomes
⁄ ⁄ e3 ⁄ 3 ⁄
1
Ê= dx = ≠t
e dt = e≠t dt,
[1,e3 ] 1 x2 0 [0,3]

as claimed. Indeed, as we have seen, orienting-preserving reparametrizations of the

integral is just the substitution formula for definite integrals.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 31

3. Let Ê = sin(x2 ) dx be a one-form on R. TRUE or FALSE:

⁄ ⁄
Ê=2 t sin(t4 ) dt.
[1,4] [≠2,≠1]

Solution. To go from the expression in x to the expression in t we need to do a change

of variables. More precisely, we consider the smooth function „ : R æ RØ0 with „(t) = t2 .
We see that „ : [≠2, ≠1] æ [1, 4], with „(≠2) = (≠2)2 = 4 and „(≠1) = (≠1)2 = 1. This
means that it changes the orientation on the interval. So we should get
⁄ ⁄
„ú Ê = ≠ Ê.
[≠2,≠1] [1,4]

By definition of pullback, we get

d„
„ú Ê = sin(t4 ) dt = 2t sin(t4 ) dt,
dt
so we conclude that ⁄ ⁄
Ê = ≠2 t sin(t4 ) dt.
[1,4] [≠2,≠1]

Therefore the statement is FALSE.

For fun, let us check that this is consistent with what we expect from the substitution
formula. Recall that the substitution formula tells us that
⁄ d ⁄ „(d)
d„
f („(t)) dt = f (x) dx.
c dt „(c)

In our case, this means that

⁄ ≠1 ⁄ 1
4
2 t sin(t ) dt = sin(x2 ) dx.
≠2 4

We recognize the left-hand-side as

⁄
„ú Ê,
[≠2,≠1]

and the right-hand-side as ⁄

≠ Ê,
[1,4]

which is consistent with what we wrote above.

3.2 Parametric curves in Rn

In the previous section we showed how one-forms can be integrated over intervals in R. Our
goal is to generalize this construction to curves in Rn (we will focus on R2 and R3 in this
course). To this end, we must first study in more details parametric curves in Rn , and the
concept of orientation of a parametric curve.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 32

Objectives
You should be able to:

• Define parametric curves in Rn .

• Determine the image of a parametric curve, or find a parametrization for a curve in R2

or R3 .

• Determine the tangent vector of a parametric curve.

• Define the orientation of a parametric curve.

• Use different parametrizations for the same curve.

• Determine whether a reparametrization preserves or reverses orientation.

3.2.1 Parametric curves

We start by recalling the definition of parametric curves.
Definition 3.2.1 Parametric curves. A parametric curve in Rn is a vector-valued
function

– : [a, b] æ Rn
t ‘æ –(t) = (x1 (t), . . . , xn (t)),

such that:

1. – can be extended to a C 1 -function1 on an open set containing [a, b];

2. –Õ (t) ”= 0 for all t œ [a, b];

3. if –(s) = –(t) for any two distinct s, t œ [a, b], then s, t œ {a, b}. 2 In other words, the
map – is injective everywhere except possibly at the end points a and b.

The image of –, which we will denote by C = –([a, b]), is a one-dimensional subspace of Rn ,

which is the curve itself. We say that the parametric curve is smooth if – can be extended
to a smooth (C Œ ) function on an open set containing [a, b]. ⌃
In words, what we are doing is mapping the interval [a, b] to a one-dimensional subspace
C µ Rn , which is the curve we are interested in. The map gives us a way of doing calculus on
the curve, by mapping points on the interval, where we can do calculus easily, to points on
the curve, which is a more difficult object to work with. This will prove important to define
integration along curves: we will pullback using the parametrization to rephrase integration
along a curve into integration along the interval [a, b], which we know how to do.
Properties 1 and 2 in Definition 3.2.1 are important to ensure that the image curve C
does not have kinks or corners (see for instance Exercise 3.2.6.6). Property 2 also ensures
1
A C 1 -function is a differentiable function whose derivative is continuous.
2
Here, the standard notation {a, b} means the set with two elements a and b. This should not be confused
with [a, b], which means the closed interval from a to b, and (a, b), which means the open interval from a to b.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 33

that a parametrization induces a well defined orientation on the curve, as we will see.
Property 3 in Definition 3.2.1 imposes that the map – is injective everywhere except
possibly at the endpoints a and b of the interval, which could be mapped to the same point.
This ensures that the image curve does not cross itself, and that the parametrization does not
go through the same path multiple times (for instance, going around a circle two times).
Because of Property 3, we can distinguish between two types of parametric curves,
depending on whether the image curve has endpoints or not.
Definition 3.2.2 Closed parametric curves. Let – : [a, b] æ Rn be a parametric curve,
with image C = –([a, b]) µ Rn .

1. If –(a) = –(b), then we say that the parametric curve is closed, as the image curve has
no endpoints (it is a loop).

2. If – : [a, b] æ Rn is injective, we call the set ˆC = {–(a), –(b)} consisting only in the
endpoints of C the boundary of the curve, which we denote by ˆC.

⌃
Example 3.2.3 Parametrizing the unit circle. Consider the function – : [0, 2fi] æ R2
given by
–(◊) = (cos ◊, sin ◊).
It is easy to see that the image –([a, b]) µ R2 is the unit circle x2 + y 2 = 1. Let us check that
the conditions in Definition 3.2.1 are satisfied:

1. – is certainly a C 1 -function on R; in fact, it is a smooth function, so this is a smooth

parametric curve.

2. –Õ (t) = (≠ sin ◊, cos ◊) is never zero over the interval [0, 2fi].

3. The only two values of ◊ œ [0, 2fi] that have the same image are ◊ = 0 and ◊ = 2fi, the
endpoints of the interval.

Because of the last statement above, this is an example of a closed parametric curve. ⇤

3.2.2 The tangent or velocity vector

The curve C itself, as a subset of Rn , is just a bunch of points. However, the parametrization
– : [a, b] æ Rn gives us more information. One can think of it as parametrizing the trajectory
of a particle moving along the curve: the particle starts at the point –(a) œ Rn , and moves
along the curve until it reaches the endpoint –(b) œ Rn . As such, we can define a velocity
vector, giving the velocity of the particle as it moves along the curve following the given
parametrization. Geometrically, the velocity vector is nothing but the tangent vector to the
curve at a given point.
Definition 3.2.4 The tangent vector to a parametric curve. Let – : [a, b] æ Rn be a
parametric curve, with the map –(t) = (x1 (t), . . . , xn (t)). The tangent vector, or velocity
vector T is a vector-valued function
T : [a, b] æ Rn
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 34

t ‘æ T(t) = –Õ (t) = (xÕ1 (t), . . . , xÕn (t)).

In other words, we simply differentiate the component functions of the vector-valued function
–. As T is a vector-valued function, it assigns a vector to every point on the curve C, namely
the tangent vector to the curve at that point. ⌃

3.2.3 Orientation of a parametric curve

If we think of the parametrization as giving the trajectory of a particle moving along the curve,
the tangent vector gives the velocity of the particle at every point on the curve. As we assume
that –Õ (t) is never zero in the definition of parametric curves Definition 3.2.1, we see that
the particle travels along the curve without stopping. Moreover, the velocity vector defines a
“direction of travel” along the curve C. Thus the parametrization of a curve naturally gives
C an orientation, corresponding to the direction of travel. More precisely, the orientation is
specified by the tangent vector.
Definition 3.2.5 Orientation of a curve. The orientation of a curve C µ Rn is given
by a choice of direction on the curve, which can be represented geometrically by an arrow on
the curve. There are two distinct choices of orientation on a curve: either the arrow is in one
direction or it is in the other. ⌃
Note that this naturally generalizes the orientation of an interval that we defined in
Definition 3.1.3.
Now we see that parametrizing a curve naturally induces a choice of orientation.
Lemma 3.2.6 Parametric curves are oriented. Let – : [a, b] æ Rn be a parametric
curve, with C = –([a, b]) µ Rn . Then the tangent vector naturally induces an orientation on
C, with the direction given by the direction of the tangent vector at each point on the curve.
Proof. The key here is that, according to the definition of parametric curves Definition 3.2.1,
the tangent vector never vanishes, and it varies continuously. In other words, the velocity
vector is never zero. Thus a particle traveling along the curve cannot turn around, as it would
first have to stop as the velocity vector varies continuously, but it cannot stop. So the particle
always travel in the same direction along the curve, which defines an orientation on the curve.
⌅
Remark 3.2.7 Another way of thinking about the fact that a parametrization induces an
orientation on the curve is that we can think of the parametrization as not only mapping
an interval [a, b] to a curve C µ Rn , but also as mapping the orientation. When we defined
parametric curves in Definition 3.2.1, we could think of the domain of – as being not only
an interval but an interval with a choice of orientation; we then always assume that the
domain of – is given the canonical orientation (in the direction of increasing real numbers).
Because of property 2 in Definition 3.2.1, the canonical orientation on the interval is then
unambiguously mapped to a choice of direction on the image curve –([a, b]) = C, which is the
induced orientation on the curve.
Example 3.2.8 Parametrizing the unit circle counterclockwise. Let us go back to
our parametrization of the unit circle in Example 3.2.3. We have the function – : [0, 2fi] æ R2
given by
–(◊) = (cos ◊, sin ◊).
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 35

We know that its image is the unit circle x2 + y 2 = 1. What is the induced orientation on the
circle? We calculate the tangent vector:

T(◊) = (xÕ (◊), y Õ (◊)) = (≠ sin ◊, cos ◊).

Now consider ◊ = 0. The parametrization maps this point to –(0) = (1, 0), so this is the point
with coordinates (1, 0) on the unit circle. As for the tangent vector, we see that T(0) = (0, 1),
and hence the tangent vector at the point (1, 0) on the unit circle is pointing upwards. This
tells us that we are moving along the curve in a counterclockwise direction, which is the
induced orientation on the unit circle. ⇤

3.2.4 Orientation-preserving reparametrizations

Given a curve C µ Rn , there isn’t a unique choice of parametrization; the curve can be
parametrized in many different ways. For instance, let – : [a, b] æ Rn be a parametric
curve, with –(t) = (x1 (t), . . . , xn (t)). Now suppose that we think of the parameter t as
a function of a new parameter u, that is, t = t(u). Our parametrization then becomes
–(t(u)) = (x1 (t(u)), . . . , xn (t(u)). Assuming that t(u) is chosen appropriately, this may define
a new parametrization —(u) = (X1 (u), . . . , Xn (u)) = (x1 (t(u)), . . . , xn (t(u)). Going from t to
u is what is called “reparametrizing the curve”.
We can be a little more precise, using the concept of pullbacks introduced in Definition 2.3.3.
Lemma 3.2.9 Reparametrizations of a curve. Let – : [a, b] æ Rn be a parametric curve,
with –(t) = (x1 (t), . . . , xn (t)) and –([a, b]) = C. Let „ : [c, d] æ [a, b] be a function that can
be extended to a C 1 -function on an open set containing [c, d]. Assume that „ is bijective, and
that „Õ (u) ”= 0 for all u œ [c, d]. The pullback

„ú – : [c, d] æ Rn
u ‘æ(„ú x1 (u), . . . , „ú xn (u)) = (x1 („(u)), . . . , xn („(u)))

is another parametrization of the same curve C.

Proof. First, it is clear that –([a, b]) = „ú –([c, d]), as we are just composing the functions x
and y with „, and thus both – and „ú – have the same image curve. But to check that „ú –
is a parametrization, we need to make sure that the three conditions in Definition 3.2.1 are
satisfied.
Property one is certainly satisfied, since it is assumed that „ can be extended to a
C 1 -function on an open set containing [c, d]. Property two is also satisfied, since
3 4
dx1 d„ dxn d„
(„ –) (u) =
ú Õ
,..., ,
d„ du d„ du

and by assumption d„du never vanishes on [c, d]. As for property three, it follows since „ is
assumed to be injective. ⌅
Remark 3.2.10 We note that since d„ du is continuous, and is never zero on [c, d], then it is
either everywhere positive on [c, d], or everywhere negative.
Since a parametric curve induces a choice of orientation on the curve, and there are only two
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 36

possible choices of orientation, it will be important to distinguish between reparametrizations

that preserve the induced orientation, and those that reverse it.
Lemma 3.2.11 Orientation-preserving reparametrizations. Consider a reparametriza-
tion as in Lemma 3.2.9, and see Remark 3.2.10. If d„ du > 0 for all u œ [c, d], then the two
parametrizations – and „ – induce the same orientation, and we call the reparametrization
ú

orientation-preserving. If d„ du < 0 for all u œ [c, d], then the two parametrizations – and
„ – induce opposite orientations, and we call the reparametrization orientation-reversing.
ú

Proof. We simply need to compare the tangent vectors. Let T– be the tangent vector associated
to the parametrization –, and T„ú – be the tangent vector associated to „ú –. We have:

d„ ! Õ " d„
T„ú – (u) = x1 („(u)), . . . , xÕn („(u)) = T– („(u)).
du du

It is then clear that if d„

du > 0, the orientation is preserved, while if d„
du < 0 is is reversed. ⌅
Example 3.2.12 Two parametrizations of the unit circle. We already saw in Ex-
ample 3.2.8 a parametrization of the unit circle that induces a counterclockwise orientation,
namely – : [0, 2fi] æ R2 with –(◊) = (cos ◊, sin ◊). Now consider a second parametrization of
the unit circle — : [≠ 3fi 2
2 , 2 ] æ R with —(t) = (sin t, cos t). What orientation is — inducing?
fi

The tangent vector reads:

T— (t) = (cos t, ≠ sin t).
Consider t = fi/2. The parametrization maps this point to the point —(fi/2) = (1, 0) on the
unit circle. The tangent vector is T— (fi/2) = (0, ≠1), and hence it points downwards. We
conclude that — is inducing a clockwise orientation on the circle, which is the opposite of our
original parametrization –.
Let us now formulate this in the language of reparametrizations as above. Consider the
function „ : [≠ 3fi
2 , 2 ] æ [0, 2fi] given by „(t) = 2 ≠ t. This function is bijective, and „ (t) = ≠1
fi fi Õ
3fi fi 2
which is of course never zero. The pullback „ – : [≠ 2 , 2 ] æ R is given by
ú

fi fi
„ú –(t) = (cos( ≠ t), sin( ≠ t)) = (sin t, cos t),
2 2
which is our second parametrization —. Since „Õ (t) < 0, we expect – and — = „ú – to induce
opposite orientation, which is exactly what we observed. ⇤

3.2.5 Piecewise parametric curves

To end this section, we note that it will sometimes be useful to consider unions of parametric
curves as defined in Definition 3.2.1. This is because our definition is fairly restrictive. It
would not allow for curves with kinks or corners, for instance, since we impose that –Õ (t)
is never zero. Also, since we require that – is injective except possibly at the endpoints, it
would not allow for curves with self-intersection. To deal with these cases, all we need to do is
consider the union C1 fi . . . fi Cn of a finite number of curves with parametrizations, and such
that two distinct components can only have one or two points in common (their endpoints).
For instance, in the case of a curve with corners, we will treat it as a union of parametric
curves, where each curve starts where the previous one ends.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 37

If a piecewise parametric curve is the union of a number of parametric curves, and each
parametric curve is smooth, we call the piecewise parametric curve piecewise smooth.
Remark 3.2.13 We add one more piece of notation. It will be useful to distinguish between
curves that have self-intersection and those that do not. We say that a curve that doesn’t
intersect itself (except possibly at the endpoints) is simple. With our definition of parametric
curves Definition 3.2.1, the image of a parametric curve will be always be simple, as it cannot
self-intersect.
Non-simple curves can be studied using piecewise parametric curves, as any non-simple
curve can be broken into a number of simple curves.
Example 3.2.14 Parametrizing a triangle. Consider the triangle with vertices A = (0, 0),
B = (0, 1) and C = (1, 0). We cannot parametrize it as a single parametric curve according to
Definition 3.2.1, since we cannot find a parametrization – that has a non-vanishing tangent
vector at the vertices of the triangle. Instead, we split it into the union of three parametric
curves, corresponding to the three edges of the triangle:

–1 :[0, 1] æ R2 , –1 (t) = (0, t),

2
–2 :[0, 1] æ R , –2 (t) = (t, 1 ≠ t),
–3 :[0, 1] æ R2 , –3 (t) = (1 ≠ t, 0).

–1 parametrizes the edge AB, –2 the edge BC, and –3 the edge CA. As each parametric
curve is smooth, the union defines a piecewise smooth parametric curve. ⇤

3.2.6 Exercises
1. Find a parametrization for the straight line between the points (0, 1, 1) and (2, 3, 3) in
R3 .
Solution. We are given two points on the line. The vector d whose direction is parallel
to the line is given by d = (2, 3, 3) ≠ (0, 1, 1) = (2, 2, 2). So we can write an equation for
the line as
r(t) = (0, 1, 1) + t(2, 2, 2), 0 Æ t Æ 1.
In the language of this section, this gives us a parametrization of the line, in the form of
a map:

– :[0, 1] æ R3
t ‘æ –(t) = (2t, 1 + 2t, 1 + 2t).
2. Express the upper half of a circle of radius 3 and centered at the point (1, 0) as a
parametric curve. Determine the orientation induced by your parametrization.
Solution. The circle of radius 3 and centered at the point (1, 0) has equation

(x ≠ 1)2 + y 2 = 9.

It is easy to find a parametrization for the circle. We can take, for instance,

x ≠ 1 = 3 cos t, y = 3 sin t, 0 Æ t Æ 2fi.

CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 38

We want only the upper half of the circle though, so we need to restrict to y Ø 0. This
amounts to restricting the domain of our parametrization to 0 Æ t Æ fi. The resulting
parametrization can be written as the map – : [0, fi] æ R2 with –(t) = (3 cos t + 1, 3 sin t).
What is the induced orientation? The tangent vector to our parametric curve is
T(t) = (≠3 sin t, 3 cos t). Pick a point on the circle, say (4, 0). This corresponds to t = 0.
At this point, the tangent vector is T(0) = (0, 3), which is pointing upwards. This means
that our parametrization induces a counterclockwise orientation around the circle.
3. Consider the parametric curve – : [0, 4fi] æ R3 , –(t) = (cos t, sin t, 4t). What is the shape
of the image curve C = –([0, 4fi]) µ R3 ? What is the induced orientation?
Solution. Let us write –(t) = (x(t), y(t), z(t)). We see that x2 (t) + y 2 (t) = cos2 t +
sin2 t = 1, so all points on the curve C lie on the cylinder x2 + y 2 = 1 with radius 1.
Moreover, as t increases, z(t) = 4t increases linearly. So the curve is an helix on the
cylinder with radius 1 centred around the z-axis.
The curve starts at the point –(0) = (1, 0, 0) on the cylinder, and ends at the point
–(4fi) = (1, 0, 16fi). As t runs from 0 to 4fi, we see that the curve goes twice around
the cylinder. Moreover, the tangent vector is T(t) = (≠ sin t, cos t, 4); in particular, its
z-coordinate is always positive, which means that the tangent vector is always pointing
upwards. We conclude that the induced orientation is going upwards along the helix.
4. Suppose that a particle is moving along the parametric curve – : [0, fi] æ R3 with
–(t) = (sin(t2 ), cos(t), t). Find its velocity at t = fi.
Solution. The velocity vector is v(t) = (2t cos(t2 ), ≠ sin(t), 1). At t = fi, we get
v(fi) = (2fi cos(fi 2 ), 0, 1). This is the velocity of the particle at t = fi, which is of course
a vector as the particle is moving in three-dimensional space.
5. Consider the curve that is the intersection of the cylinder x2 + y 2 = 1 and the surface
z = x2 ≠ y 2 in R3 . Find a parametrization for the curve. Is it a closed curve?
Solution. A point on the cylinder x2 + y 2 = 1 can be parametrized by (x, y, z) =
(cos t, sin t, z), with 0 Æ t < 2fi and z œ R. But we want the curve to lie on the
surface z = x2 ≠ y 2 , so z is fixed as z = cos2 t ≠ sin2 t. We thus get a parametric
expression for a point on the curve as (x(t), y(t), z(t)) = (cos t, sin t, cos2 t ≠ sin2 t), with
0 Æ t < 2fi. To rewrite this as a parametric curve in the language of this section,
we include the endpoint t = 2fi. We get the parametric curve – : [0, 2fi] æ R3 with
–(t) = (cos t, sin t, cos2 t ≠ sin2 t). This is a closed curve, since –(0) = (1, 0, 1) and
–(2fi) = (1, 0, 1), i.e. the starting and ending points coincide.
6. Consider the map – : [≠1, 1] æ R2 with –(t) = (x(t), y(t)), where
Y
]t2 for 0 Æ t Æ 1
x(t) = , y(t) = t2 .
[≠t2 for ≠ 1 Æ t < 0

Show that it is not a parametric curve, according to Definition 3.2.1. What does the
image C = –([≠1, 1]) µ R2 look like?
Solution. At first one may think that this is valid parametric curve. The map – is
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 39

injective, so Property 3 is satisfied. Since its derivative is –Õ (t) = (xÕ (t), y Õ (t)) with
Y
]2t for 0 Æ t Æ 1
xÕ (t) = , y Õ (t) = 2t,
[≠2t for ≠ 1 Æ t < 0

–Õ (t) exists and is continuous for all t œ R, and thus Property 1 is also satisfied (note
that – cannot be extended to a smooth function however, since xÕ (t) is not differentiable
at t = 0, but that’s ok, it doesn’t have to). However, the problem is with Property 2: we
see that –Õ (0) = (0, 0), with 0 œ [≠1, 1]; thus Property 2 is not satisfied. We conclude
that this is not a parametric curve.
The image curve C = –([≠1, 1]) is the set of points in R2 satisfying the equation
y = |x| between (≠1, 1) and (1, 1), which has a corner at the origin. This example
highlights one of the reasons why Property 2 is there in Definition 3.2.1. If it wasn’t
there, this means that we could find a parametrization for the curve y = |x|; but we do
not want the image curve to have kinks or corners. Property 2 ensures that we cannot
find a parametrization for the whole curve y = |x| between (≠1, 1) and (1, 1) at once.
This is not to say however that we cannot deal with this curve. Just like for the
triangle, the idea is to consider it as a piecewise parametric curve. I.e., we realize the
line segment from (≠1, 1) to (0, 0) as a (smooth) parametric curve, the line segment from
(0, 0) to (1, 1) as another (smooth) parametric curve, and we take the union of the two
parametric curves.

3.3 Line integrals

We are finally ready to define the integral of a one-form over a curve in Rn . Our strategy is
to start with a parametrization for the curve, and then use the parametrization to pullback
the integrand to the interval [a, b], over which we know how to integrate. In other words, we
reduce a complicated problem to something that we already know how to solve!
However, ultimately we would like our integral to be independent of our choice of
parametrization. We show that the integral is indeed invariant under orientation-preserving
reparametrizations, and thus can be understood as an object defined solely in terms of the
geometry of a curve and a choice of orientation. We also show that the integral changes sign
under orientation-reversing reparametrizations, as expected.

Objectives
You should be able to:

• Determine the pullback of a one-form along a parametric curve in R2 and R3 .

• Reformulate the pullback of a one-form along a parametric curve in terms of the

associated vector fields.

• Define the line integral of a one-form along a parametric curve in R2 and R3 , and
evaluate it.

• Rewrite the definition of line integrals in terms of the associated vector fields.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 40

• Show that line integrals are invariant under orientation-preserving reparametrizations of

the curve.

• Show that line integrals change sign under reparameterizations of the curve that reverse
its orientation.

3.3.1 The pullback of a one-form along a parametric curve

Consider a one-form Ê on an open subset U of Rn . Suppose sthat C is a curve which is
contained in U . Our goal is to define an integral of the form “ C Ê” for the integral of the
one-form Ê along the curve C (with a choice of orientation on C). However, this is not so
obvious, as it is not clear what it means to “integrate along a curve”. To make sense of this, we
use a parametrization for C, which is a map – : [a, b] æ Rn . The idea is to use the powerful
concept of pullback, which we studied in Section 2.4, to pull back the one-form from U to the
interval [a, b], and then we can integrate it, as we know how to integrate a one-form over an
interval: Definition 3.1.1. Neat!
We can apply the general expression for the pullback of a one-form obtained in Lemma 2.4.5
to the case where „ is replaced by a parametric curve – : [a, b] æ Rn . Let us write explicit
expressions for the cases of R2 and R3 .
Let – : [a, b] æ R2 be a parametric curve as in Definition 3.2.1, with –(t) = (x(t), y(t)),
and let Ê = f (x, y) dx + g(x, y) dy be a one-form on an open subset U ™ R2 containing
C = –([a, b]). Then –ú Ê is a one-form on an open subset V ™ R containing [a, b] defined by
3 4
dx dy
–ú Ê = f (–(t)) + g(–(t)) dt.
dt dt
Similarly, given a parametric curve – : [a, b] æ R3 , with –(t) = (x(t), y(t), z(t)), and a
one-form Ê = f (x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz on U ™ R3 , the pullback –ú Ê is given
by: 3 4
dx dy dz
– Ê = f (–(t))
ú
+ g(–(t)) + h(–(t)) dt.
dt dt dt
Example 3.3.1 Pulling back along a circle. Consider the counterclockwise parametriza-
tion of the unit circle introduced in Example 3.2.3, given by the function – : [0, 2fi] æ R2
with –(◊) = (cos ◊, sin ◊). Let Ê = x2 y dx + ey dy be a one-form on R2 . We can pull it back
along the parametrized unit circle to get a new one-form on R:
3 4
d d
–ú Ê = (cos ◊)2 sin ◊ cos ◊ + esin ◊ sin ◊ d◊
d◊ d◊
1 2
= ≠ cos2 ◊ sin2 ◊ + cos ◊esin ◊ d◊.

⇤
As is becoming customary, we can translate between the language of differential forms and
the language of vector fields. If F is the vector field associated to Ê, and – is a parametric
curve in Rn , then the pullback one-form –ú Ê can be written as
–ú Ê = (F(–(t)) · T(t)) dt,
where T(t) is the tangent vector to the parametric curve, and · denotes the dot product of
vectors.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 41

3.3.2 The definition of line integrals

We are now ready to define the integral of a one-form along a parametric curve: we pull back
the one-form to the interval [a, b] and integrate.
Definition 3.3.2 (Oriented) line integrals. Let Ê be a one-form on an open subset U of
Rn , and let – : [a, b] æ Rn be a parametric curve whose image C µ U . We define the line
integral of Ê along – as follows:
⁄ ⁄
Ê= –ú Ê,
– [a,b]

where the integral on the right-hand-side is defined in Definition 3.1.1.1

Explicitly, focusing on R2 for simplicity, if Ê = f (x, y) dx + g(x, y) dy, and –(t) =
(x(t), y(t)), the integral reads:
⁄ ⁄ b3 4
dx dy
Ê= f (–(t)) + g(–(t)) dt.
– a dt dt

A similar expression of course holds in R3 as well.

Such integrals are also called work integrals because of physical applications, as we will
see in Section 3.5. ⌃
What is neat is that on the right-hand-side, we end up with a one-variable definite integral,
which we certainly know how to integrate from Calculus I and II. So we have reduced the
problem of computing integrals of one-forms along curves to standard definite integrals!
Example 3.3.3 An example of a line integral. Consider the one-form Ê = y dx+cos(x) dy,
and the parametric curve – : [0, 1] æ R2 with –(t) = (t, t2 ).
First, for completeness we check that the parametric curve is well defined, according
to Definition 3.2.1. – is a smooth function on R; –Õ (t) = (1, 2t) which is never zero on
R, so in particular never zero on [0, 1]; – is injective over [0, 1]. Thus – defines a smooth
parametric curve, and its image C = –([0, 1]) has two boundary points at –(0) = (0, 0) œ R2
and –(1) = (1, 1) œ R2 . In fact, it is not difficult to see that – is a parametrization of the
parabola y ≠ x2 = 0 between (0, 0) and (1, 1).
We can compute the integral of Ê over the parametric curve –. We get:
⁄ ⁄
Ê= –ú Ê
– [0,1]
⁄ 13 4
dx dy
= f (x(t), y(t)) + g(x(t), y(t)) dt
0 dt dt
⁄ 11 2
= (t2 )(1) + cos(t)(2t) dt
A0 B
t3 -1
-
= + 2(t sin(t) + cos(t)) -
3 0
1
= + 2 sin(1) + 2 cos(1) ≠ 2
3
1
In particular, the integral on the right-hand-side uses the canonical orientation on the interval [a, b], which
is consistent with the orientation on the image curve C induced by the parametrization.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 42

5
=≠ + 2 sin(1) + 2 cos(1).
3
⇤
Remark 3.3.4 Line integrals over piecewise parametric curves. We note that we
can easily generalize the definition of line integrals to piecewise parametric curves, as in
Subsection 3.2.5. If the parametric curve is defined as a union of parametric curves, then to
integrate along the curve we simply add up the integrals over the pieces.

3.3.3 Reparametrization-invariance and orientability of line integrals

We defined line integrals in terms of a parametric curve – : [a, b] æ Rn , but in the end we
would like the integral to be defined intrinsically in terms of the geometry of the image curve
and a choice of orientation. To show that this is the case, we now show that our definition
is invariant under orientation-preserving reparametrizations. We also show that it changes
sign under orientation-reversing reparametrizations, which shows that line integrals are in
fact oriented, as we want.
Lemma 3.3.5 Line integrals are invariant under orientation-preserving reparametriza-
tions. Let Ê be a one-form on an open subset U of Rn , and let – : [a, b] æ Rn be a para-
metric curve whose image C µ U . Let „ : [c, d] æ [a, b] be as in Lemma 3.2.9, so that
„ú – = – ¶ „ : [c, d] æ Rn is a reparametrization of the curve.

1. If „ preserves orientation, as defined in Lemma 3.2.11, then

⁄ ⁄
Ê= Ê.
– „ú –

2. If „ reverses orientation, then ⁄ ⁄

Ê=≠ Ê.
– „ú –

In other words, the integral is invariant under orientation-preserving reparametrizations, and

changes sign under orientation-reversing reparametrizations. So we can really think of the
line integral as being defined intrinsically in terms of the image curve C itself, with a choice
of orientation.
Proof. To prove this statement, let us first rewrite the integrals as integrals over intervals,
using Definition 3.3.2. The integral on the left-hand-side is:
⁄ ⁄
Ê= –ú Ê.
– [a,b]

As for the second integral, we are integrating over the parametric curve – ¶ „ : [c, d] æ [a, b] æ
Rn . So we can write ⁄ ⁄
Ê= (– ¶ „)ú Ê.
„ú – [c,d]

„
However, from Exercise 2.4.3.6 we know that pulling back through the chain of maps [c, d] æ
[a, b] æ Rn is the same thing as doing it in two steps: first pulling back via –, and then via „.
–
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 43

In other words, (– ¶ „)ú Ê = „ú (–ú Ê), and we can write

⁄ ⁄
Ê= „ú (–ú Ê).
„ú – [c,d]
s s
The statement then is about the relation between [a,b] –ú Ê and [c,d] „ú (–ú Ê). But we
already studied such questions before in Lemma 3.1.5 and Lemma 3.1.6! Indeed, now that we
wrote everything in terms of integrals of one-forms over intervals, we are back in the realm of
Section 3.1. From Lemma 3.1.5 we know that the two integrals will be equal if „(c) = a and
„(d) = b, i.e. „ preserves the order of the endpoints of the interval (recall that invariance in
this case is simply the statement of the substitution formula for definite integrals). If instead
„ exchanges the order of the endpoints (i.e. „(c) = b and „(d) = a), then by Lemma 3.1.6 the
integrals differ by a sign.
But if „ú – is orientation-preserving, then „Õ (u) > 0 by Lemma 3.2.11, and so „ is a strictly
increasing function, which means that it must map c æ a and d æ b, as c Æ d and a Æ b.
Therefore the integrals are equal. While if „ú – is orientation-reversing, then „Õ (u) < 0, and so
„ is a strictly decreasing function, which means that it must map c æ b and a æ d. Therefore
the integrals differ by a sign. This completes the proof of the lemma. ⌅
Great, our two principles of orientability and reparametrization-invariance are fulfilled for
line integrals!
Example 3.3.6 How line integrals change under reparametrizations. Let us consider
the integral from Example 3.3.3 again. We consider the one-form Ê = y dx + cos(x) dy, and
the parametric curve – : [0, 1] æ R2 with –(t) = (t, t2 ).
Let us define two new parametrizations for the same curve. First, we define — : [0, ln(2)] æ
R2 with —(t) = (et ≠ 1, (et ≠ 1)2 ), and “ : [≠1, 0] æ R2 with “(t) = (≠t, t2 ). It is easy that
–, — and “ are all parametrizations of the parabola y ≠ x2 = 0 between (0, 0) and (1, 1).
Looking at the tangent vectors, we get –Õ (t) = (1, 2t), — Õ (t) = (et , 2(et ≠ 1)et ), and
“ (t) = (≠1, 2t). Are those all inducing the same orientation on the curve? Let us look
Õ

at the direction of the tangent vector at (0, 0), which corresponds to t = 0 for all three
parametrizations. We have –Õ (0) = (1, 0), — Õ (0) = (1, 0), and “ Õ (0) = (≠1, 0). So – and —
induce the same orientation on the curve (from (0, 0) to (1, 1)), while “ induces the opposite
orientation. Therefore, we expect the integral of Ê over – and — to both give the same answer,
while the integral over “ should pick a sign.
Let us do the calculation for fun. The integral over – was already performed in Exam-
ple 3.3.3. For —, we get:
⁄ ⁄
Ê= —úÊ
— [0,ln(2)]
⁄ ln(2) 3 4
dx dy
= f (x(t), y(t)) + g(x(t), y(t)) dt
0 dt dt
⁄ ln(2) 1 2
= (et ≠ 1)2 et + 2et (et ≠ 1) cos(et ≠ 1) dt
0
⁄ 11 2
= u2 + 2u cos(u) du,
0
where we did the substitution u = et ≠ 1. But this is the same integral as in Example 3.3.3,
so we get the same result indeed.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 44

As for “, we get:
⁄ ⁄
Ê= “úÊ
“ [≠1,0]
⁄ 0 3 4
dx dy
= f (x(t), y(t)) + g(x(t), y(t)) dt
≠1 dt dt
⁄ 0 1 2
= ≠t2 + 2t cos(≠t) dt
≠1
⁄ 01 2
= u2 + 2u cos(u) du,
1

where we did the substitution u = ≠t. By exchanging the limits of integration, we see that
this is minus the integral in Example 3.3.3, and thus we get a minus sign as expected. ⇤

3.3.4 Line integrals in terms of vector fields

Now that we know how to integrate one-forms along curves, we can translate the definition in
terms of the associated vector fields. This is straightforward, since we saw in Subsection 3.3.1
how to rephrase the pullback of a one-form along a curve in terms of the associated vector
field.
Lemma 3.3.7 Line integrals in terms of vector fields. Let Ê be a one-form on an
open subset U of Rn , and let – : [a, b] æ Rn be a parametric curve whose image C µ U . Let
F : U æ Rn be the vector field associated to Ê. Then we can write the line integral as follows:
⁄ ⁄ b
Ê= F(–(t)) · T(t) dt.
– a

Proof. This is clear, using the translation established in Subsection 3.3.1. ⌅

This is how line integrals, or “work integrals”, are generally defined in standard vector
calculus textbooks.2 The integrand is justified in the context of the calculation of work in
physics (as we will see in Section 3.5, such line integrals can be used to calculate work), but it
is not clear why taking the dot product between the vector field and the tangent vector to
the parametric curve is the right thing to do in general. In our context, the integrand arises
naturally by pulling back the one-form in order to be able to integrate over an interval. So it
gives a natural geometric interpretation to these work integrals.
Remark 3.3.8 In standard vector calculus textbooks, such as CLP4, the following notation
is often used. Instead of writing – : [a, b] æ Rn for the parametric curve, the symbol r = r(t)
is often used, with a Æ t Æ b, as if it was the position function of an object moving along the
image curve C. Then the tangent (or velocity) vector is written as

dr
T= ,
dt
2
Note that the symbol T(t) is sometimes used to denote the normalized or unit tangent vector (i.e. our
tangent vector divided by its norm |T(t)|), in which case dt should be replaced by ds = |T(t)| dt.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 45

standing for the velocity of the object moving along the curve, and the notation
dr
dr = dt
dt
is used. With this notation, one can rewrite the line integral of a vector field along the curve
as ⁄
F · dr.
C
The notation makes sense, as we know that the integral is invariant under orientation-
preserving reparametrizations, so we can rewrite it in terms of the image curve C itself, with
the orientation specified by the direction of travel along the curve. With this notation however
one needs to keep in mind that to evaluate the integral we need to compose the vector field
with the parametrization to rewrite F as a function of the parameter t before integrating in t
over the interval [a, b].

3.3.5 Exercises
1. Consider the one-form Ê = xy dx + z 2 dy + z dz on R3 . Find its pullback along the
helix centered around the z-axis and with tangent vector T(t) = (≠3 sin t, 3 cos t, 4),
0 Æ t Æ 2fi, and initial position r(0) = (3, 0, 1).
Solution. First, we find a parametrization for the helix. Integrating the tangent
vector, we know that the parametrization must be given by – : [0, 2fi] æ R3 with
–(t) = (3 cos t + A, 3 sin t + B, 4t + C) for some constants A, B, C. Using the fact that
–(0) = (3, 0, 1), we conclude that A = 0, B = 0, and C = 1. Thus the parametrization is
–(t) = (3 cos t, 3 sin t, 4t + 1).
We can then calculate the pullback –ú Ê. We get:
1 2
–ú Ê = (3 cos t)(3 sin t)(≠3 sin t) + (4t + 1)2 (3 cos t) + (4t + 1)(4) dt
1 2
= ≠27 cos t sin2 t + 3(4t + 1)2 cos t + 4(4t + 1) dt.
2. Consider the one-form Ê = x2 +y
x
2 dx + x2 +y 2 dy + z dz. Explain why you cannot pull it
y

back along the parametric curve – : [0, 2] æ R3 with –(t) = (t ≠ 1, t ≠ 1, t2 ).

Solution. The key here is to be careful about the domain of definition of the one-
form Ê. The largest open subset U ™ R3 over which Ê is defined is all points in R3
such that x2 + y 2 ”= 0, so that the denominators in the component functions do not
vanish. But x2 + y 2 = 0 if and only if (x, y) = (0, 0), so the domain of definition of Ê
is U = {(x, y, z) œ R3 | (x, y) ”= (0, 0)}. In other words, it consists of all points in R3
except the points on the z-axis.
To be able to pull back our one-form along the parametric curve –, we must make
sure that its image C = –([0, 2]) lies within U . Unfortunately, we see that at t = 1,
–(1) = (0, 0, 1), so the parametric curve intersects the z-axis, which is not in U ! Thus we
can’t pull back along –.
Just for fun, let’s see what would happen if we had naively tried to pull back along –.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 46

What we would get is:

3 4 3 4
t≠1 t≠1 1
–ú Ê = + + t2 (2t) dt = + 2t3 dt.
2(t ≠ 1)2 2(t ≠ 1)2 t≠1

The problem is that this is not defined at t = 1, which is part of the interval [0, 2] for the
parametric curve. In other words, the result of the pullback is not actually a well defined
one-form on an open subset containing [0, 2], since it is not defined at t = 1.
3. Find the integral of the one-form Ê = x dx + x dy + y dz in R3 along one turn clockwise
around the circle of radius two in the xy-plane and centered at the origin.
Solution. The circle has equation x2 + y 2 = 4. We parametrize one turn clockwise
around the circle with – : [0, 2fi] æ R3 , –(t) = (2 sin t, 2 cos t, 0) (recall that we are in
R3 ). The pullback of the one-form is

–ú Ê = ((2 sin t)(2 cos t) + (2 sin t)(≠2 sin t) + (2 cos t)(0)) dt = 2 (sin(2t) + cos(2t) ≠ 1) dt.

The line integral becomes

⁄ ⁄ 2fi
–ú Ê =2 (sin(2t) + cos(2t) ≠ 1) dt
[0,2fi] 0
=(≠ cos(4fi) + cos(0)) + (sin(4fi) ≠ sin(0)) ≠ 2(2fi)
= ≠ 4fi.
4. Consider the vector field F(x, y) = (ey , ex ) in R2 . Find its line integral along the oriented
curve obtained by first moving from the origin to the point (0, 1), and then from (0, 1) to
the point (4, 0) along straight lines.
Solution. This is a piecewise parametric curve, so we need to split it into line segments.
For the first segment from (0, 0) to (0, 1), we can write a parametrization as –1 : [0, 1] æ R2
with –1 (t) = (0, t). For the second line segment from (0, 1) to (0, 4), we write a
parametrization as –2 : [0, 1] æ R2 with –2 (t) = (4t, 1 ≠ t). The tangent vectors are:
T1 (t) = (0, 1), T2 (t) = (4, ≠1).
We thus get:
F(x(t), y(t)) · T1 (t) = 1, F(x(t), y(t)) · T2 (t) = 4e1≠t ≠ e4t .
The line integral becomes:
⁄ ⁄
F(x(t), y(t)) · T1 (t) dt + F(x(t), y(t)) · T2 (t) dt
[0,1] [0,1]
⁄ 1 ⁄ 11 2
= dt + 4e1≠t ≠ e4t dt
0 0
3 41
1≠t 1
=1 + ≠4e ≠ e4t
4 0
1 1
=1 ≠ 4 ≠ e4 + 4e +
4 4
1 4 11
=4e ≠ e ≠ .
4 4
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 47

5. Let C be the curve from (0, 0, 0) to (1, 1, 1) along the intersection of the surfaces y = x2
and z = x3 . Find the integral of the vector field F(x, y, z) = (x2 , xy, z 2 ) along this curve.
Solution. We first need to parametrize the curve. A point in R3 on the surface y = x2
has coordinates (t, t2 , z) for (z, t) œ R2 . At the intersection with the surface z = x3 , we
must also have z = t3 . It then follows that points on the intersection of the two surfaces
have coordinates (t, t2 , t3 ) with t œ R. Now we want our curve to start at (0, 0, 0) and
end at (1, 1, 1). So our parameter must go from t = 0 to t = 1. We thus end up with the
parametrization
– : [0, 1] æ R3 , –(t) = (t, t2 , t3 ).
The tangent vector is T(t) = (1, 2t, 3t2 ). The line integral becomes:
⁄ ⁄ 11 2
F · T dt = (t2 , t3 , t6 ) · (1, 2t, 3t2 ) dt
C 0
⁄ 11 2
= t2 + 2t4 + 3t8 dt
0
1 2 3
= + +
3 5 9
16
= .
15

3.4 Fundamental Theorem of line integrals

In Section 2.2 we studied an important class of one-forms called exact, which arise as
differentials of functions. Their associated vector fields are called conservative, and can be
expressed as the gradient of a potential function. In this section we see that line integrals of
such one-forms are very nice and satisfy beautiful properties. This leads us to the Fundamental
Theorem of line integrals, which is a natural generalization of the Fundamental Theorem of
Calculus.

Objectives
You should be able to:

• State the Fundamental Theorem of line integrals for line integrals of exact one-forms,
and use it to evaluate line integrals.

• Show that the Fundamental Theorem of line integrals implies that line integrals of exact
one-forms only depend on the starting and ending points of the curve.

• Show that the Fundamental Theorem of line integrals implies that line integrals of exact
one-forms over closed curves vanish.

• State these results in terms of conservative vector fields and their associated potential
functions.

• Use the Fundamental Theorem of line integrals and its consequences to show that a
given one-form cannot be exact.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 48

3.4.1 The Fundamental Theorem of line integrals

Recall from Section 2.2 that an exact one-form is a one-form that can be written as the
differential of a function: Ê = df . Conversely, its associated vector field F can be written as
the gradient of a function, F = Òf ; we say that F is conservative and that f is its associated
potential.
Theorem 3.4.1 The Fundamental Theorem of line integrals. Let Ê = df be an exact
one-form on an open subset U ™ Rn , and – : [a, b] æ Rn be a parametric curve whose image
–([a, b]) = C µ U . Then:
⁄ ⁄
Ê= df = f (–(b)) ≠ f (–(a)).
– –

The integral thus only depends on the starting and ending points of the image curve C.
Proof. You have probably noticed that this theorem is similar in flavour to the Fundamental
Theorem of Calculus for definite integrals; in fact it follows from it, as we will see.
First, by the definition of line integrals, we have:
⁄ ⁄
df = –ú (df ).
– [a,b]

Next, we can use one of the fundamental properties of the pullback, which is that –ú (df ) =
d(–ú f ). So we can write: ⁄ ⁄
df = d(–ú f ).
– [a,b]

If we introduce a parameter t for the parametric curve, i.e. –(t) = (x1 (t), . . . , xn (t)), then
–ú f (t) = f (–(t)), and we can write the integral as:
⁄ ⁄ b
d
df = (f (–(t)) dt.
– a dt
But then, the right-hand-side is just a standard definite integral of the derivative of a function.
By the Fundamental Theorem of Calculus (part 2), we know that the right-hand-side is simply
equal to f (–(b)) ≠ f (–(a)). We thus get:
⁄
df = f (–(b)) ≠ f (–(a)).
–

⌅
This result makes it very easy to evaluate line integrals for exact one-forms. But it also
has deeper implications. Since the integral only depends on the starting and ending points on
the image curve, this means that it does not actually depend on the choice of curve itself!
Pick any two parametric curves whose images start and end at the same place: the integral
will be the same. This is rather striking!
Corollary 3.4.2 The line integrals of an exact form along two curves that start
and end at the same points are equal. If – and — are two parametric curves whose
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 49

image share the same starting and ending points, and Ê = df is exact, then:
⁄ ⁄
Ê= Ê.
– —
Another direct consequence of the Fundamental Theorem of line integrals is that the
integral of an exact one-form over a closed curve always vanishes! Indeed, the curve is closed if
–(b) = –(a) (so that the image curve is a “loop”), and so the right-hand-side in Theorem 3.4.1
vanishes.
Corollary 3.4.3 The line integral of an exact one-form along a closed curve
vanishes. Let Ê = df be an exact one-form on an open subset U ™ Rn , and – be any closed
parametric curve whose image C µ U . Then
⁄
Ê = 0.
–
i
We sometimes write –Ê for the line integral of a one-form along a closed parametric curve.
Example 3.4.4 An example of a line integral of an exact one-form. Suppose that
you want to integrate the one-form Ê = y 2 z dx + 2xyz dy + xy 2 dz over the line segment
joining the origin to the point (1, 1, 1) œ R3 . In principle, to evaluate the line integral, you
would need to find a parametrization for the line, and use the definition of line integrals
Definition 3.3.2 to evaluate the integral. However, we notice here that Ê is exact! Indeed, if
you pick the function f (x, y, z) = xy 2 z, then
ˆf ˆf ˆf
df = dx + dy + dz = y 2 z dx + 2xyz dy + xy 2 dz,
ˆx ˆy ˆz
which is Ê. Thus we can use the Fundamental Theorem of line integrals to evaluate the line
integral. Let – be any parametrization of the line segment joining the origin to the point
(1, 1, 1). We get: ⁄
Ê = f (1, 1, 1) ≠ f (0, 0, 0) = 1 ≠ 0 = 1.
–
What is neat as well is that you know that the line integral of Ê along any curve joining
the origin to the point (1, 1, 1) will be equal to 1! The curve does not have to be a line. It
could be a parabola, the arc of a circle, anything! For instance, just for fun let us pick the
following parametric curve — : [0, 1] æ R3 with —(t) = (t, t2 , t3 ), whose image curve joins the
origin to (1, 1, 1). Let us show that this works. By definition of line integrals,
⁄ ⁄ 11 2
Ê= (t2 )2 t3 (1) + 2(t)(t2 )(t3 )(2t) + t(t2 )2 (3t2 ) dt
— 0
⁄ 11 2
= t7 + 4t7 + 3t7 dt
0
⁄ 1
=8 t7 dt
0
-1
8-
=t -
0
=1.

Neat!
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 50

One thing that we did not explain however here: how did we know that Ê was exact? This
is not always so easy to figure out. We will discuss this further in Section 3.6. ⇤

3.4.2 The Fundamental Theorem of line integrals for vector fields

To end this section, let us rewrite the Fundamental Theorem for line integrals in terms of the
associated vector fields.
Theorem 3.4.5 The Fundamental Theorem of line integrals for vector fields. Let
F = Òf be a conservative vector field on an open subset U ™ Rn , and – : [a, b] æ Rn a
parametric curve whose image –([a, b]) = C µ U . Then:
⁄ b ⁄ b
F(–(t)) · T(t) dt = Òf (–(t)) · T(t) dt = f (–(b)) ≠ f (–(a)).
a a

In the notation introduced in Remark 3.3.8, we can rewrite this integral as

⁄ ⁄
F · dr = Òf · dr = f (P1 ) ≠ f (P0 ),
C C

where P0 = r(a) and P1 = r(b) are the starting and ending points respectively of the curve C
with direction of travel specified by the position function r.
Note that from Corollary 3.4.2 and Corollary 3.4.3, we know that:

• the line integral of a conservative vector field does not depend on the path chosen
between two points;

• the line integral of a conservative vector field along a closed curve is always zero.

3.4.3 Exercises
1. Consider the one-form Ê = (y + zex ) dx + (x + ey sin z) dy + (z + ex + ey cos z) dz on R3 .
Show that Ê is an exact form, and use this fact to evaluate the integral of Ê along the
parametric curve – : [0, fi] æ R3 with –(t) = (t, et , sin t).
Solution. To show that it is exact, we simply find a function f (x, y, z) such that Ê = df
by inspection (we can also do that by integrating the partial derivatives as we did a
number of times already). We guess that f (x, y, z) = xy + zex + ey sin z + 12 z 2 , and check
that it works. Its differential is:

df = (y + zex ) dx + (x + ey sin z) dy + (z + ex + ey cos z) dz,

which is indeed Ê. So our guess is correct, and we have shown that Ê is exact.
Using the Fundamental Theorem of line integrals, we can integrate Ê directly along
–:
⁄
Ê =f (–(fi)) ≠ f (–(0))
–
=f (fi, efi , 0) ≠ f (0, 1, 0)
=fiefi .
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 51

2. Recall from Example 2.2.13 (see also Example 3.6.5) that the one-form Ê = ≠ x2 +y
y
2 dx +
x 2
dy on R \ {(0, 0)} is closed. However, we said that it was not exact. Use the
x2 +y 2
integral of Ê along one turn counterclockwise around the unit circle to show that Ê
cannot be exact.
Solution. We parametrize the unit circle as usual by – : [0, 2fi] æ R2 with –(t) =
(cos t, sin t). The pullback one-form is
3 4
sin t cos t
– Ê= ≠ 2
ú
2 (≠ sin t) + (cos t) dt
cos t + sin t cos t + sin2 t
2

=(sin2 t + cos2 t) dt
=dt.

The line integral thus simply becomes:

⁄ ⁄ 2fi
Ê= dt = 2fi.
– 0

In particular, it is non-zero. This proves that Ê cannot be exact on R2 \ {(0, 0)}, since if
it was exact its line integral along a closed curve would have to vanish.
3. Suppose that F is a conservative vector field in R2 and that its integral from point (1, 0)
to (≠1, 0) along the upper half of the unit circle is 5. What should the integral from
(1, 0) to (≠1, 0) but along the lower half of the circle be?
Solution. It should be 5! Indeed, since F is conservative, we know that its line integral
does not depend on the path chosen between two points. Since both paths here start
and end at the same points, the line integrals along these paths must be equal.
4. Consider the one-form Ê = df on R2 with f (x, y) = sin(x + y). Find a parametric curve
– that is not closed but such that ⁄
Ê = 0.
–

Solution. The one-form Ê = df is obviously exact. By the Fundamental Theorem for

line integrals, we know that
⁄
Ê = f (–(b)) ≠ f (–(a)).
–

If we write –(t) = (x(t), y(t)), then this becomes

⁄
Ê = sin(x(b) + y(b)) ≠ sin(x(a) + y(a)).
–

Now we want this integral to be zero. Thus we want

sin(x(b) + y(b)) = sin(x(a) + y(a)).

But we don’t want a closed curve, so we must choose our curve such that –(b) ”= –(a).
There are of course many possible choices. Here is one example:

–(a) = (0, 0), –(b) = (fi, 0).

CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 52

Then
sin(x(a) + y(a)) = sin(0) = 0, sin(x(b) + y(b)) = sin(fi) = 0.
Thus ⁄
Ê=0
–
for any parametric curve that starts at (0, 0) and ends at (fi, 0). For instance, we could
pick a straight line between the two points.
5. Let Ê be a one-form that is defined on all of R2 . Let P0 , P1 , P0Õ , P1Õ be any four points in
R2 . Suppose that ⁄ ⁄
Ê= Ê
C1 C2

for any two curves C1 and C2 that start at P0 and end at P1 . Show that it implies that
⁄ ⁄
Ê= Ê
C1Õ C2Õ

for any two curves C1Õ and C2Õ that start at P0Õ and end at P1Õ .
In other words, if the line integral of a one-form between two given points is path
independent, then it is path independent everywhere.
Solution. The proof is fairly intuitive. Fix P0 , P1 œ R2 , and pick any two other points
P0Õ , P1Õ œ R2 . Let D0 be a fixed curve from P0 to P0Õ , and D1 a fixed curve from P1Õ to P1 .
Suppose that C1Õ and C2Õ are two curves from P0Õ to P1Õ .
On the one hand, the curve C1 = D0 fi C1Õ fi D1 is curve from P0 to P1 . The line
integral of Ê along C1 is
⁄ ⁄ ⁄ ⁄
Ê= Ê+ Ê+ Ê.
C1 D0 C1Õ D1

On the other hand, the curve C2 = D0 fi C2Õ fi D1 is also a curve from P0 to P1 . The line
integral of Ê along C2 is
⁄ ⁄ ⁄ ⁄
Ê= Ê+ Ê+ Ê.
C2 D0 C2Õ D1

But we know that ⁄ ⁄

Ê= Ê.
C1 C2

Equating the two expressions for these line integrals, and simplifying, we end up with
the statement that ⁄ ⁄
Ê= Ê.
C1Õ C2Õ

Since this must be true for any points P0Õ and P1Õ , and any curves C1Õ and C2Õ from P0Õ to
P1Õ , we conclude that the line integral of Ê is path independent everywhere.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 53

3.5 Applications of line integrals

We mentioned in Subsection 3.3.2 that line integrals are sometimes called “work integrals”. In
this section we explain why, and work through an example where line integrals can be used to
calculate the work done. In this section we use vector field notation instead of one-forms, as
this is what is most commonly encountered in such applications.

Objectives
You should be able to:

• Determine and evaluate appropriate line integrals in the context of applications in

science, in particular for evaluating the work done while moving an object in a force
field.

3.5.1 Work
If you pick something off the ground, you expend energy. In physics, this is called “work”,
because you move an object that is under the influence of a force field. If you move an object
along a given path in a force field, how can you find the work done?
The idea is to use the well known “slicing” principle that turns a problem into an integration
question. Suppose that there is a force field F(x, y, z) in R3 , and that we move an object
along a path specified by a position vector r(t) with a Æ t Æ b. If the force field was constant
and did not depend on the position (x, y, z) of the object, from general physics principles
the work done on the object moving along the path would be F · (r(b) ≠ r(a)), i.e. the dot
product of the force and the displacement (in other words, it is the magnitude of the force
times the displacement in the direction of the force). However, as the force field depends on
the position of the object, we cannot easily calculate the work directly. But we can slice the
problem and sum over slices to rewrite the calculation as an integral.
We slice our time interval [a, b] into small slices of width t. Over a small time interval
t, the object moves from position r(t) to position r(t) + r, where r = ( x, y, z). If
the time interval is small, we can assume that the force is constant, and the work done during
this time interval can be calculated, to first order, by F(r(t)) · r. Now we sum over slices,
and take the limit where we have an infinite number of infinitesimal time intervals; this turns
the calculation into a definite integral:
⁄ b ⁄ b
dr
W = F · dr = F·
dt.
a a dt
We of course recognize the line integral of a vector field as defined in Lemma 3.3.7. This is
why line integrals are called work integrals: if the vector field is a force field, the line integral
over a parametrized curve calculates the work done when the objects moves along this curve.
Example 3.5.1 Work done by a (non-conservative) force field. Consider the force
field in R2 given by F(x, y) = (y, 5x). We will calculate the work done when moving an object
along two closed curves:

1. going once around the unit circle counterclockwise, starting and ending at (1, 0);

2. going once around a square counterclockwise, with vertices (1, 1), (1, ≠1), (≠1, ≠1), (≠1, 1),
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 54

and starting and ending at (1, 1).

Let us start with the circle (call it C). We parametrize the circle by r(t) = (cos t, sin t),
with 0 Æ t Æ 2fi. The velocity vector is then dr dt = (≠ sin t, cos t). To calculate the work done,
we evaluate the line integral along the circle:
⁄ ⁄ 2fi
dr
W = F· dt = ((sin t)(≠ sin t) + (5 cos t)(cos t)) dt
C dt 0
⁄ 2fi 1 2
= ≠ sin2 t + 5 cos2 t dt
0
⁄ 2fi 1 2
= 6 cos2 t ≠ 1 dt
0
⁄ 2fi
= (3(1 + cos(2t)) ≠ 1) dt
0
⁄ 2fi
= (2 + 3 cos(2t)) dt
0
3 3
=2(2fi) + sin(4fi) ≠ sin(0)
2 2
=4fi.
We see that even if the curve is closed (i.e. it starts and ends at the same point), the work
done is non-zero: this is because the force field is not conservative. If it was conservative, by
Corollary 3.4.3 the work would have been zero.
Now consider the square (call it S). It is a piecewise parametric curve, so we need to
parametrize the four line segments separately. We use the following parametrizations:
• (L1 ) From (1, 1) to (≠1, 1) : r1 (t) = (1 ≠ t, 1), 0 Æ t Æ 2;
• (L2 ) From (≠1, 1) to (≠1, ≠1): r2 (t) = (≠1, 1 ≠ t), 0 Æ t Æ 2;
• (L3 ) From (≠1, ≠1) to (1, ≠1): r3 (t) = (≠1 + t, ≠1), 0 Æ t Æ 2;
• (L4 ) From (1, ≠1) to (1, 1): r4 (t) = (1, ≠1 + t), 0 Æ t Æ 2.
The work done is then calculated by summing the four line integrals:
⁄
dr
W = F· dt
S dt
⁄ ⁄ ⁄ ⁄
dr1 dr2 dr3 dr4
= F· dt + F· dt + F· dt + F· dt
L1 dt L2 dt L3 dt L4 dt
⁄ 2 ⁄ 2
= ((1)(≠1) + 5(1 ≠ t)(0)) dt + ((1 ≠ t)(0) + 5(≠1)(≠1)) dt
0 0
⁄ 2 ⁄ 2
+ ((≠1)(1) + 5(≠1 + t)(0)) dt + ((≠1 + t)(0) + 5(1)(1)) dt
0 0
= ≠ 2 + 10 ≠ 2 + 10
=16.
We see that the work is again non-zero, and in fact it is not the same as the work done when
going around the unit circle. This is as expected: as the force is non-conservative, the work
done should depend on the path chosen. ⇤
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 55

3.5.2 Conservation of energy

A force field that can be written as the gradient of a potential is called “conservative” for a
reason. The name comes from physics, as it is related to conservation of energy, as we now
see.
Let F be a force field in R3 , and suppose that an object moves along a parametrized path
r(t) from t = a to t = b (let’s call this parametric curve –). By Newton’s law, we know that

d2 r
F=m ,
dt2
where m is the mass of the object. From the discussion above, we see that the work done by
the force on the object is:
⁄
W = F · dr
–
⁄ b 2
d r dr
=m · dt
a dt2 dt
⁄ b 3 4 ⁄ b
d dr dr dr d2 r
=m · dt ≠ m · 2 dt.
a dt dt dt a dt dt

We notice that the second integral on the last line is the same as the integral on the previous
line, so we end up with the statement that
⁄ b 2 ⁄ 3 4
d r dr 1 b d dr dr
m · dt = m · dt
a dt2 dt 2 a dt dt dt
m m
= |rÕ (b)|2 ≠ |rÕ (a)|2 .
2 2
Since rÕ (t) = v(t) is the velocity of the object, we recognize the expression m 2
2 |r (t)| as being
Õ

the kinetic energy of the object at the point parametrized by t on its path C, which we will
denote by T (t). Thus we conclude that the work is the difference in kinetic energy

W = T (b) ≠ T (a)

between the kinetic energy T (b) of the object at the ending point of the path and its kinetic
energy T (a) at the starting point.
Now assume that F is a conservative force field. It can then be written as the gradient
of a potential. We now change our conventions (only for this section), to be consistent with
the physics literature, and introduce a minus sign. We write: F = ≠ÒV for some potential
function V . By the Fundamental Theorem of line integrals, the work can then be evaluated
as follows:
⁄
W = F · dr
–
⁄ b
=≠ ÒV · dr
a
= ≠ (V (r(b)) ≠ V (r(a)).
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 56

Comparing with our previous calculation of the work, we conclude that

T (b) ≠ T (a) = ≠(V (r(b)) ≠ V (r(a))),
or, rearranging,
T (b) + V (r(b) = T (a) + V (r(a)).
This statement is the well known law of energy conservation in physics! Indeed, the potential
V is interpreted as the potential energy, and the equality says that the total sum of the
object’s kinetic and potential energies remains constant as the object moves along the path.
This is why such forces are called “conservative”!

3.5.3 Exercises
1. The force exerted by an electric charge at the origin on a charged particle at a point
(x, y, z) œ R3 is
K
F(x, y, z) = 2 (x, y, z),
(x + y + z 2 )3/2
2

where K is a constant. Find the work done as the particle moves along a straight line
from (1, 0, 0) to (2, 2, 3).
Solution. We parametrize the line as – : [0, 1] æ R3 with –(t) = (1 + t, 2t, 3t). The
tangent vector is T(t) = (1, 2, 3). The line integral thus becomes
⁄ ⁄ 1
1
F · dr = K ((1 + t) + 2(2t) + 3(3t)) dt.
– 0 ((1 + t)2 + (2t)2 + (3t)2 )3/2

We do the substitution u = (1+t)2 +(2t)2 +(3t)2 , with du = 2 ((1 + t) + 2(2t) + 3(3t)) dt,
and u(0) = 1, u(1) = 4 + 4 + 9 = 17. The integral becomes
⁄ ⁄ 17
K
F · dr = u≠3/2 du
– 2 1
3 4
1
=K 1 ≠ Ô .
17
2. True or False. A force field F(x, y, z) = k(x, y, z), with k any constant, does no work on
a particle that moves once around the unit circle in the xy-plane.
Solution. This is true, since the force field is conservative, and the integral of a
conservative vector field around a closed curve is always zero. To show that the force
field is conservative, consider the potential f (x, y, z) = k2 (x2 + y 2 + z 2 ). Then
Òf = k(x, y, z) = F.
While the above is a sufficient solution, let us compute the line integral for fun, to
see that we get zero indeed. We parametrize the circle as – : [0, 2fi] æ R3 , –(t) =
(cos t, sin t, 0), with tangent vector T(t) = (≠ sin t, cos t, 0). The line integral becomes
⁄ ⁄ 2fi
F · dr =k (≠ sin t cos t + sin t cos t + 0) dt
– 0
=0,
as expected.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 57

3. Find the work done by the force field F(x, y, z) = (x2 + y, x + y, 0) when moving an
object from (1, 1, 1) to (0, 0, 0).
Solution. The question does not specify the path taken between (1, 1, 1) and (0, 0, 0);
so we can only calculate the work if the force is conservative, in which case its line integral
does not depend on the path chosen.
3 2
Fortunately, the force is conservative. Pick the potential f (x, y, z) = x3 + xy + y2 .
Then 1 2
Òf = x2 + y, x + y, 0 = F.
Then, using the Fundamental Theorem for line integrals, we calculate the work:
⁄ ⁄
F · dr = Òf · dr
C C
=f (0, 0, 0) ≠ f (1, 1, 1)
1 1
=≠ ≠1≠
3 2
11
=≠ .
6

3.6 Poincare’s lemma for one-forms

We left one question unanswered: how do we determine whether a one-form is exact? How
do we know whether a vector field is conservative? With the Fundamental Theorem of line
integrals under our belts, we will be able to answer this question, which is known as Poincare’s
lemma.

Objectives
You should be able to:

• Show that a one-form defined on all of R2 or R3 is exact if and only if it is closed

(Poincare’s lemma). Understand the extension to simply connected domains.

• Rephrase Poincare’s lemma in terms of conservative vector fields.

• Use Poincare’s lemma to determine whether a one-form is exact or a vector field

conservative.

3.6.1 One-forms defined on all of Rn

Recall the definition of closed one-forms from Definition 2.2.9 and Definition 2.2.14. We know
that exact one-forms are necessarily closed, see Lemma 2.2.10 and Lemma 2.2.15. But is the
converse statement true? Are closed one-forms necessarily exact? We know that this cannot
always be true, as we have already studied an example of a closed one-form that was not
exact (see Exercise 3.4.3.2). So when are closed forms necessarily exact?
This is an important question, because showing that a one-form is closed is much easier
than showing that it is exact: one only needs to calculate the partial derivatives of its
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 58

cooordinate functions and show that they satisfy the requirements in Definition 2.2.9 and
Definition 2.2.14.)
It turns out that the answer to the question is fairly subtle. There is one simple case
however when the statement is always true: it is when the one-form is defined (and smooth,
by definition) on all of Rn .
Theorem 3.6.1 Poincare’s lemma, version I. Let Ê be a one-form defined on all of Rn .
Then Ê is exact if and only if Ê is closed.1
The corresponding statement in terms of the associated vector field F is that, if F is defined
and has continuous partial derivatives on all of R2 or R3 , then F is conservative if and only
if it passes the screening test from Lemma 2.2.11 and Lemma 2.2.16. In the language that we
will introduce in Section 4.4, in the case of R3 one can say that F is conservative if and only
if it is curl-free:
Ò ◊ F = 0.
Proof. The proof is rather interesting, and in fact constructive, as it provides a way of
calculating the function f such that Ê = df if Ê is closed. We will write the proof only for R2 ,
but a similar proof works in Rn .
First, we notice that one direction of implication is clear: we already know from Lemma 2.2.10
that exact one-forms are closed. So all we need to show is the other direction of implication,
namely that closed one-forms are exact.
Assume that Ê = f (x, y) dx + g(x, y) dy is defined on all of R2 , and that it is closed. From
Definition 2.2.9, this means that ˆf ˆy = ˆx .
ˆg

Let us now construct a function q as follows. We take our one-form Ê, and we integrate it
along a curve in R2 which consists of, first, a horizontal line from the origin (0, 0) to the point
(x0 , 0) for some fixed x0 > 0, and then a vertical line from the point (x0 , 0) to the point (x0 , y0 )
for some fixed y0 > 0. This is a piecewise-parametric curve, but we can easily parametrize
each line segment. For the line segment from (0, 0) to (x0 , 0), we use the parametrization
–1 : [0, x0 ] æ R2 with –1 (t) = (t, 0), and for the second line segment from (x0 , 0) to (x0 , y0 ),
we use the parametrization –2 : [0, y0 ] æ R2 with –2 (t) = (x0 , t). The pullbacks of the
one-form Ê are –1ú Ê = f (t, 0) dt and –2ú Ê = g(x0 , t) dt. We construct our new function as q as
the line integral of Ê along this curve:
⁄ ⁄ ⁄ x0 ⁄ y0
q(x0 , y0 ) = Ê+ Ê= f (t, 0) dt + g(x0 , t) dt.
–1 –2 0 0

q is a function of (x0 , y0 ). Now we rename the variables (x0 , y0 ) to be (x, y), and extend the
function to all (x, y) œ R2 , not just positive numbers, as the integrals on the right-hand-side
remain well defined. So we get the function
⁄ x ⁄ y
q(x, y) = f (t, 0) dt + g(x, t) dt
0 0

defined on R2 .
Our claim is that this new function q(x, y) is in fact the potential function for Ê, i.e.,
Ê = dq, which would of course show that Ê is exact. So let us compute dq. To do so, we need
ˆx and ˆy . First,
ˆq ˆq

⁄ x ⁄ y
ˆq ˆ ˆ
= f (t, 0) dt + g(x, t) dt
ˆy ˆy 0 ˆy 0
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 59

=g(x, y)

where we used the Fundamental Theorem of Calculus part 1 for the second integral (recalling
that x is kept fixed when we evaluate the partial derivative with respect to y) and the fact
that the first integral does not depend on y at all. As for the partial derivative with respect
to x, we get:
⁄ x ⁄
y
ˆq ˆ ˆ
= f (t, 0) dt + g(x, t) dt
ˆx ˆx 0 ˆx 0
⁄ y
ˆg(x, t)
=f (x, 0) + dt by FTC Part 1 for the first term
0 ˆx
⁄ y
ˆf (x, t) ˆg(x, t) ˆf (x, t)
=f (x, 0) + dt since = , as Ê is closed,
0 ˆt ˆx ˆt
=f (x, 0) + f (x, y) ≠ f (x, 0)
=f (x, y).

Therefore,
ˆq ˆq
dq = dx + dy = f (x, y) dx + g(x, y) dy = Ê,
ˆx ˆy
and we have shown that Ê is exact. Moreover, we found an explicit expression for the potential
function as a line integral of Ê. ⌅
So we now have a clear criterion to determine whether a one-form on R is exact: we
n

simply need to show that is closed. In terms of vector fields, all we have to do is show that it
passes the screening test.
Example 3.6.2 Closed forms are exact. Consider the example from Example 3.4.4. The
one-form was Ê = y 2 z dx + 2xyz dy + xy 2 dz, and we know that it is exact, as it can be
written as Ê = df for f (x, y, z) = xy 2 z. But suppose that we don’t know that. How can we
determine quickly whether it is exact or not?
First, we notice that Ê is well defined on all of R3 . So to determine that it is exact, all
that we need to do is show that it is closed.
Let us write Ê = f1 dx + f2 dy + f3 dz. We calculate partial derivatives:
ˆf1 ˆf1 ˆf2
= 2yz, = y2, = 2xy,
ˆy ˆz ˆz
and
ˆf2 ˆf3 ˆf3
= 2yz, = y2, = 2xy.
ˆx ˆx ˆy
The statement that Ê is closed is just that the partial derivatives in the first line are equal
to the partial derivatives in the second line, which is indeed true. Thus Ê is closed, and by
Poincare’s lemma we can conclude that it must be exact.
That doesn’t tell us how to find the potential function f though. To find f , we proceed as
usual. Let us do it here for completeness.
1
To be precise, we only defined closeness for one-forms in R2 and R3 at this stage. But the statement holds
in Rn using the general definition of closed one-forms in Definition 4.6.1 -- see Theorem 4.6.4.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 60

We want to find a function f = f (x, y, z) such that df = ˆf

ˆx dx + ˆf
ˆy dy + ˆf
ˆz dz =
y z dx + 2xyz dy + xy 2 dz. First, we want:
2

ˆf
= y 2 z.
ˆx
We can integrate the partial derivative -- the “constant of integration” here will be any function
g(y, z) that is independent of x. Thus we get:
⁄ x
f= y 2 z dx + g(y, z) = xy 2 z + g(y, z).
0

Next, we want
ˆf
= 2xyz.
ˆy
Using the fact that f = xy 2 z + g(y, z), this equation reads

ˆg ˆg
2xyz + = 2xyz … = 0.
ˆy ˆy
Integrating, we get:
g(y, z) = h(z),
where h(z) is a function of z alone. Putting this together, we have f = xy 2 z + h(z). Finally,
we must satisfy the remaining equation:
ˆf
= xy 2 .
ˆz
Using the fact that f = xy 2 z + h(z), this becomes

dh dh
xy 2 + = xy 2 … =0 … h = C,
dz dz
for some constant C. As we are only interested in one function f such that df = Ê, we can set
the constant C = 0. We obtain that Ê = df with f (x, y, z) = xy 2 z, as stated in Example 3.4.4.
⇤
In fact, we can go a little further and state the following theorem, which gives equivalent
formulations of what it means for a one-form to be exact (or a vector field to be conservative)
on all of Rn .
Theorem 3.6.3 Equivalent formulations of exactness on Rn . Let Ê be a one-form
defined on all of Rn , and F its associated vector field. Then the following statements are
equivalent:

1. Ê is exact (F is conservative).

2. Ê is closed (F passes the screening test).

s
3. The line integral –Ê = 0 for any closed parametric curve –.

4. Line integrals of Ê are path independent.

CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 61

In other words, if one of these statements is true, then all four statements are true.
Proof. We want to prove equivalence of the four statements. To do so, it is sufficient to prove
that (1) ∆ (2), (2) ∆ (3), (3) ∆ (4), and (4) ∆ (1). We will write a proof only for R2 .
(1) ∆ (2). All exact one-forms are closed, see Lemma 2.2.10.
(2) ∆ (3). By Poincare’s lemma, Theorem 3.6.1, we know that closed one-forms defined on
all of Rn are exact, so (1) … (2). We also know that if Ê is exact, then its line integral along
closed curves always vanishes: this is Corollary 3.4.3, which follows from the Fundamental
Theorem of line integrals. So (2) ∆ (3).
(3) ∆ (4). This follows from Exercise 3.4.3.5. Indeed, suppose that P0 is on the closed
curve that you are integrating along. Pick P1 = P0 . Then we know that the line integral of
Ê is path independent for all curves starting at P0 and ending at P1 = P0 , since by (3) the
line integrals all vanish. It then follows from Exercise 3.4.3.5 that the line integrals are path
independent everywhere, which is (4).
(4) ∆ (1). For this one we need to do a bit more work. We want to show that if the line
integrals of Ê = f dx + g dy are path independent, then Ê is exact. We proceed like in the
proof of Theorem 3.6.1. First, consider a curve C1 which consists in a horizontal line from
(0, 0) to a fixed point (x0 , 0), and then a vertical line from (x0 , 0) to a fixed point (x0 , y0 ), with
x0 , y0 > 0. We parametrize it by –1 : [0, x0 ] æ R2 with –1 (t) = (t, 0), and –2 : [0, y0 ] æ R2
with –2 (t) = (x0 , t). The pullbacks are –1ú Ê = f (t, 0) dt, –2ú Ê = g(x0 , t) dt. The line integral
then reads ⁄ ⁄ x0 ⁄ y0
q(x0 , y0 ) := Ê= f (t, 0) dt + g(x0 , t) dt.
C1 0 0
Next, we consider a second curve C2 which consists in a vertical line from (0, 0) to (0, y0 ),
and then a horizontal line from (0, y0 ) to (x0 , y0 ). A parametrization is —1 : [0, y0 ] æ R2 with
—1 (t) = (0, t), and —2 : [0, x0 ] æ R2 with —2 (t) = (t, y0 ). The pullbacks are —1ú Ê = g(0, t) dt,
and —2ú Ê = f (t, y0 ) dt. The line integral reads
⁄ ⁄ y0 ⁄ x0
p(x0 , y0 ) := Ê= g(0, t) dt + f (t, y) dt.
C2 0 0

Note that the two curves C1 and C2 start at (0, 0) and end at the same point (x0 , y0 ). Since
by (4) we know that the line integrals are path independent, we know that
q(x0 , y0 ) = p(x0 , y0 ).
As in the proof of Theorem 3.6.1, we then rename the variables (x0 , y0 ) æ (x, y) and extend
the domain of definition of the function q(x, y) = p(x, y) to all (x, y) œ R2 , since the integrals
on the right-hand-side remain well defined.
Since q(x, y) = p(x, y), the partial derivatives of q and p are also equal. In particular,
⁄ y
ˆq ˆ
= g(x, t) dt = g(x, y),
ˆy ˆy 0

and x ⁄
ˆq ˆp ˆ
= = f (t, y) dt = f (x, y),
ˆx ˆx ˆx 0
where in both cases we used FTC part 1. We conclude that
ˆq ˆq
dq = dx + dy = f (x, y) dx + g(x, y) dy = Ê,
ˆx ˆy
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 62

and hence Ê is exact, which is (1). ⌅

3.6.2 One-forms on simply connected subsets of Rn

Going back to Poincare’s lemma, the proof of Theorem 3.6.1 relied on the fact that we could
take x and y to be any two real numbers, which was possible because Ê was assumed to be a
(smooth) one-form on all of R2 . But if the one-form is defined on an open subset U ™ R2 ,
does the proof still work? The answer is: not always. For instance, it would work if U is an
open rectangle in R2 ; however, it wouldn’t work if U is R2 \ {(0, 0)}, that is R2 minus the
origin.
In other words, it isn’t always true that closed one-forms are exact. The precise statement
is that it is true if Ê is a one-form defined on an open subset U ™ Rn that is “simply connected”.
What does this mean?
We say that a set U is path connected if any two points can be connected by a path (or
a parametric curve). In other words, the set contains only one piece. Then, we say that it is
simply connected if it is path connected, with the extra property that any simple closed
curve (loop) in U can be continuously contracted to a point. Intuitively, a simply connected
region in R2 consists of only one piece and has no holes.
For instance, open rectangles and open disks in R2 are simply connected. However, if you
consider U = R2 \ {(0, 0)}, while it is path connected as you can connect any two points by
a path, it is not simply connected, since loops around the origin cannot be contracted to a
point within U (there is a hole at the origin).
We will state the more general Poincare’s lemma here for completeness, but without a
proof, as it would go beyond the scope of this course.
Theorem 3.6.4 Poincare’s lemma, version II. Let Ê be a one-form defined on an open
subset U ™ Rn that is simply connected. Then Ê is exact if and only if it is closed.
Example 3.6.5 An example of a closed one-form that is not exact. We saw in
Example 2.2.13 an example of a one-form that is closed but not exact. The one-form was
y x
Ê=≠ dx + 2 dy.
x2 +y 2 x + y2
We showed that it is closed. But we also showed in Exercise 3.4.3.2 that it is not exact,
since its line integral around a closed curve is non-vanishing. Does that contradict Poincare’s
lemma? No. The reason is that Ê is not defined on all of R2 . Indeed, its coefficient functions
are only defined for (x, y) œ R2 \ {(0, 0)}, as at the origin one would be dividing by zero. As
(x, y) œ R2 \ {(0, 0)} is not simply connected, Poincare’s lemma does not apply. ⇤

3.6.3 Exercises
1. Determine whether the one-form Ê = y 2 exy dx + (1 + xy)exy dy is exact. If it is, find a
function f such that Ê = df .
Solution. First, we notice that the component functions are smooth on R2 , so we know
that Ê is exact if and only if it is closed. We calculate the partial derivatives of the
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 63

component functions:
ˆ 2 xy ˆ
(y e ) = 2yexy + xy 2 exy , ((1 + xy)exy ) = yexy + y(1 + xy)exy .
ˆy ˆx
We see that the two expressions are equal. Thus Ê is closed, and hence it is also exact.
We are looking for a function f such that
ˆf ˆf
df = dx + dy = y 2 exy dx + (1 + xy)exy dy.
ˆx ˆy
Integrating the partial derivative in x, we get

f = yexy + g(y).

Substituting in the partial derivative for y, we get

exy + xyexy + g Õ (y) = (1 + xy)exy ,

from which we conclude that g Õ (y) = 0, i.e. g(y) = C, which we set to zero. We thus
have found a function f such that Ê = df :

f (x, y) = yexy .
2. Determine whether the field F(x, y) = (ex + ey , xey + x) is conservative. If it is, find a
potential function.
Solution. The component functions are smooth on R2 , so the vector field is conservative
if and only if it passes the screening test. We calculate the partial derivatives:
ˆ x ˆ
(e + ey ) = ey , (xey + x) = ey + 1.
ˆy ˆx
As these two expressions are not equal, we conclude that the vector field is not conservative
on R2 .
3. Determine whether or not the following sets are (a) open, (b) path connected, and (c)
simply connected:

(a) S = {(x, y) œ R2 | y Ø 0}

(b) U = {(x, y) œ R2 | (x, y) ”= (1, 1)}

(c) T = {(x, y) œ R2 | y ”= 0}

(d) The unit circle in R2

(e) The unit sphere in R3

(f) V = R3 \ {(0, 0, 0)}

Solution. Recall that a set is open if for all points in the set, there is an open ball
centered at that point that lies within the set. It is path connected if any two points in
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 64

the set can be connected by a path. It is simply connected if it is path connected, and
all closed curves can be contracted to a point within the set.

(a) S consists in the upper half of the xy-plane, including the x-axis. First, it is not
open, since any point on the x-axis cannot be the centre of an open disk within S
(as points below the x-axis are not in S). It is however path connected, as any two
points can be connected by a path, and it is simply connected, as all closed curves
can be contracted to a point within S.

(b) U is the xy-plane minus the point (1, 1). It is certainly open and path connected,
but it is not simply connected as any closed curve surrounding (1, 1) cannot be
contracted to a point in U (as there is a hole at (1, 1)).

(c) T is the xy-plane with the x-axis removed. It is an open set. However, it is not
path connected, since two points on both sides of the x-axis cannot be connected
by a path within T . It then follows that it is also not simply connected.

(d) The unit circle in R2 , i.e. the solutions to the equation x2 + y 2 = 1, is not open
in R2 . Indeed, there is no point on the unit circle that can be surrounded by an
open disk within the circle itself (note that we are considering only the circle itself
here, not the disk). It is path connected, as you can connect any two points on the
circle by a path on the circle (the arc between the two points). But it is not simply
connected, as the circle itself (which is a closed loop) cannot be contracted to a
point (recall that the interior of the circle is not part of our set).

(e) The unit sphere in R3 is not open, just as for the circle in R2 . It is path connected,
and in this case it is also simply connected, as all closed loops on the sphere can be
contracted to a point within the sphere.

(f) V is R3 minus the origin. It is certainly open, as we are just removing one point
from R3 , and it is path connected, as any two points can be connected by a path
within the set V . Is it also simply connected? The answer is yes, it is simply
connected. Indeed, pick any closed curve within V ; you can always contract it
to a point in V . The hole at the origin does not create any issue here, because
we are in R3 ; informally, the point is that even if you pick, say, a closed curve in
the xy-plane that surrounds the origin, then you can still contract it to a point
within the set V , since you can move up in the z-direction while you contract (you
don’t have to stick to the xy-plane in the contraction process, as V is a subset in
R3 ). The upshot here is that it’s important to keep in mind that the interpretation
of simply-connectedness as meaning “no holes” is only true in R2 . For instance,
in R3 , you can convince yourself that the requirement that all closed curves can
be contracted to a point within the set could instead be interpreted as meaning
that there are no “missing lines” in the set. Ultimately, it is easier to just use the
definition of simply connectedness, i.e. that the set is path connected and that all
closed curves can be contracted to a point within the set, to check whether a set is
simply connected.
CHAPTER 3. INTEGRATING ONE-FORMS: LINE INTEGRALS 65

2
4. Consider the one-form Ê = 2 |y|
x
dx ≠ xy2 dy on the open subset U = {(x, y) œ R2 | y < 0}.
Determine whether Ê is exact, and if it is find a function f such that Ê = df .
Solution. Since we are restricting to y < 0, we can replace |y| = ≠y. The one-form is
then
x x2
Ê = ≠2 dx ≠ 2 dy.
y y
Since U is simply connected, Poincare’s lemma applies. We calculate partial derivatives:
3 4 A B
ˆ x x ˆ x2 x
≠2 = 2 2, ≠ 2 = ≠2 .
ˆy y y ˆx y y2

As the two expressions are not equal, we conclude that Ê is not closed on U , and thus it
cannot be exact.
We remark here that the choice of U was very important. If we had instead defined
the one-form on the open subset V = {(x, y) œ R2 | y > 0}, then we would have obtained
a different result! Indeed, on V , |y| = y. It then follows that Ê is closed, and since
Poincare’s lemma applies as V is simply connected, it is also exact on V . Indeed, one
2
can check that Ê = df with f (x, y) = xy on V .
In fact, the largest domain of definition for Ê would be U fi V , i.e. the set {(x, y) œ
R2 | y ”= 0}. However, this is not a simply connected set, so we cannot apply Poincare’s
lemma. In any case, Ê is not exact on this larger set, as it cannot be written as Ê = df
on all of U fi V .
Chapter 4

k-forms

In this section we go beyond one-forms and introduce the general theory of differential forms,
with a focus on R3 .

4.1 Differential forms revisited: an algebraic approach

Now that we are familiar with one-forms, we take a step back, and revisit the definition. We
introduce a more algebraic approach to one-forms, which will allow us to generalize it and
introduce the concept of k-forms.

Objectives
You should be able to:

• Define basic one-forms as linear maps, and basic two- and three-forms as multilinear
maps.

• Define k-forms in R3 .

• Prove the antisymmetry propeties of basic k-forms.

• Relate two-forms to vector fields.

4.1.1 An algebraic approach to one-forms

When we introduced one-forms in Definition 2.1.1, we said that the objects dx, dy and dz
could be understood a placeholders. We paused briefly in Remark 2.2.4 to relate these objects
to the differentials of “projection functions” on R3 , but this wasn’t entirely satisfactory. Now
we go back and give a more rigorous definition of what these objects stand for.
Definition 4.1.1 The basic one-forms. The basic one-form dxi , with i œ {1, . . . , n}, is
a linear map dxi : Rn æ R which takes a vector u = (u1 , . . . , un ) œ Rn and projects it onto
the xi -axis:
dxi (u1 , . . . , un ) = ui .
⌃

66
CHAPTER 4. k-FORMS 67

Remark 4.1.2 When we are working on R3 , it is customary to write (x, y, z) for (x1 , x2 , x3 ),
and we write the basic one-forms as dx, dy, and dz (instead of dx1 , dx2 and dx3 ).
This gives a rigorous meaning of these placeholders. Using this definition, we can write a
general linear map M : R3 æ R as

M = A dx + B dy + C dz,

where A, B, C œ R are just constants. In other words, it is an arbitrary linear combination

of the three projection operators. In general, given an abstract vector space V , the set of
linear maps M : V æ R forms a vector space itself, which is called the “dual vector space”
and denoted by V ú (see for instance https://en.wikipedia.org/wiki/Dual_space).
Let us now look back at the definition of one-forms in Definition 2.1.1, focusing on R3 for
simplicity. A (differential) one-form was defined as a linear combination of basic one-forms
with coefficients that are smooth functions on an open subset U ™ R3 . That is, we can write
a one-form as:
Ê = f dx + g dy + h dz,
for smooth functions f, g, h : U æ R. With our new understanding of the placeholders
dx, dy, dz, we can make sense of this object. For any point P œ U , the one-form Ê defines a
linear map R3 æ R (or equivalently an element of the vector space dual to R3 ). This is the
dual concept to vector fields: a vector field is a rule that assigns to all points on U a vector in
R3 , while a one-form is a rule that assigns to all points on U a linear map R3 æ R (that is, a
“dual vector”). Nice!

4.1.2 Basic k-forms

This algebraic understanding of the basic one-forms as linear maps allows us to define a
natural generalization. Instead of looking only at linear maps Rn æ R, we now also define
multilinear maps Rn ◊ Rn æ R, and Rn ◊ Rn ◊ Rn æ R, and so on. The definition relies on
the determinant.
Definition 4.1.3 Basic two-forms. The basic two-form dxi · dxj , for i, j œ {1, . . . , n} is
a multilinear map1 dxi · dxj : Rn ◊ Rn æ R which takes two vectors (u, v) œ Rn ◊ Rn , with
u = (u1 , . . . , un ) and v = (v1 , . . . , vn ), and maps them to the following determinant:
A B
ui vi
dxi · dxj (u, v) = det .
uj vj

⌃
In the same way we can define the notion of basic three-forms.
Definition 4.1.4 Basic three-forms. The basic three-form dxi · dxj · dxk , for
i, j, k œ {1, . . . , n} is a multilinear map dxi · dxj · dxk : Rn ◊ Rn ◊ Rn æ R which takes
three vectors (u, v, w) œ Rn ◊ Rn ◊ Rn , with u = (u1 , . . . , un ) and v = (v1 , . . . , vn ), and
1
A multilinear map is a function of several variables that is linear separately in each variable.
CHAPTER 4. k-FORMS 68

w = (w1 , . . . , wn ), and maps them to the following determinant:

Q R
ui vi wi
c d
dxi · dxj · dxk (u, v, w) = det a uj vj wj b .
uk vk wk

⌃
We wrote the definition of basic one-, two-, and three-forms explicitly for clarity, but in
fact they are just special cases of the more general definition of basic k-forms. We now define
basic k-forms, for completeness, but don’t worry if the notation is confusing you: as we will
see, in R3 all basic k-forms with k Ø 4 automatically vanish, so the above definitions are
sufficient.
Definition 4.1.5 Basic k-forms. The basic k-form dxi1 · . . . · dxik , with i1 , . . . , ik œ
{1, . . . , n}, is a multilinear map dxi1 · . . . · dxik : (Rn )k æ R which takes k vectors
(u1 , . . . , uk ) œ (Rn )k , with uj = (uj1 , . . . , ujn ), and maps them to the following determinant:
Q R
u1i1 . . . uki1
c . .. . d.
a ..
dxi1 · . . . · dxik (u1 , . . . , uk ) = det c . .. d b
u1ik k
. . . uik

⌃
It looks like there are many possibilities here, but many of them either vanish or are related
to each other. More precisely, basic k-forms satisfy the following antisymmetry relations,
which significantly reduce the number of non-zero basic forms R3 .
Lemma 4.1.6 Antisymmetry of basic k-forms. The basic two-forms satisfy the following
properties. For any i, j œ {1, . . . , n},

dxi · dxj = ≠dxj · dxi .

In particular,
dxi · dxi = 0, i œ {1, . . . , n}.
Similarly, the basic k-forms pick a sign whenever we exchange the order of two of the dxi ’s,
and vanish if two of the dxi ’s are the same.
It follows that the only non-vanishing basic k-forms in Rn are for 1 Æ k Æ n. In particular,
in R3 we only have basic one-, two-, and three-forms. The independent basic two-forms in R3
are (using the x, y, z notation for R3 ):

dy · dz, dz · dx, dx · dy,

and the only independent basic three-form in R3 is

dx · dy · dz.
Proof. This follows directly from the property of the determinant. If we exchange a dxi with
dxj , we exchange two rows in the matrix that we take the determinant of, and hence the
determinant picks a sign. Similarly, if two of the dxi ’s are the same, then the matrix that we
CHAPTER 4. k-FORMS 69

take the determinant of has two equal rows, and hence the determinant is zero. ⌅
Remark 4.1.7 We note here that the basic k-forms can be given a geometric interpretation.
We focus on R3 for simplicity. It’s easier to start with the basic three-form in R3 . The three
vectors u, v and w in R3 span a three-dimensional parallelepiped. Then the basic three-
form dx · dy · dz(u, v, w) calculates its oriented volume, since this is what the determinant
calculates.
As for the basic two-forms, the two vector u and v in R3 span a parallelogram. The basic
two-form dx · dy(u, v) calculates the oriented area of its projection on the xy-plane, while
the two-forms dx · dz(u, v) and dy · dz(u, v) calculate the oriented area of its projection on
the xz- and yz-planes respectively.

4.1.3 k-forms in R3
We are now ready to introduce the concept of k-forms, with k œ {0, 1, 2, 3}, in R3 , which
naturally generalizes the one-forms introduced in Definition 2.1.1.
Definition 4.1.8 k-forms in R3 . Let U ™ R3 be an open subset.

1. A zero-form is a smooth function f : U æ R.

2. A one-form is an expression of the form

f dx + g dy + h dz,

for smooth functions f, g, h : U æ R.

3. A two-form is an expression of the form

f dy · dz + g dz · dx + h dx · dy,

for smooth functions f, g, h : U æ R.

4. A three-form is an expression of the form

f dx · dy · dz,

for a smooth function f : U æ R.

⌃
Of course there is no point in defining k-forms with k Ø 4 in R3 ,
as those would necessarily
vanish, by Lemma 4.1.6. But we note that this definition of k-forms can naturally be
generalized to Rn using the general definition of basic k-forms in Definition 4.1.5.
Remark 4.1.9 Our definition of two- and three-forms involves a specific choice of basic two-
and three-forms. For instance, we used dx · dy · dz instead of dz · dx · dy. When we express
the differential forms with the choice of basic form in Definition 4.1.8, we say that the k-forms
are in standard form. We generally want to present differential forms in standard form, to
simplify things.
With this being said, you may wonder why we chose this particular choice of ordering for
CHAPTER 4. k-FORMS 70

the basic two-forms, i.e. why we wrote

f dy · dz + g dz · dx + h dx · dy

instead of, say,

f dx · dy + g dx · dz + h dy · dz.
The reason behind this choice will become clear in the next section, when we relate the wedge
product of differential forms to the cross-product of vector fields.
An easy way to remember this choice of ordering is to rename in your head (x, y, z) ‘æ
(x1 , x2 , x3 ), and to denote the component functions by f1 , f2 , f3 . Then the proper choice of
ordering is
f1 dx2 · dx3 + f2 dx3 · dx1 + f3 dx1 · dx2 ,
which runs through the three cyclic permutations of (1, 2, 3)), namely (1, 2, 3), (2, 3, 1) and
(3, 1, 2).
Remark 4.1.10 Using the algebraic interpretation of the basic one-, two-, and three-forms
in the previous subsections, we can give a geometric meaning to k-forms: a 0-form assigns a
number to all points in U , while a k-form (with k Ø 1) assigns a multilinear map (R3 )k æ R
to all points in U . In other words, if we act on a given set of vectors, a k-form assigns a
notion of k-dimensional oriented volume for the corresponding projection of the k-dimensional
parallepiped generated by the k vectors.

4.1.4 k-forms and vector calculus

As has become customary, we end this section by relating our construction in the world of
differential forms to the traditional concepts in vector calculus. In Principle 2.1.3, we saw that
we can naturally associate to a one-form a corresponding vector field. This correspondence
can be generalized to k-forms in R3 . We get the following table, which provides a dictionary
between differentials forms in R3 and vector calculus concepts. Note that to establish the
dictionary, we write the k-forms on the left-hand-side of the table in standard form (see
Remark 4.1.9).
Table 4.1.11 Dictionary between k-forms in R3 and vector calculus concepts
Differential form concept Vector calculus concept
0-form f function f
1-form f dx + g dy + h dz vector field (f, g, h)
2-form f dy · dz + g dz · dx + h dx · dy vector field (f, g, h)
3-form f dx · dy · dz function f

4.1.5 Exercises
1. Show that dz · dx · dy = dx · dy · dz.
Solution. We know that dz · dx = ≠dx · dz, and dz · dy = ≠dy · dz. So we can write

dz · dx · dy = ≠dx · dz · dy = dx · dy · dz.
CHAPTER 4. k-FORMS 71

2. List the independent non-vanishing basic k-forms in R4 .

Solution. Because of anti-symmetry, we know that the only non-vanishing basic k-forms
in R4 are for k Æ 4. Let us write the basic one-forms by

dx1 , dx2 , dx3 , dx4 .

Then the basic two-forms are obtained by pairing those two-by-two, up to anti-symmetry.
We get the basic two-forms:

dx1 · dx2 , dx1 · dx3 , dx1 · dx4 , dx2 · dx3 , dx2 · dx4 , dx3 · dx4 .

For the basic three-forms, we pair the one-forms three-by-three, without repeated factors
(otherwise they would vanish). We get:

dx1 · dx2 · dx3 , dx1 · dx2 · dx4 , dx1 · dx3 · dx4 , dx2 · dx3 · dx4 .

Finally, there is only one independent basic four-form, since there cannot be repeated
dxi factors. (There is always only one independent basic “top-form”, i.e. basic n-form in
Rn ). It reads:
dx1 · dx2 · dx3 · dx4 .
3. Write down the vector field F : R3 æ R3 associated to the two-form

Ê = xy dx · dy + xyz dx · dz + x2 dy · dz.

Solution. Before we extract the vector field we need to make sure that we write the
one-form in the correct form according to the dictionary Table 4.1.11. We have:

Ê = x2 dy · dz ≠ xyz dz · dx + xy dx · dy.

Then the associated vector field is:

F(x, y, z) = (x2 , ≠xyz, xy).

4. Let Ê = dx · dz be a basic two-form on R3 , and u = (1, 2, 3) and v = (3, 2, 1) be vectors
in R3 . Evaluate
Ê(u, v).

Solution. By definition of a basic two-form (and recalling that we use the standard
notation here that dx · dz = dx1 · dx3 ), we have:
A B
u v
dx · dz(u, v) = det 1 1 .
u3 v3

Substituting the entries for the vectors u and v, we get:

A B
1 3
dx · dz(u, v) = det = 1 ≠ 9 = ≠8.
3 1
CHAPTER 4. k-FORMS 72

5. Let Ê = dx · dy · dz be the basic three-form on R3 .

(a) Show that Ê(u, v, w) = 1, with u = (1, 0, 0), v = (0, 1, 0) and w = (0, 0, 1) basis
vectors in R3 .

(b) Show that Ê(v, u, w) = ≠1.

(c) In general, show that there are three choices of ordering of the basis vectors for
which Ê evaluates to 1, and three choices for which it evaluates to ≠1.

Solution. (a) By definition of a basic three-form, we get:

Q R
1 0 0
c d
dx · dy · dz(u, v, w) = det a0 1 0b = 1.
0 0 1

(b) We changed the ordering of the basis vectors here. We get:

Q R
0 1 0
c d
dx · dy · dz(v, u, w) = det a1 0 0b = ≠1.
0 0 1

(c) In general, it is easy to see that the following three orderings give 1:

dx · dy · dz(u, v, w) = dx · dy · dz(w, u, v) = dx · dy · dz(v, w, u) = 1,

while the following three orderings give ≠1:

dx · dy · dz(v, u, w) = dx · dy · dz(w, v, u) = dx · dy · dz(u, w, v) = ≠1.

The reason is that whenever we permute two basis vectors, we exchange two columns in
the matrix that we are taking the determinant of. But we know from properties of the
determinant that swapping two columns of matrix changes its detereminant by ≠1. So
we conclude that doing an even number of two-by-two swaps of basis vectors does not
change the determinant, while doing an odd numbers of two-by-two swaps changes the
determinant by ≠1.
FYI: in the language of group theory, the group of permutations of three objects
is called the “symmetric group” S3 , whose elements are the permutations. We call a
permutation that is a swap of two objects a “transposition”. All permutations can be
obtained by doing a finite number of transpositions (i.e. by swapping objects two-by-two
a finite number of time). We define the “sign” of a permutation as being positive if
the permutation can be obtained by an even number of transpositions, and negative if it
requires an odd number of transpositions. The statement above would then be that the
determinant is unchanged if the basis vectors are related by a positive permutation, and
picks a sign if they are related by a negative permutation.
CHAPTER 4. k-FORMS 73

4.2 Multiplying k-forms: the wedge product

With the introduction of k-forms we can now introduce a new operation: we can multiply
differential forms. This operation is called the “wedge product”, which is what we study in
this section.

Objectives
You should be able to:

• Determine the wedge product of two k-forms in R3 .

• Relate the wedge product of two one-forms to the cross-product of the associated vector
fields, and the wedge product of a one-form and a two-form to the dot product of the
associated vector fields.

• State and use general properties of the wedge product for k-forms.

4.2.1 Multiplying k-forms: the wedge product

When we introduced one-forms in Section 2.1, we explained how we can add two one-forms
to get another one-form, and how we can multiply a one-form by a function to get another
one-form. Let us first state that the addition property obviously holds for two- and three-forms
as well: the sum of two two-forms is a two-form, and the sum of two three-forms is a three-form.
Note that it doesn’t really make sense to add a one-form with a two-form, etc.
When we introduced one-forms we did not however talk about multiplying two one-forms
together. To do this, we need the full theory of k-forms, as multiplying two one-forms will
give us a two-form. The operation of multiplying k-forms is called the “wedge product”, and
denoted by ·, to which we now turn to.
Definition 4.2.1 The wedge product. Consider a simple k-form Ê and a simple m-form
÷ on an open subset U ™ Rn of the form:

Ê = f dxi1 · . . . · dxik , ÷ = g dxj1 · . . . · dxjm ,

for smooth functions f, g : U æ R. Then the wedge product Ê · ÷ is a (k + m)-form defined

by:
Ê · ÷ = f g dxi1 · . . . · dxik · dxj1 · . . . · dxjm .
Note that the order is important here on the right, since we know that exchanging two dxi ’s
will pick a sign, by the definition of basic forms.
The wedge product of arbitrary k- and m-forms is then defined by linearity, i.e. by
distributing the wedge product term by term in the expression of the forms as sums of terms.
⌃
This definition looks complicated, but in the end it is fairly simple. It is easier to understand
it by working through some examples.
Example 4.2.2 The wedge product of two one-forms. Consider the wedge product of
the two one-forms Ê = x dx + x dy + ez dz and ÷ = y dx + x dy + z dz on R3 . By definition,
CHAPTER 4. k-FORMS 74

the result is the two-form given by:

Ê · ÷ = (x dx + x dy + ez dz) · (y dx + x dy + z dz)
=xy dx · dx + x2 dx · dy + xz dx · dz + xy dy · dx + x2 dy · dy + xz dy · dz
+ yez dz · dx + xez dz · dy + zez dz · dz
=(xz ≠ xez )dy · dz + (≠xz + yez )dz · dx + (x2 ≠ xy) dx · dy,

where in the last line we used the fact that dx · dx = dy · dy = dz · dz = 0 and dx · dy =

≠dy · dx, dx · dz = ≠dz · dx, and dy · dz = ≠dz · dy. ⇤
Example 4.2.3 The wedge product of a one-form and a two-form. Consider the
wedge product of the one-form Ê = xy dx + y dy and the two-form ÷ = z dy · dz on R3 . The
result is the three-form:

Ê · ÷ = (xy dx + y dy) · (z dy · dz)

=xyz dx · dy · dz + yz dy · dy · dz
=xyz dx · dy · dz,

where we used the fact that dy · dy · dz = 0 since it has repeated factors. ⇤

Example 4.2.4 The wedge product of a zero-form and a k-form. The wedge product
with a zero-form is just a standard product, since a zero-form is just a function. For instance,
given the zero-form f = x and the two-form Ê = ydy · dz on R3 , the wedge product is the
two-form:
f · Ê = xy dy · dz.
We usually write f Ê, without the · symbol, when we multiply with a zero-form, since it is
just a function.
Note that this generalizes the statement that a one-form mutiplied by a function is another
one-form; this is also true for a two-form and a three-form. ⇤
Remark 4.2.5 These examples pretty much exhaust the possible non-zero wedge products in
R3 . Indeed, since we know that k-forms with k Ø 4 necessarily vanish in R3 , this means that
the only possible non-zero wedge products are:

• A zero-form (i.e. a function) with a k-form, with k œ {0, 1, 2, 3};

• A one-form with a one-form, which gives a two-form;

• A one-form with a two-form, which gives a three-form.

All other wedge products will necessarily vanish.

Now that we are familiar with the wedge product, a natural question is whether it is
“commutative”. Pick two differential forms Ê and ÷. Is Ê · ÷ equal to ÷ · Ê? The answer is
no, not quite! The precise statement is the following lemma.
Lemma 4.2.6 Comparing Ê · ÷ to ÷ · Ê. Let Ê be a k-form and ÷ be an m-form. Then
Ê · ÷ = (≠1)km ÷ · Ê.
In other words:
CHAPTER 4. k-FORMS 75

• If either k or m is even, then Ê · ÷ = ÷ · Ê, and the wedge product is commutative.

• If both k and m are odd, then Ê · ÷ = ≠÷ · Ê, and the wedge product is anti-commutative.

In particular, if Ê is a k-form with k odd, then Ê · Ê = 0.

Proof. This statement follows from the fact that exchanging two dxi ’s in a basic form picks a
sign. More precisely, let us first assume that Ê and ÷ take the simple forms:

Ê = f dxi1 · . . . dxik , ÷ = g dxj1 · . . . · dxjm .

Then
Ê · ÷ = f g (dxi1 · . . . dxik ) · (dxj1 · . . . · dxjm ) ,
while
÷ · Ê = f g (dxj1 · . . . · dxjm ) · (dxi1 · . . . dxik ) .
To relate the second expression to the first, we need to move the dxj ’s to the left of the dxi ’s.
We first move dxi1 to the left. Each time we pass a dxj , we pick a sign. So in the end we get:

÷ · Ê = f g(≠1)m dxi1 · (dxj1 · . . . · dxjm ) · (dxi2 · . . . dxik ) .

We do the same thing with dxi2 , moving it pass all the dxj ’s, and so on all the way to dxik .
The final result is

÷ · Ê =f g(≠1)km dxi1 · . . . dxik · (dxj1 · . . . · dxjm )

=(≠1)km Ê · ÷.

Finally, the statement is proved for general differential forms by doing this manipulation term
by term after distributing the wedge product. ⌅

4.2.2 The wedge product and vector calculus

In Table 4.1.11 we established a dictionary between differential forms in R3 and vector calculus
concepts. We can extend this dictionary to understand the concept of wedge product in vector
calculus.
As we saw, there are really only three types of non-zero wedge products in R3 :
1. The wedge product of a zero-form with a k-form with k œ {0, 1, 2, 3};
2. A one-form with a one-form;
3. A one-form with a two-form.
Let us now translate each of those operations in the language of vector calculus.
First, multiplying a zero-form by a k-form is just multiplying the k-form by a function.
So the same is true in vector calculus: we multiply the corresponding vector field or function
by a function. No big deal.
Multiplying two one-forms is interesting though. Let Ê = f1 dx + f2 dy + f3 dz and
÷ = g1 dx + g2 dy + g3 dz. Then the wedge product is
Ê · ÷ = (f2 g3 ≠ f3 g2 )dy · dz + (f3 g1 ≠ f1 g3 )dz · dx + (f1 g2 ≠ f2 g1 )dx · dy.
CHAPTER 4. k-FORMS 76

In terms of the associated vector fields

F = (f1 , f2 , f3 ), G = (g1 , g2 , g3 ),

what we are doing is constructing a new vector field, let’s call it H for the time being,
associated to Ê ·÷, with component functions given by (according to the dictionary established
in Table 4.1.11 for relating a two-form to a vector field):

H = (f2 g3 ≠ f3 g2 , f3 g1 ≠ f1 g3 , f1 g2 ≠ f2 g1 ) .

What is this vector field? It is nothing but the cross-product of the vector fields F and G!
Indeed,
F ◊ G = (f2 g3 ≠ f3 g2 , f3 g1 ≠ f1 g3 , f1 g2 ≠ f2 g1 ) .
Therefore, we end up with the following statement:
Lemma 4.2.7 The wedge product of two one-forms is the cross-product of the
associated vector fields. If Ê and ÷ are two one-forms on U ™ R3 , with associated vector
fields F and G, then the vector field associated to the two-form Ê · ÷ via the dictionary in
Table 4.1.11 is the cross-product F ◊ G.
Finally, we consider the wedge product of a one-form and a two-form. Let Ê = f1 dx +
f2 dy + f3 dz and ÷ = g1 dy · dz + g2 dz · dx + g3 dx · dy. Then the wedge product is the
three-form:
Ê · ÷ = (f1 g1 + f2 g2 + f3 g3 ) dx · dy · dz.
According to the dictionary in Table 4.1.11, we thus conclude that the function associated
to the three-form Ê · ÷ is nothing but the dot product of the two vector fields F and G
associated to Ê and ÷ respectively:

F · G = f1 g1 + f2 g2 + f3 g3 .

Neat! So we get the following result:

Lemma 4.2.8 The wedge product of a one-form and a two-form is the dot product
of the associated vector fields. If Ê is a one-form and ÷ a two-form on U ™ R3 , with
associated vector fields F and G via the dictionary in Table 4.1.11, then the function associated
to the three-form Ê · ÷ is the dot product F · G.
These two lemmas justify the ordering in the definition of two-form mentioned in Re-
mark 4.1.9.
We can summarize the dictionary between the wedge product and vector products in the
following two tables:
Table 4.2.9 Dictionary between the wedge product of two one-forms in R3 and
vector calculus concepts
Differential form concept Vector calculus concept
1-form Ê vector field F
1-form ÷ vector field G
2-form Ê · ÷ vector field F ◊ G
CHAPTER 4. k-FORMS 77

Table 4.2.10 Dictionary between the wedge product of a one-form and a two-form
in R3 and vector calculus concepts
Differential form concept Vector calculus concept
1-form Ê vector field F
2-form ÷ vector field G
3-form Ê · ÷ vector field F · G

4.2.3 Exercises
1. Let Ê = ex dx + y dz and ÷ = xy dx + z dy + y dz. Find Ê · ÷, and write your result in
standard form.
Solution. We find (using the fact that dx · dx = dy · dy = dz · dz = 0):

Ê · ÷ =(ex dx + y dz) · (xy dx + z dy + y dz)

=zex dx · dy + yex dx · dz + xy 2 dz · dx + yz dz · dy
= ≠ yz dy · dz + (xy 2 ≠ yex ) dz · dx + zex dx · dy.
2. Let Ê = x dy · dz + y dz · dx + z dx · dy and ÷ = x dx + y dy + z dz. Find Ê · ÷, and
write your result in standard form.
Solution. We find (using the fact that any wedge product with two repeated dx, dy or
dz is zero):

Ê · ÷ =(x dy · dz + y dz · dx + z dx · dy) · (x dx + y dy + z dz)

=x2 dy · dz · dx + y 2 dz · dx · dy + z 2 dx · dy · dz
=(x2 + y 2 + z 2 ) dx · dy · dz,

where we used the fact that dy · dz · dx = dz · dx · dy = dx · dy · dz.

3. Let Ê = f1 dx + f2 dy and ÷ = g1 dx + g2 dy be one-forms on R2 , with associated vector
fields F = (f1 , f2 ) and G = (g1 , g2 ). Show that
A B
f g
Ê · ÷ = det 1 1 dx · dy.
f2 g2

Solution. Let Ê = f1 dx + f2 dy, and ÷ = g1 dx + g2 dy. The associated vector fields

are F = (f1 , f2 ), G = (g1 , g2 ). We calculate the wedge product:

Ê · ÷ =(f1 dx + f2 dy) · (g1 dx + g2 dy)

=f1 g2 dx · dy + f2 g1 dy · dx
=(f1 g2 ≠ f2 g1 )dx · dy
A B
f g
= det 1 1 dx · dy.
f2 g2

We note that this determinant could be taken as a “definition” of what “cross-product”

of vector fields means in R2 . Note that the result however is a function, not a vector
field. From the point of view of differential forms, we could define the “cross-product” of
CHAPTER 4. k-FORMS 78

vector fields in Rn as follows: take two one-forms Ê and ÷ on Rn , with associated vector
fields F and G. We would define the “cross-product” of the vector fields as being given
by the two-form Ê · ÷ in Rn . Note that only on R3 can we associate to the result a new
vector field (this is because of Hodge duality between one-forms and two-forms in R3 ,
see Section 4.8). For instance, in R4 a two-form has 6 component functions (there are 6
independent non-vanishing basic two-forms in R4 ), so it cannot be associated to a vector
field.
4. Let Ê = f dx + g dy + h dz be an arbitrary one-form on R3 . By doing an explicit
calculation, show that Ê · Ê = 0.
Solution. We find:

Ê · Ê =(f dx + g dy + h dz) · (f dx + g dy + h dz)

=f g dx · dy + f h dx · dz + gf dy · dx + gh dy · dz + hf dz · dx + hg dz · dy
=(gh ≠ hg) dy · dz + (hf ≠ f h) dz · dx + (f g ≠ gf ) dx · dy.

Since functions commute with each other, i.e. gh = hg, hf = f h, f g = gf , we conclude

that Ê · Ê = 0.
5. Find the cross-product of the vectors F = (1, 1, 0) and G = (0, 2, 3) in R3 by computing
the wedge product of the associated one-forms.
Solution. The one-forms associated to F and G are Ê = dx + dy and ÷ = 2dy + 3dz.
We calculate the wedge product:

Ê · ÷ =(dx + dy) · (2dy + 3dz)

=2dx · dy + 3dx · dz + 3dy · dz
=3dy · dz ≠ 3dz · dx + 2dx · dy

According to the dictionary Table 4.1.11, the vector field associated to this two-form is

H = F ◊ G = (3, ≠3, 2).

This is indeed the cross-product, as you can calculate using standard formulae from
linear algebra. For instance, you may have seen the formula:
Q R
i j k
c d
F ◊ G = det a F1 F2 F3 b
G 1 G2 G3
Q R
i j k
c d
= det a1 1 0 b
0 2 3
=3i ≠ 3j + 2k,

which, in component notation, reads (3, ≠3, 2).

6. Let A, B, C be vectors in R3 . In your linear algebra course you may have seen that

A · (B ◊ C) = B · (C ◊ A) = C · (A ◊ B).
CHAPTER 4. k-FORMS 79

Prove this property by looking at the wedge product of the three one-forms Ê, ÷, ⁄
associated to the vectors A, B, C.
Solution. As discussed in Lemma 4.2.7 and Lemma 4.2.8, we know that

Ê · ÷ · ⁄ = A · (B ◊ C)dx · dy · dz,

÷ · ⁄ · Ê = B · (C ◊ A)dx · dy · dz,
and
⁄ · Ê · ÷ = C · (A ◊ B)dx · dy · dz.
But since ⁄, Ê, ÷ are one-forms, we know that Ê · ÷ = ≠÷ · Ê, ÷ · ⁄ = ≠⁄ · ÷, and
Ê · ⁄ = ≠⁄ · Ê. Therefore,

Ê · ÷ · ⁄ = ÷ · ⁄ · Ê = ⁄ · Ê · ÷,

and the statement is proved. In other words, it follows directly from anti-commutativity
of the wedge product of one-forms.

4.3 Differentiating k-forms: the exterior derivative

We know how to add k-forms, and now also how to mutiply k-forms, thanks to the notion
of wedge product. In this section we study how we can “differentiate” k-forms, using the
notion of exterior derivative, which generalizes the differential of a function introduced in
Definition 2.2.1.

Objectives
You should be able to:

• Define the exterior derivative of a k-form, focusing on zero-, one- and two-forms in R3 .

• State and use the graded product rule for the exterior derivative.

• Show that applying the exterior derivative twice always gives zero.

4.3.1 The exterior derivative

Let us start by recalling the definition of the differential of a function f : U æ R, with U ™ R3 ,
from Definition 2.2.1. The differential df is the one-form on U given by
ˆf ˆf ˆf
df = dx + dy + dz.
ˆx ˆy ˆz
Since we now think of f as a zero-form, we see that the operator d takes a zero-form and
outputs a one-form. Our goal is to generalize this operation, which we will now call the
“exterior derivative”, to take a k-form and ouput a (k + 1)-form.
Definition 4.3.1 The exterior derivative of a k-form. The exterior derivative of a
CHAPTER 4. k-FORMS 80

zero-form f on U µ Rn is the one-form df on U given by:

n
ÿ ˆf
df = dxi ,
i=1
ˆxi

which is the same thing as the differential introduced

ÿin Definition 2.2.1.
The exterior derivative of a k-form Ê = fi1 ···ik dxi1 · · · · · dxik on U µ Rn
1Æi1 <···<ik Æn
is the (k + 1)-form dÊ on U given by:
ÿ
dÊ = d(fi1 ···ik ) · dxi1 · · · · · dxik ,
1Æi1 <···<ik Æn

where d(fi1 ···ik ) means the exterior derivative of the zero-form (function) fi1 ···ik . In other
words, we are applying the exterior derivative d to the component functions of Ê. ⌃
This definition may seem a little daunting because of the summations, so let us be more
explicit for k-forms in R3 .
Lemma 4.3.2 The exterior derivative in R3 .

1. If f is a zero-form on U ™ R3 , then its exterior derivative df is the one-form:

ˆf ˆf ˆf
df = dx + dy + dz.
ˆx ˆy ˆz

2. If Ê = f dx + g dy + h dz is a one-form on U ™ R3 , then its exterior derivative dÊ is

the two-form:

dÊ =d(f ) · dx + d(g) · dy + d(h) · dz

3 4 3 4 3 4
ˆh ˆg ˆf ˆh ˆg ˆf
= ≠ dy · dz + ≠ dz · dx + ≠ dx · dy.
ˆy ˆz ˆz ˆx ˆx ˆy

3. If ÷ = f dy · dz + g dz · dx + h dx · dy is a two-form on U ™ R3 , then its exterior

derivative d÷ is the three-form:

d÷ =d(f ) · dy · dz + d(g) · dz · dx + d(h) · dx · dy.

3 4
ˆf ˆg ˆh
= + + dx · dy · dz.
ˆx ˆy ˆz

And that’s it. In particular, the exterior derivative of a three-form on R3 must vanish, since it
would give a four-form, but all k-forms with k Ø 4 necessarily vanish on R3 by Lemma 4.1.6.
Proof. We start with Definition 4.3.1 restricted to the case with n = 3. For the exterior
derivative of a zero-form, the statement is obvious. For the exterior derivatives of a one-form
and a two-form, all we have to do is evaluate the exterior derivatives d(f ), d(g) and d(h) of
the zero-form f, g, h, and rearrange terms using Lemma 4.1.6. ⌅
Note that you certainly should not aim at learning these formulae by heart. The whole
point is precisely that you don’t need to learn these formulae! All you need to remember is
CHAPTER 4. k-FORMS 81

that, to evaluate the exterior derivative of a k-form, you act with the exterior derivative on
the component functions of the k-form. This may be clearer with examples.
Example 4.3.3 The exterior derivative of a zero-form on R3 . We are already familiar
with the calculation of the exterior derivative of a zero-form, since this is the same thing as
the calculation of the differential of a function that we defined in Definition 2.2.1. But let us
give an example here for completeness.
Let f (x, y, z) = y ln(x) + z be a smooth function on U = {(x, y, z) œ R3 | x > 0}. Its
exterior derivative is the one-form df on U given by:
ˆf ˆf ˆf
df = dx + dy + dz
ˆx ˆy ˆz
y
= dx + ln(x) dy + dz.
x
⇤
Example 4.3.4 The exterior derivative of a one-form on R3 . Let Ê = xy dx + (z +
y) dy + xyz dz be a one-form on R . Its exterior derivative is the two-form dÊ on R3 given by:
3

dÊ =d(xy) · dx + d(z + y) · dy + d(xyz) · dz

= (y dx + x dy) · dx + (dy + dz) · dy + (yz dx + xz dy + xy dz) · dz
=x dy · dx + dz · dy + yz dx · dz + xz dy · dz
=(xz ≠ 1) dy · dz ≠ yz dz · dx ≠ x dx · dy.

⇤
Example 4.3.5 The exterior derivative of a two-form on R3 . Let Ê = (x2 + y 2 ) dy ·
dz + sin(z) dz · dx + cos(xy) dx · dy be a two-form on R3 . Its exterior derivative is the
three-form dÊ on R3 given by:

dÊ =d(x2 + y 2 ) · dy · dz + d(sin(z)) · dz · dx + d(cos(xy)) · dx · dy

=(2x dx + 2y dy) · dy · dz + (cos(z) dz) · dz · dx + (≠y sin(xy) dx ≠ x sin(xy) dy) · dx · dy
=2x dx · dy · dz.

We see that going from the second line to the third line, most terms vanish, since anytime we
take the wedge of dx with itself we get zero, and same for dy and dz. ⇤
To end this section, we note that the exterior derivative is linear. If Ê and ÷ are two
k-forms, and a, b œ R, then
d(aÊ + b÷) = adÊ + bd÷.
This is proven in Exercise 4.3.4.3.

4.3.2 The graded product rule

One of the most fundamental properties of the derivative is the product rule:
d df dg
(f g) = g+f .
dx dx dx
CHAPTER 4. k-FORMS 82

Now that we have defined the exterior derivative for differential forms, and that we know how
to multiply differential forms using the wedge product, we could ask whether the exterior
derivative satisfies a similar product rule with respect to the wedge product. It turns out that
it does, but with a twist (or more precisely a sign). We call this the “graded product rule” for
the exterior derivative.
Lemma 4.3.6 The graded product rule for the exterior derivative. Let Ê be a k-form
and ÷ be an l-form on U ™ Rn . Then:

d(Ê · ÷) = d(Ê) · ÷ + (≠1)k Ê · d(÷).

Proof. First, we show that it is true for zero-forms. If Ê and ÷ are zero-forms (that is
k = l = 0), that is functions Ê = f and ÷ = g, then

d(Ê · ÷) =d(f g)
n
ÿ ˆ
= (f g) dxi
i=1
ˆxi
ÿn 3 4
ˆf ˆg
= g+f dxi
i=1
ˆxi ˆxi
=d(f )g + f d(g).

To prove the general statement we unfortunately need to use lots of summations. Let us
introduce the following notation for the k-form Ê and the l-form ÷:
ÿ ÿ
Ê= wi1 ···ik dxi1 · · · · · dxik , ÷= hj1 ···jl dxj1 · · · · · dxjl .
1Æi1 <···<ik Æn 1Æj1 <···<jl Æn

The exterior derivative of the wedge product is:

ÿ ÿ
d(Ê · ÷) = d(wi1 ···ik hj1 ···jl ) · dxi1 · · · · · dxik · dxj1 · · · · · dxjl
1Æi1 <···<ik Æn 1Æj1 <···<jl Æn
ÿ ÿ
= (d(wi1 ···ik )hj1 ···jl + wi1 ···ik d(hj1 ···jl )) · dxi1 · · · · · dxik · dxj1 · · · · · dxjl .
1Æi1 <···<ik Æn 1Æj1 <···<jl Æn

To go from the first to the second line, we used the fact that the product rule is satisfied for
the exterior derivative of the product of zero-forms, as shown above.
Now let us study the terms on the right-hand-side of the equation we just obtained. First,
we get:
ÿ ÿ
d(wi1 ···ik )hj1 ···jl · dxi1 · · · · · dxik · dxj1 · · · · · dxjl
1Æi1 <···<ik Æn 1Æj1 <···<jl Æn
Q R Q R
ÿ ÿ
=a d(wi1 ···ik ) · dxi1 · · · · · dxik b · a hj1 ···jl dxj1 · · · · · dxjl b
1Æi1 <···<ik Æn 1Æj1 <···<jl Æn

= d(Ê) · ÷

All that we did to go from the first line to the second line is move the zero-form (or function)
hj1 ···jl to the right of the dxi ’s, which we can do since it is a zero-form and hence commutes
with the dxi ’s by Lemma 4.2.6.
CHAPTER 4. k-FORMS 83

That takes care of the first set of terms. The remaining ones take the form
ÿ ÿ
wi1 ···ik d(hj1 ···jl ) · dxi1 · · · · · dxik · dxj1 · · · · · dxjl .
1Æi1 <···<ik Æn 1Æj1 <···<jl Æn

We would like to do the same and commute d(hj1 ···jl ) to the right of the dxi ’s, so that we can
identify this term as Ê · d(÷). However, d(hj1 ···jl ) is a one-form, and thus by Lemma 4.2.6
d(hj1 ···jl ) · dxi = ≠dxi · d(hj1 ···jl ). So every time we commute d(hj1 ···jl ) past a dxi , we pick a
sign. Since there are k dxi ’s in this expression, this tells us that it is equal to

(≠1)k Ê · d(÷).

Putting all this together, we end up with the statement that

d(Ê · ÷) = d(Ê) · ÷ + (≠1)k Ê · d(÷),

which is the graded product rule stated in the lemma. ⌅

It is very important not to forget the sign in the graded product rule!
Example 4.3.7 The exterior derivative of the wedge product of two one-forms.
Let Ê = xy dz and ÷ = (y + z) dx + 2dy be two one-forms on R3 . Suppose that we want
to calculate the exterior derivative of the wedge product Ê · ÷. There are two ways we can
do that: we can first find an explicit expression for the wedge product, and then take the
exterior derivative, or we can use the graded product rule. In this example we show that both
calculations give the same result, as they should.
Let us first calculate the wedge product explicitly and take its exterior derivative. We
have:

Ê · ÷ =(xy dz) · ((y + z) dx + 2 dy)

= ≠ 2xydy · dz + xy(y + z)dz · dx.

We then calculate the exterior derivative, which gives the following three-form:

d(Ê · ÷) =d(≠2xy) · dy · dz + d(xy(y + z)) · dz · dx

= (≠2y dx ≠ 2x dy) · dy · dz + (y(y + z) dx + (2xy + xz) dy + xy dz) · dz · dx
=(≠2y + (2xy + xz))dx · dy · dz,

where we used the fact that dy · dz · dx = dx · dy · dz.

Let us now calculate the same exterior derivative but using the graded product rule. Since
Ê is a one-form, we have:
d(Ê · ÷) = d(Ê) · ÷ ≠ Ê · d(÷).
We calculate:

d(Ê) =d(xy) · dz
=ydx · dz + xdy · dz,

and

d(÷) =d(y + z) · dx + d(2) · dy

CHAPTER 4. k-FORMS 84

=dy · dx + dz · dx,

since d(2) = 0. Putting this together, we get:

d(Ê · ÷) =(ydx · dz + xdy · dz) · ((y + z) dx + 2 dy) ≠ (xy dz) · (dy · dx + dz · dx)
=(≠2y + x(y + z) + xy)dx · dy · dz,

which is indeed the same result as obtained above. ⇤

Remark 4.3.8 In R3 , the graded product rule can be split into the four following non-vanishing
cases.

1. If Ê = f is a zero-form (in which case we write f · ÷ = f ÷ as usual when multiplying

with a function) and ÷ = g is a zero-form, then

d(f g) = d(f )g + f d(g).

2. If Ê = f is a zero-form and ÷ is a one-form, then

d(f ÷) = d(f ) · ÷ + f d(÷).

3. If Ê = f is a zero-form and ÷ is a two-form, then

d(f ÷) = d(f ) · ÷ + f d(÷).

4. If Ê is a one-form and ÷ is a one-form, then

d(Ê · ÷) = d(Ê) · ÷ ≠ Ê · d(÷).

4.3.3 d2 = 0
There’s another fundamental property of the exterior derivative: if we apply the exterior
derivative twice, we always get zero. This may seem surprising, as this is certainly not true for
the ordinary derivative d/dx, but it is true for the exterior derivative because of antisymmetry
of the wedge product. More precisely:
Lemma 4.3.9 d2 = 0. Let Ê be a k-form on U ™ Rn . Then

d(d(Ê)) = 0.

In other words, if we apply the exterior derivative twice on any differential form, we always
get zero. We often abbreviate this statement as d2 = 0, meaning that applying the exterior
derivative twice always gives zero.
q
Proof. We write Ê = 1Æi1 <···<ik Æn wi1 ···ik dxi1 · · · · · dxik . We have:
Q R
ÿ
d(d(Ê)) =d a d(wi1 ···ik ) · dxi1 · · · · · dxik b
1Æi1 <···<ik Æn
CHAPTER 4. k-FORMS 85
Q R
ÿ n
ÿ ˆwi1 ···ik
=d a dx– · dxi1 · · · · · dxik b
1Æi1 <···<ik Æn –=1
ˆx–
ÿ ÿn 3 4
ˆwi1 ···ik
= d · dx– · dxi1 · · · · · dxik
1Æi1 <···<ik Æn –=1
ˆx–
Q R
ÿ n ÿ
ÿ n
ˆ 2 wi1 ···ik
= a dx— · dx– b · dxi1 · · · · · dxik .
1Æi1 <···<ik Æn –=1 —=1
ˆx— ˆx–

If we can show that the term in brackets in the last line is zero, then clearly d(d(Ê)) = 0. We
have:
n ÿ
ÿ n
ˆ 2 wi1 ···ik n ÿ
ÿ n
ˆ 2 wi1 ···ik
dx— · dx– = dx— · dx–
–=1 —=1
ˆx— ˆx– –=1 —=1
ˆx– ˆx—
ÿn ÿ n
ˆ 2 wi1 ···ik
=≠ dx– · dx— .
–=1 —=1
ˆx– ˆx—

ˆ2w ˆ2w
In the first, line, we used the fact that ˆx—i1ˆx···ik
–
= ˆx–i1ˆx
···ik
—
by the Clairaut-Schwarz theorem,
since the coefficient functions wi1 ···ik are assumed to be smooth. In the second line, we used
the fact that dx— · dx– = ≠dx– · dx— , by Lemma 4.1.6. Finally, in the summation on the
right-hand-side, we can simply rename – to be —, and — to be –, since those are indices
that are summed over, and hence we can give them the name we want. We end up with the
statement that
n ÿ
ÿ n
ˆ 2 wi1 ···ik n ÿ
ÿ n
ˆ 2 wi1 ···ik
dx— · dx– = ≠ dx— · dx– .
–=1 —=1
ˆx— ˆx– –=1 —=1
ˆx— ˆx–

But the only two-form that is equal to minus itself is the zero two-form, that is
n ÿ
ÿ n
ˆ 2 wi1 ···ik
dx— · dx– = 0.
–=1 —=1
ˆx— ˆx–

Therefore we conclude that d(d(Ê)) = 0. ⌅

4.3.4 Exercises
1. Let Ê = xy dx + xy dy + xz dz on U = {(x, y, z) œ R3 | x ”= 0}. Find the two-form dÊ and
write your result in standard form. What is the vector field associated to dÊ?
Solution. From the definition, we find:
3 4 3 4
y z
dÊ =d(xy) · dx + d · dy + d · dz
x x
1 y 1 z
=y dx · dx + x dy · dx +dy · dy ≠ 2 dx · dy + dz · dz ≠ 2 dx · dz
3 4 x x x x
z y
= 2 dz · dx ≠ x + 2 dx · dy.
x x
CHAPTER 4. k-FORMS 86

Using the dictionary Table 4.1.11, the vector field associated to the two-form dÊ is
3 4
z y
F(x, y, z) = 0, 2
, ≠x ≠ 2 .
x x
2. Find the three-form dÊ if Ê = xyz(dy · dz + dz · dx + dx · dy), and write your result in
standard form.
Solution. We find:

dÊ =d(xyz) · (dy · dz + dz · dx + dx · dy)

= (yz dx + xz dy + xy dz) · (dy · dz + dz · dx + dx · dy)
=yz dx · dy · dz + xz dy · dz · dx + xy dz · dx · dy
= (yz + xz + xy) dx · dy · dz.
3. Show that the exterior derivative is linear. That is, if Ê and ÷ are two k-forms, and
a, b œ R, then
d(aÊ + b÷) = adÊ + bd÷.

Solution. First, we show that the property holds if Ê and ÷ are 0-forms. Since those
are simply functions, let us write them as f and g. Then:
n
ÿ ˆ
d(af + bg) = (af + bg) dxi
i=1
ˆxi
n 3
ÿ 4
ˆf ˆg
= a +b dxi
i=1
ˆxi ˆxi
ÿn ÿn
ˆf ˆg
=a dxi + b dxi
i=1
ˆxi i=1
ˆxi

=adf + bdg.

For the second equality we used the fact that partial derivatives are linear.
Now we can prove the general case. Suppose that Ê and ÷ are k-forms on U ™ Rn .
Let ÿ
Ê= fi1 ···ik dxi1 · · · · · dxik
1Æi1 <···<ik Æn

and ÿ
÷= gi1 ···ik dxi1 · · · · · dxik .
1Æi1 <···<ik Æn

Then
ÿ
d(aÊ + b÷) = d(afi1 ···ik + bgi1 ···ik ) · dxi1 · · · · · dxik
1Æi1 <···<ik Æn
ÿ
= (a dfi1 ···ik + b dgi1 ···ik ) · dxi1 · · · · · dxik
1Æi1 <···<ik Æn
ÿ
=a dfi1 ···ik · dxi1 · · · · · dxik
1Æi1 <···<ik Æn
CHAPTER 4. k-FORMS 87
ÿ
+b dgi1 ···ik · dxi1 · · · · · dxik
1Æi1 <···<ik Æn
=adÊ + bd÷.

In the second equality we used the calculation above that showed that linearity holds for
0-forms.
4. Let Ê = xey dx + z dy + yex dz. Show by explicit calculation that d2 Ê = 0.
Solution. Of course, we know that d2 Ê = 0 since this is true for any differential form
Ê, as proven in Lemma 4.3.9. But let us show that it is true by explicit calculation for
this particular one-form Ê.
We first calculate the two-form dÊ:

dÊ =d(xey ) · dx + dz · dy + d(yex ) · dz
=xey dy · dx + ey dx · dx + dz · dy + yex dx · dz + ex dy · dz
=(ex ≠ 1) dy · dz ≠ yex dz · dx ≠ xey dx · dy.

We then calculate the three-form d2 Ê:

d2 Ê =d(dÊ)
=d(ex ≠ 1) · dy · dz ≠ d(yex ) · dz · dx ≠ d(xey ) · dx · dy
=ex dx · dy · dz ≠ yex dx · dz · dx ≠ ex dy · dz · dx ≠ xey dy · dx · dy
≠ ey dx · dx · dy
=(ex ≠ ex )dx · dy · dz
=0,

as expected.
5. Let Ê = (x2 ≠ y 2 ) dx + y dz and ÷ = (x2 + y 2 ) dy + y dz. By explicit calculation, show
that
d(Ê · ÷) = dÊ · ÷ ≠ Ê · d÷,
which is consistent with the graded product rule Lemma 4.3.6 since Ê is a one-form.
Solution. We need to calculate d(Ê · ÷), dÊ · ÷, and Ê · d÷. First, we calculate Ê · ÷:
Ê · ÷ =((x2 ≠ y 2 ) dx + y dz) · ((x2 + y 2 ) dy + y dz)
=(x4 ≠ y 4 )dx · dy + y(x2 ≠ y 2 ) dx · dz + y(x2 + y 2 ) dz · dy
= ≠ y(x2 + y 2 ) dy · dz ≠ y(x2 ≠ y 2 ) dz · dx + (x4 ≠ y 4 )dx · dy.
Then
d(Ê · ÷) = ≠ d(y(x2 + y 2 )) · dy · dz ≠ d(y(x2 ≠ y 2 )) · dz · dx + d(x4 ≠ y 4 ) · dx · dy
= ≠ 2xy dx · dy · dz ≠ (x2 ≠ 3y 2 ) dy · dz · dx
=(3y 2 ≠ 2xy ≠ x2 )dx · dy · dz.
Next, we calculate dÊ · ÷. We have:
dÊ =d(x2 ≠ y 2 ) · dx + dy · dz
CHAPTER 4. k-FORMS 88

= ≠ 2ydy · dx + dy · dz
=dy · dz + 2ydx · dy.

Then

dÊ · ÷ =(dy · dz + 2ydx · dy) · ((x2 + y 2 ) dy + y dz)

=2y 2 dx · dy · dz.

Finally, we calculate Ê · d÷. We have:

d÷ =d(x2 + y 2 ) · dy + dy · dz
=2x dx · dy + dy · dz
=dy · dz + 2x dx · dy.

Then

Ê · d÷ =((x2 ≠ y 2 ) dx + y dz) · (dy · dz + 2x dx · dy)

=(x2 ≠ y 2 )dx · dy · dz + 2xy dz · dx · dy
=(x2 + 2xy ≠ y 2 ) dx · dy · dz.

Putting all this together, we conclude that

dÊ · ÷ ≠ Ê · d÷ =(3y 2 ≠ 2xy ≠ x2 )dx · dy · dz

=d(Ê · ÷),

as expected from the graded product rule Lemma 4.3.6.

6. Let Ê be a k-form, ÷ an m-form, and ⁄ a ¸-form, all on U µ Rn . Show that

d(Ê · ÷ · ⁄) = dÊ · ÷ · ⁄ + (≠1)k Ê · d÷ · ⁄ + (≠1)k+m Ê · ÷ · d⁄.

Solution. First, using the graded product rule Lemma 4.3.6, we get that

d(Ê · ÷ · ⁄) = dÊ · (÷ · ⁄) + (≠1)k Ê · d(÷ · ⁄).

Next, again from the graded product rule we know that

d(÷ · ⁄) = d÷ · ⁄ + (≠1)m ÷ · d⁄.

Putting this together, we get:

d(Ê · ÷ · ⁄) = dÊ · ÷ · ⁄ + (≠1)k Ê · d÷ · ⁄ + (≠1)k+m Ê · ÷ · d⁄.

7. Let Ê = (x + y) dx + xy dy be a one-form on R2 , and let „ : R2 æ R2 be given by
„(u, v) = (eu+v , eu≠v ). Find the two-form d(„ú Ê).
Solution. We first calculate the pullback one-form „ú Ê. We get:
1 2 ! "
„ú Ê =(eu+v + eu≠v ) eu+v du + eu+v dv + eu+v eu≠v eu≠v du ≠ eu≠v dv
CHAPTER 4. k-FORMS 89
1 2 1 2
= e2(u+v) + e2u + e3u≠v du + e2(u+v) + e2u ≠ e3u≠v dv.

We can then calculate its exterior derivative. We get:

1 2 1 2
d(„ú Ê) =d e2(u+v) + e2u + e3u≠v · du + d e2(u+v) + e2u ≠ e3u≠v · dv
1 2 1 2
= 2e2(u+v) ≠ e3u≠v dv · du + 2e2(u+v) + 2e2u ≠ 3e3u≠v du · dv
1 2
=2 e2u ≠ e3u≠v du · dv.
8. Let f, g : U æ R be smooth functions with U ™ Rn , and – : [a, b] æ Rn be a parametric
curve whose image is in U . Using the product rule for the exterior derivative d(f g), show
that ⁄ ⁄
f dg = f (–(b))g(–(b)) ≠ f (–(a))g(–(a)) ≠ gdf.
– –
One can think of this as the natural generalization of “integration by parts” to line
integrals of one-forms. Indeed, if the curve is just an interval in R, this reduces to the
standard statement of integration by parts for definite integrals.
Solution. From the graded product rule, we know that

d(f g) = f dg + gdf,

since f is a 0-form (here we omit the wedge product symbol, since we are either multiplying
two functions or a function with a one-form). By the Fundamental Theorem of line
integrals, we know that
⁄
d(f g) = f (–(b))g(–(b)) ≠ f (–(a))g(–(a)),
–

since d(f g) is an exact one-form. Using the graded product rule, we thus conclude that
⁄ ⁄ ⁄
d(f g) = f dg + gdf = f (–(b))g(–(b)) ≠ f (–(a))g(–(a)).
– – –
s
Solving for – f dg, we get the desired statement.
9. Let Ê and ÷ be one-forms that differ by the exterior derivative of a 0-form, that is,
÷ = Ê + df
for some function f . Show that
d(Ê · ÷) = dÊ · df.

Solution. We have:
d(Ê · ÷) = d(Ê · (Ê + df )) = d(Ê · Ê) + d(Ê · df ).
The first term on the right-hand-side is zero, since Ê · Ê = 0 for a one-form (see
Lemma 4.2.6; the wedge product is anti-commutative for odd forms). As for the second
term, we use the graded product rule and the fact that d2 = 0 to get:
d(Ê · ÷) = dÊ · df ≠ Ê · d2 f = dÊ · df.
CHAPTER 4. k-FORMS 90

4.4 The exterior derivative and vector calculus

In this section we continue developing our dictionary between differential forms and standard
vector calculus concepts. We introduce the vector calculus operations corresponding to the
exterior derivative.

Objectives
You should be able to:

• Relate the exterior derivative of a zero-form in R3 to the gradient of a function.

• Relate the exterior derivative of a one-form in R3 to the curl of a vector field.

• Relate the exterior derivative of a two-form in R3 to the div of a vector field.

• Derive various vector calculus identities from the graded product rule for the exterior
derivative and the statement that d2 = 0.

4.4.1 Grad, div and curl

In R3 , we saw that there are three possibilities to get a non-zero differential forms as a result
of acting with the exterior derivative: either we take the exterior derivative of a zero-form, a
one-form, or a two-form. All three of these operations are given separate names and notation
in the standard vector calculus language.
Definition 4.4.1 The gradient of a function. Let f be a zero-form (a function) on
U ™ R3 . Its exterior derivative df is the one-form
ˆf ˆf ˆf
df = dx + dy + dz.
ˆx ˆy ˆz
We define the gradient of f , and denote it by Òf , to be the vector field associated to the
one-form df according to Table 4.1.11:
3 4
ˆf ˆf ˆf
Òf = , , .
ˆx ˆy ˆz
Note that the input of the gradient is a function, and the output is a vector field. ⌃
Of course, we already knew the definition of the gradient of a function, and how it is the
vector field associated to the exterior derivative of a zero-form: this was already stated in
Fact 2.2.2. We include the statement here for completeness. From our point of view, we could
take this as the definition of the gradient of a function: it is the vector field associated to the
exterior derivative of a zero-form.
Definition 4.4.2 The curl of a vector field. Let Ê = f1 dx + f2 dy + f3 dz be a one-form
on U ™ R3 , with its associated vector field F = (f1 , f2 , f3 ). Its exterior derivative dÊ is the
two-form
3 4 3 4 3 4
ˆf3 ˆf2 ˆf1 ˆf3 ˆf2 ˆf1
dÊ = ≠ dy · dz + ≠ dz · dx + ≠ dx · dy.
ˆy ˆz ˆz ˆx ˆx ˆy
CHAPTER 4. k-FORMS 91

We define the curl of F, and denote it by Ò ◊ F, to be the vector field associated to the
two-form dÊ according to Table 4.1.11:
3 4
ˆf3 ˆf2 ˆf1 ˆf3 ˆf2 ˆf1
Ò◊F= ≠ , ≠ , ≠ .
ˆy ˆz ˆz ˆx ˆx ˆy
Note that input of the curl is a vector field, and the output is also a vector field. ⌃
Finally, we can apply the exterior derivative to a two-form to get a three-form.
Definition 4.4.3 The divergence of a vector field. Let ÷ = f1 dy · dz + f2 dz · dx +
f3 dx · dy be a two-form on U ™ R3 , with its associated vector field F = (f1 , f2 , f3 ). Its
exterior derivative d÷ is the three-form
3 4
ˆf1 ˆf2 ˆf3
d÷ = + + dx · dy · dz.
ˆx ˆy ˆz
We define the divergence of F, and denote it by Ò · F, to be the function associated to the
three-form d÷ according to Table 4.1.11:
ˆf1 ˆf2 ˆf3
Ò·F= + + .
ˆx ˆy ˆz
Note that the input of the diveregence is a vector field, and the output is a function. ⌃
Remark 4.4.4 Now you can start to see the power of developing the framework of differential
forms. These three operators, namely grad, div, and curl, which appear as independent
operators in vector calculus, are just the action of the same operator, namely the exterior
derivative, but on zero-, one-, and two-forms respectively. Moreover, we don’t need to
remember these definitions by heart: all we need to remember is how to act with the exterior
derivative on k-forms, which simply amounts to acting with the exterior derivative on the
component functions. So much simpler!
Even more powerful is the fact that the framework of differential forms naturally extend
to any dimension, not only R3 . However, the defintions of grad, curl, div, the cross-product,
etc. rely on the geometry of R3 . The natural generalization to higher dimensions is just the
action of the exterior derivative as we defined it.
Remark 4.4.5 The introduction of the curl of a vector field allows us to rephrase the
screening test for conservative vector fields in R3 in a nicer way. Looking at the screening test
in Lemma 2.2.16, it is clear that the screening test for a vector field F is satisfied if and only if

Ò ◊ F = 0.

In other words, the screening test was simply saying that the curl of the vector field vanishes.
Remark 4.4.6 Just as for the cross product of two vectors, in standard vector calculus
textbooks a determinant formula is usually given to remember how to calculate the curl of a
vector field F = (f1 , f2 , f3 ) in R3 :
Q R
i j k
c ˆ d
Ò ◊ F = det a ˆx
ˆ ˆ
ˆy ˆz b ,
f1 f2 f3
CHAPTER 4. k-FORMS 92

where i, j, k are the unit vectors in the x, y, z directions. You can use this formula if you
want. Or you can remember that the curl is obtained by taking the exterior derivative of the
one-form associated to F.
Example 4.4.7 Maxwell’s equations. Maxwell’s equations form the foundations of
electromagnetism. It turns out that they are written in terms of the divergence and the curl.
More precisely, if E is the electric vector field, and B is the magnetic vector field, both defined
on R3 (our space), Maxwell’s equations state that

Ò · E =4fifl,
Ò · B =0,
1 ˆB
Ò◊E+ =0,
c ˆt
1 ˆE 4fi
Ò◊B≠ = J,
c ˆt c
where c is the speed of light, fl is the total electric charge density, and J is the total electric
current density (which is a vector field). In particular, the equations simplify when there is
no charge or current (such as in vacuum), with fl = J = 0.
Note that we are abusing notation a little bit here. Those equations are the “time-
dependent” Maxwell’s equations. What this means is that we think of E and B as vector
fields in R3 (in space), but that also depend on another variable t corresponding to time. This
is why the equations above include partial derivatives of E and B with respect to t. The
“time-independent” Maxwell’s equations, in which E and B are true vector fields on R3 (with
no time dependence), would correspond to setting the two terms involving partial derivatives
with respect to t to zero. ⇤
3
We can summarize the dictionary between the exterior derivative in R and vector calculus
operations in the following table.
Table 4.4.8 Dictionary between the exterior derivative in R3 and vector calculus
concepts
Differential form concept Vector calculus concept
d of a 0-form df gradient Òf
d of a 1-form dÊ curl Ò◊F
d of a 2-form d÷ divergence Ò · F

4.4.2 The graded product rule and vector calculus identities

The power of the formalism of differential forms is further highlighted by the following lemma.
We saw above that the standard concepts of grad, curl, and div, are just reformulations of the
exterior derivative. We showed in Lemma 4.3.6 that the exterior derivative satisfies a graded
product rule. In R3 , this graded product rule can be split into two cases, as in Remark 4.3.8,
depending on whether Ê is a zero- or a one-form. Using the definition of grad, curl, and div
in Definition 4.4.1, Definition 4.4.2 and Definition 4.4.3, those statements can be translated
into vector calculus identities. The result is the following lemma.
CHAPTER 4. k-FORMS 93

Lemma 4.4.9 Vector calculus identities, part 1. Let f, g be smooth functions on

U ™ R3 , and F, G be smooth vector fields on U . Then the following identities are satisfied:

1.
Ò(f g) = (Òf )g + f (Òg),

2.
Ò ◊ (f F) = (Òf ) ◊ F + f Ò ◊ F,

3.
Ò · (f F) = (Òf ) · F + f Ò · F,

4.
Ò · (F ◊ G) = (Ò ◊ F) · G ≠ F · (Ò ◊ G).
Proof. This is just the reformulation in terms of vector fields of the four different non-vanishing
cases of the graded product rule for differential forms in R3 (see Remark 4.3.8). ⌅
Now you may be starting to like this. Learning this kind of vector calculus identities by
heart is frustrating. But these are just reformulations of the one and only graded product
rule for the exterior derivative Lemma 4.3.6, which is all that you have to remember (sure,
there is an annoying sign in the graded product rule, but it’s much better than learning vector
calculus identities!).

4.4.3 d2 = 0 and vector calculus identities

Another key property of the exterior derivative is that d2 = 0, see Lemma 4.3.9. In R3 ,
this corresponds to two separate statements, namely that d(d(f )) = 0 with f a zero-form (a
function), and d(d(Ê)) = 0 with Ê a one-form. The corresponding vector calculus identities
are the following:
Lemma 4.4.10 Vector calculus identities, part 2. Let f be a smooth function on
U ™ R3 and F a smooth vector field on U . Then the following identities are satisfied:

1.
Ò ◊ (Òf ) = 0 (curl of grad is zero),

2.
Ò · (Ò ◊ F) = 0 (div of curl is zero).

Proof. This is just the reformulation of the statement that d2 = 0 for a zero-form and a
one-form in R3 . ⌅
The game of translating easy statements for the exterior derivative into complicated
statements for grad, curl, and div is fun, isn’t it? Let’s prove one more vector calculus identity
for now, which follows by combining the statement that d2 = 0 with the graded product rule.
Lemma 4.4.11 Vector calculus identities, part 3. Let f, g, h be smooth functions on
U ™ R3 . Then:
Ò · (f (Òg ◊ Òh)) = Òf · (Òg ◊ Òh).
CHAPTER 4. k-FORMS 94

Proof. Consider the action of the exterior derivative on the two-form f dg · dh:

d(f dg · dh) = df · (dg · dh) + f d(dg · dh),

where we used the graded product rule and the fact that f is a zero-form. Using the graded
product rule again, we know that

d(dg · dh) =d(dg) · dh ≠ dg · d(dh)

=0,

since d(dg) = d(dh) = 0. Thus

d(f dg · dh) = df · (dg · dh).

The translation for the associated vector fields is:

Ò · (f (Òg ◊ Òh)) = Òf · (Òg ◊ Òh),

as claimed. ⌅

4.4.4 Two more vector calculus identities

We end this section by noting that there are two more vector calculus identities involving grad,
curl and div. We will present the identities without proof here. To get them from differentials
forms, we would need to introduce the concept of Lie derivatives, which is beyond the scope
of this course.
Keep in mind that you certainly do not need to learn these identities by heart! We are
presenting them here just so that you are aware of them.
Lemma 4.4.12 Vector calculus identities, part 4. Let F and G be smooth vector fields
on U ™ R3 . Then the following identities are satisfied:

1.
Ò(F · G) = F ◊ (Ò ◊ G) + (Ò ◊ F) ◊ G + (G · Ò)F + (F · Ò)G,

2.
Ò ◊ (F ◊ G) = F(Ò · G) ≠ (Ò · F)G + (G · Ò)F ≠ (F · Ò)G.

Here (G · Ò)F means

ˆ ˆ ˆ
(G · Ò)F = G1 F + G2 F + G3 F.
ˆx ˆy ˆz
Finally, there are a few more vector calculus identities that involve the Laplacian operator,
which in the language of differential forms requires the introduction of the Hodge star operator.
We will come back to this in Section 4.8.
CHAPTER 4. k-FORMS 95

4.4.5 Exercises
1. Let F = (xy, yz, xz) be a vector field on R3 . Find its curl Ò ◊ F and divergence Ò · F.
Solution. The curl of the vector field is given by:
Q R
i j k
c d
Ò ◊ F = det a ˆx
ˆ ˆ
ˆy
ˆ
ˆz b
xy yz xz
= ≠ yi ≠ zj ≠ xk.

In component notation, this reads Ò ◊ F = (≠y, ≠z, ≠x).

As for the div, we get:
ˆ ˆ ˆ
Ò·F= (xy) + (yz) + (xz)
ˆx ˆy ˆz
=y + z + x.

Note that we could have done these calculations using differential forms. To get the
curl, we associate to F a one-form Ê = xy dx + yz dy + xz dz and calculate its exterior
derivative:

dÊ =xdy · dx + ydz · dy + zdx · dz

= ≠ ydy · dz ≠ zdz · dx ≠ xdx · dy

The curl Ò ◊ F is then the vector field associated to this two-form, that is Ò ◊ F =
(≠y, ≠z, ≠x).
To calculate the divergence Ò · F, we associate to F a two-form ÷ = xy dy · dz +
yz dz · dx + xz dx · dy and calculate its exterior derivative:

d÷ =ydx · dy · dz + zdy · dz · dx + xdz · dx · dy

=(x + y + z)dx · dy · dz.

Therefore Ò · F = x + y + z.
2. For the following two vector fields, find their curl and divergence:
1
(a) F(x, y, z) =  2 (x, y, 0).
x + y2
1
(b) G(x, y, z) =  (≠y, x, 0).
x2+ y2

Solution. Let’s solve this one using differential forms. You can do it directly using the
formulae for curl and div as well.
(a) To find the curl, we associated a one-form Ê to F:
1
Ê= 2 (x dx + y dy).
x + y2
CHAPTER 4. k-FORMS 96

We calculate its exterior derivative:

xy xy
dÊ = ≠ 2 dy · dx ≠ 2 dx · dy
(x + y )2 3/2 (x + y 2 )3/2
=0.
Thus we conclude that Ò ◊ F = 0.
To find the divergence, we associate a two-form ÷ to F:
1
÷= 2 (x dy · dz + y dz · dx).
x + y2
We calulate its exterior derivative:
1 2
d÷ = (x2 + y 2 )≠1/2 ≠ x2 (x2 + y 2 )≠3/2 dx · dy · dz
1 2
+ (x2 + y 2 )≠1/2 ≠ y 2 (x2 + y 2 )≠3/2 dy · dz · dx
A B
2 x2 + y 2
=  2 ≠ dx · dy · dz
x + y 2 (x2 + y 2 )3/2
1
= 2 dx · dy · dz.
x + y2
We conclude that
1
Ò·F=  2 .
x + y2
(b) To find the curl, we associate a one-form Ê to G:
1
Ê= 2 (≠y dx + x dy).
x + y2
We calculate its exterior derivative:
1 2
dÊ = ≠(x2 + y 2 )≠1/2 + y 2 (x2 + y 2 )≠3/2 dy · dx
1 2
+ (x2 + y 2 )≠1/2 ≠ x2 (x2 + y 2 )≠3/2 dx · dy
A B
2 x2 + y 2
=  2 ≠ dx · dy
x + y 2 (x2 + y 2 )3/2
1
= 2 dx · dy.
x + y2
Thus we conclude that A B
1
Ò ◊ G = 0, 0,  2 .
x + y2
To find the divergence, we associate to G the two-form:
1
÷= 2 (≠y dy · dz + x dz · dx).
x + y2
We calculate its exterior derivative:
xy xy
d÷ = 2 dx · dy · dz ≠ 2 dy · dz · dx
(x + y 2 )3/2 (x + y 2 )3/2
=0
Therefore Ò · G = 0.
CHAPTER 4. k-FORMS 97

3. Find a vector field F = (0, f2 , f3 ) such that Ò ◊ F = (0, z, y).

Solution. Since F = (0, f2 , f3 ), and using the definition of the curl, we know that
3 4
ˆf3 ˆf2 ˆf3 ˆf2
Ò◊F= ≠ ,≠ , .
ˆy ˆz ˆx ˆx
Thus we need to solve the equations
ˆf3 ˆf2 ˆf3 ˆf2
≠ = 0, ≠ = z, = y.
ˆy ˆz ˆx ˆx
Integrating the last two equations, we get:

f2 = xy + g(y, z), f3 = ≠xz + h(y, z),

for some functions g(y, z), h(y, z). The first condition then imposes that

ˆ ˆ
h(y, z) = g(y, z).
ˆy ˆz

There are many possible choices, but the simplest would be g(y, z) = h(y, z) = 0. We
would then conlude that
F = (0, xy, ≠xz)
is a vector field such that Ò ◊ F = (0, z, y).
4. Is there a vector field F on R3 such that Ò ◊ F = (x, y + xz, z)? Justify your answer.
Solution. We know that Ò · (Ò ◊ F) = 0 for any vector field F. So if there is a
vector field F such that Ò ◊ F = (x, y + xz, z), then the divergence of the vector on the
right-hand-side (let’s call it G) must be zero. But

Ò · G = 1 + 1 + 1 = 3,

which is obviously non-zero. Therefore, we conclude that there does not exist a vector
field F such that Ò ◊ F = (x, y + xz, z).
5. Let F = (xy, y 2 , xy + z) and G = (xyz, yz, z 2 ) be smooth vector fields on R3 , and
– : [0, 2fi] æ R3 be the parametric curve given by –(t) = (sin(t), cos(t), t(t ≠ 2fi)). Show
that the line integral of the vector field
F ◊ (Ò ◊ G) + (Ò ◊ F) ◊ G + (G · Ò)F + (F · Ò)G
along – is zero.
Solution. Well, you certainly do not want to evaluate this line integral, it would be
painful!
First, we notice that the parametric curve – is closed, since
–(0) = (0, 1, 0) = –(2fi).
Next, we notice that we can use the vector calculus identity 1 from Lemma 4.4.12, which
states that
Ò(F · G) = F ◊ (Ò ◊ G) + (Ò ◊ F) ◊ G + (G · Ò)F + (F · Ò)G.
CHAPTER 4. k-FORMS 98

So the vector field that we want to evaluate the line integral of is Ò(F · G). As this is
the gradient of a function, this means that the vector field is conservative. Therefore, its
integral along any closed curve vanishes. We conclude that the integral along – is zero!
6. Suppose that you study a vector field F in a lab. You measure that

F(x, y, z) = (xz + yz + x2 y, –(yz + x2 z), —(xyz + y)),

for some constants –, — that you are not able to determine experimentally. However, from
theoretical considerations you know that F must be divergence-free (i.e., its divergence is
zero). Find the values of – and —.
Solution. We know that
ˆ ˆ ˆ
Ò·F= (xz + yz + x2 y) + – (yz + x2 z) + — (xyz + y)
ˆx ˆy ˆz
=z + 2xy + –z + —xy.

Since we know that Ò · F = 0, and that – and — are constants (i.e. do not depend on
x, y, z), we conclude that we must have

– = ≠1, — = ≠2.
7. Let F(x, y, z) = (f (x), g(y), h(z)) for some smooth functions f (x), g(y), h(z) on R (note
that those are functions of a single variable), and let q(x, y, z) be an arbitrary smooth
function on R3 . Show that
Ò · (F ◊ Òq) = 0.
Solution. Let us first solve it using vector calculus identities, and then provide an
alternative but equivalent solutions using differential forms. Identity 4 of Lemma 4.4.9
states that
Ò · (F ◊ G) = (Ò ◊ F) · G ≠ F · (Ò ◊ G).
Applying this to the case at hand, we get:
Ò · (F ◊ Òq) = (Ò ◊ F) · Òq ≠ F · (Ò ◊ Òq).
We then calculate the curl of F. We get:
3 4
ˆh(z) ˆg(y) ˆf (x) ˆh(z) ˆg(y) ˆf (x)
Ò◊F= ≠ , ≠ , ≠
ˆy ˆz ˆz ˆx ˆx ˆy
=0.
Moreover, from Identity 1 of Lemma 4.4.10, we know that
Ò ◊ Òq = 0.
Therefore
Ò · (F ◊ Òq) = 0.
Let us now solve the question using differential forms. Let Ê = f (x) dx + g(y) dy +
h(z) dz be the one-form associated to F. Then Ò · (F ◊ Òq) is the function associated
to d(Ê · dq). So we want to show that
d(Ê · dq) = 0.
CHAPTER 4. k-FORMS 99

Using the graded product rule, we have:

d(Ê · dq) =dÊ · dq ≠ Ê · d2 q

=dÊ · dq,

where we used the fact that d2 = 0. But

dÊ =df (x) · dx + dg(y) · dy + dh(z) · dz

=0,

since by evaluating the differentials we only get terms involving the vanishing basic
two-forms dx · dx = dy · dy = dz · dz = 0. We thus conclude that

d(Ê · dq) = 0.

4.5 Physical interpretation of grad, curl, div

In this section we pause for a moment and explore further the physical interpretation of the
vector calculus operations of gradient, curl, and divergence.

Objectives
You should be able to:

• Interpret the gradient of a vector field as giving the direction and magnitude of fastest
increase.

• Interpret the curl of a vector field in terms of a rotational motion in a fluid.

• Interpret the divergence of a vector field in terms of expansion and contraction of a fluid.

4.5.1 The gradient of a function

Let f : U æ R be a smooth function with U ™ R3 . We start by reviewing the interpretation
of the gradient 3 4
ˆf ˆf ˆf
Òf = , , ,
ˆx ˆy ˆz
which as we saw is the vector field associated to the exterior derivative df . At a point p œ U :
• The direction of the gradient is the direction in which the function f increases most
quickly at p;
• The magnitude of the gradient is the rate of fastest increase at p.
This may be most easily understood with an example.
Example 4.5.1 The direction of steepest slope. In this example we work with a function
on R2 instead of R3 , but the interpretation is the same.
Suppose that the function H(x, y) = ≠x2 ≠ y 2 + 1000 gives the height above sea level
at point (x, y) on a surface. The height is highest at the origin, where H(0, 0) = 1000, and
CHAPTER 4. k-FORMS 100

decreases
Ô as we move away from the origin, until it reaches sea level on the circle with radius
1000. So we can think of this altitude function as representing in a circular mountain centred
at the origin.
The level curves of H are the curves of constant elevation, i.e.

H(x, y) = ≠x2 ≠ y 2 + 1000 = C … x2 + y 2 = 1000 ≠ C

for constant elevations 0 Æ C Æ 1000. Those would correspond to the level curves of elevation
on a topographical map. In this case, they are all circles centred at the origin.
The gradient of H is 3 4
ˆH ˆH
ÒH = , = ≠2(x, y).
ˆx ˆy
For any point (x, y) on the surface, this is a vector that points towards the origin, which says
that the direction of steepest increasing slope is towards the origin, as expected from a circular
mountain centred at the origin. Moreover, we see that the magnitude of the gradient vector is
Ò
|ÒH| = 2 x2 + y 2 .

This says that the slope is steeper far away from the origin, and becomes less and less steep
as we get closer to the origin.
Here is below the contour map for this function (the graph of its level curves), and also a
3D plot (both produced using Mathematica).

Figure 4.5.2 A contour map of the function H(x, y) = ≠x2 ≠ y 2 + 1000, for ≠100 Æ x Æ 100
and ≠100 Æ y Æ 100.
CHAPTER 4. k-FORMS 101

Figure 4.5.3 A 3d plot of the function z = ≠x2 ≠ y 2 + 1000 (i.e. thinking of H(x, y) as
representing the height of the surface above sea level), for ≠100 Æ x Æ 100 and ≠100 Æ y Æ 100.
⇤

4.5.2 The curl of a vector field

Let F : U æ R3 be a smooth vector field with U ™ R3 , with component functions F =
(f1 , f2 , f3 ). Its curl is given by the vector field
3 4
ˆf3 ˆf2 ˆf1 ˆf3 ˆf2 ˆf1
Ò◊F= ≠ , ≠ , ≠ .
ˆy ˆz ˆz ˆx ˆx ˆy
What is it actually computing? How can we interpret the vector field Ò ◊ F?
It is easiest to understand the meaning of the curl of a vector field by thinking of F(x, y, z)
as the velocity field of a moving fluid in three dimensions. So let us call it v(x, y, z) instead.
The curl is related to rotational motion induced by the fluid.
Consider an infinitesimally small ball (sphere) located within the fluid, and centered at a
point p œ U . Assume that the ball has a rough surface. The fluid moving around the ball will
generally make it rotate. Then:
• The direction of the curl Ò ◊ v at p gives the axis of rotation (according to the right
hand rule);1
• Half the magnitude of the curl Ò ◊ v at p gives the angular speed of rotation.
In particular, if Ò ◊ v = 0 at a point p, the fluid does not cause the sphere centered at p to
rotate. Because of this interpretation, we say that the velocity field of a fluid is irrotational
if it is curl-free at all points on U , that is Ò ◊ v = 0 on U .

1
The “right hand rule” means that if you align the thumb of your right hand along the vector, the fingers of
your right hand will “curl” around the axis of rotation in the direction of rotation.
CHAPTER 4. k-FORMS 102

Example 4.5.4 The curl of the velocity field of a moving fluid. Suppose that a
moving fluid has velocity field given by

v(x, y, z) = (≠y, x, 0).

Sketching the vector field (see Exercise 2.1.3.2), one sees that at any point, the fluid is rotating
counterclockwise around the z-axis. Using the right-hand-rule, we thus expect the curl Ò ◊ v
to point in the positive z-direction, since it is the axis of rotation. We calculate:

Ò ◊ v = (0, 0, 2) .

Indeed, it points in the positive z-direction, as expected. Furthermore, half the magnitude of
the curl is
1
|Ò ◊ v| = 1,
2
which means that the angular speed of rotation of the ball would be 1 radian per unit of time.
⇤
Example 4.5.5 An irrotational velocity field. Conside a moving fluid with velocity field
given by
v(x, y, z) = (x, y, z).
Sketching the vector field, we see that this would be a fluid in expansion. As the fluid is
expanding, regardless of where the small sphere is located, it should not cause it to rotate (try
to visualize this yourself). So we expect the velocity field to be curl-free. From the definition
of the curl, we calculate:
Ò ◊ v = (0, 0, 0).
Thus it is curl-free, as expected, and this is an example of an irrotational velocity field. ⇤
Example 4.5.6 Another irrotational velocity field. Irrotational fluids do not have to
be necessarily spherically symmetric. Consider for instance a moving fluid with velocity field

v(x, y, z) = (0, 0, 1).

This would be a fluid that is moving uniformly in the positive z-direction. If you think about
it for a little bit, it should not induce any rotation either, regardless of where the small sphere
is located. So we expect the velocity field to be curl-free again. Indeed, from the definition we
Ò ◊ v = (0, 0, 0), which says that the velocity field is irrotational. ⇤

4.5.3 The divergence of a vector field

Let F : U æ R3 be a smooth vector field with U ™ R3 , with component functions F =
(f1 , f2 , f3 ). Its divergence is the function:
ˆf1 ˆf2 ˆf3
Ò·F= + + .
ˆx ˆy ˆz
What is its intepretation?
As for the curl, it is easiest to interpret the divergence by thinking of the vector field as
being the velocity field v(x, y, z) of a moving fluid or gas. The divergence is then related to
CHAPTER 4. k-FORMS 103

expansion and compression of the fluid.

More precisely, consider an infinitesimally small sphere around a point p œ U :
• The divergence Ò · v at p measures the rate at which the fluid is exiting the small sphere
at p (per unit of time and unit of volume).
The key here is that if the divergence at a point is positive, it means that more fluid is
exiting the small sphere centered at this point then entering the sphere, and vice-versa if the
divergence is negative. If it is zero, this means that the fluid may still be moving, but there is
exactly the same amount of fluid entering and existing the sphere.
For instance, if a gas is heated, it will expand. This means that for any small sphere
in the gas, there will be more gas exiting then entering. So the divergence will be positive
everywhere. Conversely, if a gas is cooled, it will contract, and its divergence will be negative
everywhere.
Motivated by this interpretation, we say that the velocity field of a fluid is incompressible
if it is divergence-free everywhere, i.e. Ò · v = 0 everywhere on U . We also call such vector
fields solenoidal.
Example 4.5.7 The divergence of the velocity field of an expanding fluid. Consider
a fluid/gas with velocity field
v(x, y, z) = (x, y, z).
The velocity field is pointing outwards in all directions. This corresponds to an expanding fluid.
If we first think of a small sphere centered at the origin, then the fluid is moving outwards in
all directions, exiting the sphere, and thus we expect the divergence to be positive. In fact,
even if the small sphere is located elsewhere, we still expect the divergence to be positive, as
there will be more fluid exiting the sphere than entering the sphere. From the definition, we
calculate
Ò · v = 1 + 1 + 1 = 3,
which is indeed positive everywhere, as expected. ⇤
Example 4.5.8 An imcompressible velocity field. Consider a fluid with velocity field

v(x, y, z) = (≠y, x, 0).

As we have seen (see Exercise 2.1.3.2), the fluid is rotating counterclockwise around the z-axis.
It is not so obvious to see whether more fluid is existing or entering small spheres in the fluid.
It is easiest to consider first a sphere centered around the origin. Because of the rotational
motion, we see that actually no fluid is entering or leaving the sphere at all. So we expect the
divergence to be zero, at least at the origin.
It is not so obvious to see why the same should be true for all spheres not centered at the
origin, but you can try to visualize it. In the end, through direct calculation, we get that
Ò · v = 0, and hence the velocity field is divergence-free (that is, incompressible). ⇤
Example 4.5.9 Another incompressible velocity field. Consider the fluid with velocity
field
v(x, y, z) = (0, 0, 1),
which has uniform velocity in the positive z-direction. Here, if you pick a small sphere centered
anywhere, there is certainly fluid entering and exiting the sphere because of the linear motion
CHAPTER 4. k-FORMS 104

of the fluid, but the exact same amount of fluid will enter and exit the sphere. Therefore we
expect the divergence to be zero everywhere, and indeed, Ò · v = 0. Another example of an
incompressible velocity field.
Comparing with Example 4.5.6, we see that this velocity field is both irrotational and
incompressible, since it is both curl-free and divergence-free. ⇤

4.5.4 Exercises
1. Show that any vector field of the form

F(x, y, z) = (f (x), g(y), h(z))

is irrotational.
Solution. To show that F is irrotational, we need to show that

Ò ◊ F = 0.

We can use directly the formula for the curl, or we can use the language of differential
forms. In the latter, we associate a one-form Ê to the vector field F:

Ê = f (x) dx + g(y) dy + h(z) dz.

We want to show that dÊ = 0. But this is clear true, as taking the exterior derivative
gives
df dg dh
dÊ = dx · dx + dy · dy + dz · dz = 0,
dx dy dz
since f (x), g(y), h(z) are only functions of x, y, z respectively. Therefore F is irrotational.
2. Show that any vector field of the form

F(x, y, z) = (f (y, z), g(x, z), h(x, y))

is incompressible.
Solution. To show that F is incompressible, we need to show that

Ò · F = 0.

This is the statement that

ˆf ˆg ˆh
+ + = 0.
ˆx ˆy ˆz
But each of those partial derivatives vanishes, since f (y, z) does not depend on x, g(x, z)
does not depend on y, and h(x, y) does not depend on z. Therefore F is incompressible.
3. Consider a vector field F(x, y, z) = (f1 (x, y), f2 (x, y), 0) on R3 ; it is independendent of z,
and its z-component is zero. A sketch of the vector field in the xy-plane is shown in the
figure below; as it is independent of z, it looks the same in all other horizontal planes.

(a) Is Ò · F positive, negative, or zero at the origin? What about at the point (5, 0, 0)?
And (≠5, 0, 0)?
CHAPTER 4. k-FORMS 105

(b) Is Ò ◊ F = 0 at the origin? If not, in what direction does it point?

Figure 4.5.10 A sketch of the vector field F in the xy-plane.

Solution. (a) First we see that all arrows point in the positive x-direction. The lengths
of the arrows appear to be symmetric about the y-axis. The arrows get longer and longer
as you move away from the y-axis.
If you picture a small sphere around the origin, then there is fluid entering and leaving
the sphere, and the arrows of the vectors entering and exiting have the same length. So
you expect the divergence to be zero at the origin. In fact, the same is true for all points
on the y-axis, i.e. with x = 0.
At the point (5, 0, 0), there are arrows entering and leaving the sphere as well. But
the arrows that leave the sphere are longer than those that enter, so there should be
more fluid exiting than entering. So we expect the divergence to be positive. This should
be the case for all points (x, y) with x > 0.
As for (≠5, 0, 0), the opposite happens; the arrows entering the sphere are longer
than those leaving. So there should be more fluid entering than leaving, and we expect
the divergence to be negative. The same is true for all points (x, y) with x < 0.
(b) At the origin, the arrows are all pointing horizontally. As such, it will not induce
any rotation on a sphere centered at the origin, and we expect the curl to be zero at the
origin. In fact, this will be the case everywhere, so the curl should be zero everywhere.
For your interest, this is the vector field F(x, y, z) = (x2 , 0, 0). Its divergence is
Ò · F = 2x. We see that it is zero when x = 0, positive for all x > 0, and negative for all
x < 0, as expected. The curl is Ò ◊ F = (0, 0, 0), which vanishes as expected (the vector
field is irrotational).
4. Consider a vector field F(x, y, z) = (f1 (x, y), f2 (x, y), 0) on R3 ; it is independendent of z,
and its z-component is zero. A sketch of the vector field in the xy-plane is shown in the
figure below; as it is independent of z, it looks the same in all other horizontal planes.
CHAPTER 4. k-FORMS 106

(a) Is Ò · F positive, negative, or zero at the origin? What about at (0, 5, 0)?

(b) Is Ò ◊ F = 0 at the origin? If not, in what direction does it point?

Figure 4.5.11 A sketch of the vector field F in the xy-plane.

Solution. (a) We look at a small sphere around the origin, and we want to know
whether there is more or less fluid entering the sphere versus exiting the sphere. Around
the origin, we see that the arrows appear to be the same length on both sides of the y-axis,
and on both sides of the x-axis. The arrows are all pointing the positive x-direction.
Furthermore, the arrows above the x-axis point downwards, while the arrows below the
x-axis point upwards, in a symmetric way. If you think about it, this means that there is
more fluid entering the sphere than exiting (because the arrows above and below the
x-axis are pointing towards the x-axis). We thus expect the divergence to be negative at
the origin.
At the point (0, 5, 0), the fluid is moving downwards, and the arrows appear to be
bigger for y > 5 than for y < 5. So it looks like there is still more fluid entering the
sphere than exiting the sphere. So we expect the divergence to be negative at this point
as well.
(b) At the origin, all arrows point in the positive x-direction. Furthermore, the arrows
above the x-axis point downwards, while the arrows below the x-axis point upwards, in a
symmetric way. Therefore, if you imagine a small sphere around the origin, the motion
of the fluid would not induce any rotation. So we expect its curl to vanish.
For your interest, this is the vector field F(x, y, z) = (10, ≠y, 0). Its divergence is
Ò · F = ≠1, which is negative everywhere. As for the curl, we get Ò ◊ F = (0, 0, 0), so it
is zero everywhere (the vector field is irrotational).
CHAPTER 4. k-FORMS 107

5. Consider a vector field F(x, y, z) = (f1 (x, y), f2 (x, y), 0) on R3 ; it is independendent of z,
and its z-component is zero. A sketch of the vector field in the xy-plane is shown in the
figure below; as it is independent of z, it looks the same in all other horizontal planes.

(a) Is Ò · F positive, negative, or zero at the origin? What about at (0, 5, 0)?

(b) Is Ò ◊ F = 0 at the origin? If not, in what direction does it point?

Figure 4.5.12 A sketch of the vector field F in the xy-plane.

Solution. (a) Arrows that are diametrically opposite around the origin appear to have
the same length but point in opposite directions. As a result, the same amount of fluid
should be entering and exiting a small sphere around the origin, and hence we expect
the divergence to be zero at the origin.
As for the point (0, 5, 0), we see that at this point (just like for any point with x = 0)
all arrows are pointing horizontally. Furthermore, the length of the arrows on both sides
of the y-axis appear to be the same, so we expect the same amount of fluid entering and
exiting the sphere. Again, the divergence should be zero at this point.
(b) It is clear from the figure that the moving fluid would make a sphere centered at
the origin rotate clockwise. We thus expect the curl to be non-zero at the origin. Using
the right hand rule, we expect it to point in the negative z-direction. In fact, looking at
the figure, we expect this to be true at all points. So we expect the divergence to be zero
everywhere.
For your interest, this is the vector field F(x, y, z) = (y 5 , ≠x5 , 0). Its divergence is
Ò · F = 0,, which is zero everywhere, as expected (the vector field is incompressible). Its
curl is Ò ◊ F = (0, 0, ≠5x4 ≠ 5y 4 ). We see that it points in the negative z-direction for
all values of x, y, as expected.
CHAPTER 4. k-FORMS 108

6. Consider a vector field F(x, y, z) = (f1 (x, y), f2 (x, y), 0) on R3 ; it is independendent of z,
and its z-component is zero. A sketch of the vector field in the xy-plane is shown in the
figure below; as it is independent of z, it looks the same in all other horizontal planes.

(a) Is Ò · F positive, negative, or zero at the origin? What about at (0, 5, 0)?

(b) Is Ò ◊ F = 0 at the origin? If not, in what direction does it point?

Figure 4.5.13 A sketch of the vector field F in the xy-plane.

Solution. (a) We see that all arrows point away from the origin, in a way that is
spherically symmetric (the lengths of all arrows on a circle of a fixed radius about the
origin appear to be the same). As such, the fluid is all exiting the sphere, so we expect
the divergence to be positive at the origin.
At the point (0, 5, 0), there are arrows coming in and arrows coming out, but the
arrows coming out are longer than the arrows coming out, so we expect again the
divergence to be positive since more fluid is exiting the sphere than entering. In fact,
this will be the case at all points in the figure. So we expect the divergence to be always
positive.
(b) At the origin, the fluid is all pushing outwards in spherically symmetric way, so it
will induce no rotation on a sphere centered at the origin. We thus expect the curl to be
zero at the origin. While it may not be as obvious, you can probably convince yourself
that this should be true at all points in the figure, so the curl should be zero everywhere.
For your interest, this is the vector field F(x, y, z) = (x, y, 0). Its divergence is
Ò · F = 2, which is positive everywhere, as expected. Its curl is Ò ◊ F = (0, 0, 0), which
is zero everywhere as expected (the fuild is irrotational).
CHAPTER 4. k-FORMS 109

4.6 Exact and closed k-forms

We define exact and closed k-forms using the exterior derivative. We show that exact forms
are always closed, and determine when closed forms are exact. We rephrase these statements
in the language of vector calculus.

Objectives
You should be able to:

• Define closed and exact k-forms using the exterior derivative.

• Determine when a k-form is closed or exact, focusing on one-, two-, and three-forms in
R3 .

• Show that the general definition of closeness for one-forms reduces to our previous
definition in terms of partial derivatives.

• Show that exact k-forms are always closed.

• Recall the statement of Poincare’s lemma.

• Rephrase and use these statements in the language of vector calculus.

4.6.1 Exact and closed k-forms

We introduced the notion of exact one-forms in Definition 2.2.5, using the concept of differential.
We also introduced closed one-forms in R2 in Definition 2.2.9 and in R3 in Definition 2.2.14,
but our definition was rather ad hoc. These concepts are much more natural now that we
have introduced the exterior derivative.
Definition 4.6.1 Exact and closed k-forms. Let Ê be a k-form on U ™ Rn . We say that
Ê is closed if dÊ = 0. We say that it is exact if there exists a (k ≠ 1)-form ÷ on U such that
Ê = d÷. ⌃
Example 4.6.2 Exact and closed one-forms in R3 . Before we move on, let us show
that this reproduces the definitions that we used in Definition 2.2.5 and Definition 2.2.14 for
one-forms in R3 .
First, if Ê is a one-form, according to Definition 4.6.1 is it exact if there exists a zero-form
f such that Ê = df . As we know that the exterior derivative of a zero-form is the same thing
as the differential of a function introduced in Definition 2.2.1, this is precisely the definition
of an exact one-form that we gave in Definition 2.2.5.
As for closeness, according to Definition 4.6.1 a one-form Ê is closed if dÊ = 0. If Ê is a
one-form on U µ R3 , we can write Ê = f dx + g dy + h dz. Then
3 4 3 4 3 4
ˆh ˆg ˆf ˆh ˆg ˆf
dÊ = ≠ dy · dz + ≠ dz · dx + ≠ dx · dy.
ˆy ˆz ˆz ˆx ˆx ˆy
Thus dÊ = 0 if and only if
ˆh ˆg ˆf ˆh ˆg ˆf
= , = , = ,
ˆy ˆz ˆz ˆx ˆx ˆy
CHAPTER 4. k-FORMS 110

which is precisely the condition stated in Definition 2.2.14. In fact, this explains where this
strange condition comes from! ⇤
As for one-forms, it is easy to show that exact forms are always closed: it follows directly
from the fact that d2 = 0, which is a key property of the exterior derivative proved in
Lemma 4.3.9.
Lemma 4.6.3 Exact k-forms are closed. If a k-form Ê on U ™ Rn is exact, then it is
closed.
Proof. This is a direct consequence of the fact that d2 = 0. If Ê is exact, then there exists a
(k ≠ 1)-form ÷ such that Ê = d÷. But then

dÊ = d(d÷) = 0

by Lemma 4.3.9. Therefore Ê is closed. ⌅

As for one-forms, the converse statement is much more subtle. When are closed k-forms
exact? The answer depends on the domain of definition of the k-form. However, there is a
simple case for which closed k-forms are always exact, as in Theorem 3.6.1. This is called
“Poincare’s lemma” for k-forms. We will state the result here without proof.
Theorem 4.6.4 Poincare’s lemma for k-forms, version 1. Let Ê be a k-form defined
on all of Rn . Then Ê is exact if and only if it is closed.
In fact, the theorem can be generalized slightly, as in Theorem 3.6.4. What matters in the
proof is not so much that Ê is defined on all of Rn , but rather that it is defined on a domain
U that is contractible, which intuitively means that it can be continuously shrunk to a point
within U . A more precise statement of Poincare’s lemma could then be formulated as follows.
Theorem 4.6.5 Poincare’s lemma for k-forms, version II. Let Ê be a k-form defined
on an open ball U ™ Rn . Then Ê is exact if and only if it is closed.
It is important to note however that if U is not an open ball or the whole of Rn , then
closed forms may not necessarily be exact.
Remark 4.6.6 We should note that contrary to Poincare’s lemma for one-forms in The-
orem 3.6.4, for k Ø 2 it is not true that closed k-forms on simply connected open subsets
U ™ Rn are necessarily exact. This is only true for one-forms; simple-connectedness is not
sufficient for k Ø 2. What we need is a higher-dimensional analog of simple-connectedness;
this is why it was replaced by the statement that U is an open ball in Theorem 4.6.5. (Slightly
more generally, one could say that U must be “contractible”. Every contractible space is
simply connected, but not the other way around. This is the kind of thing that is studied in
cohomology and homology, see Remark 4.6.7.)
Just to highlight the subtlety here, consider the two-form
1
Ê= (x dy · dz + y dz · dx + z dx · dy),
(x2 + y 2 + z 2 )3/2

which is defined on U = R3 \ {(0, 0, 0)}, which is simply connected. One can check that Ê is
closed, but not exact. Does that contradict Poincare’s lemma? Fortunately it doesn’t, as U
is not an open ball in R3 (as it does not contain the origin). But it shows that there exists
k-forms with k Ø 2 defined on simply connected open sets that are closed but not exact.
CHAPTER 4. k-FORMS 111

Remark 4.6.7 The world of cohomology (this is just for fun and beyond the
scope of this class!). In fact, the relation between closed and exact forms is quite deep.
As we have seen, it is closely connected to the existence of “holes” in a space, which is the
subject of topology. In fact, studying when closed forms are not exact gives rise to the topic of
cohomology, which is an important branch of geometry and topology. Believe me, cohomology
is all over the place. You wouldn’t believe it, but it is even used to describe gauge theories in
physics!
While describing cohomology is way beyond the scope of this course, let me explain in a few
words what it is about, just for fun, in the context of differential forms. Suppose that Ê and ÷
are closed k-forms on U . If they differ from each other by an exact form, that is Ê ≠ ÷ = dfl
for some (k ≠ 1)-form fl, then we say that Ê and ÷ are “equivalent” (or “cohomologous”).
In this way, we construct equivalence classes of closed k-forms that differ by an exact form.
Those equivalence classes (which are called “cohomology classes”) form a vector space, which
is called the “de Rham cohomology space” H k (U ).1
If U is the whole of Rn , or an open ball in Rn , then all closed k-forms are exact, and hence
they are all equivalent to the zero k-form. The de Rham cohomology spaces H k (U ) (with
k Ø 1) are then all trivial (the zero vector space). So the de Rham cohomology spaces H k (U )
with k Ø 1 control, in a sense, how topologically non-trivial U is.
For instance, if we consider U = R2 \ {(0, 0)}, one can show that

H 1 (U ) = R, H 2 (U ) = 0.

The fact that H 2 (U ) = 0 says that all (closed) two-forms on U are exact. However, the
interesting fact here is that H 1 (U ) = R, which says that not all closed one-forms are exact;
indeed, we saw one example of such a one-form in Example 2.2.13, which was closed but not
exact. Since H 1 (U ) is one-dimensional, this means that all closed one-forms that are not
exact differ from the one we saw in that example by an exact form (they are in the same
cohomology class), up to overall rescaling.

4.6.2 Translation into vector calculus for R3

Focusing on R3 , we can rephrase the notions of exactness and closeness in terms of vector
calculus objects, using Table 4.1.11. In R3 , we focus on zero-, one-, two-, and three-forms. In
fact, the interesting statements are for one-forms and two-forms.
We already translated the exactness and closeness statements for one-forms in the language
of vector calculus in Section 2.2, but let summarize the statements here.

1. A one-form Ê on R3 is exact if and only if its associated vector field F is conservative,

which is the statement that
F = Òf
for some potential function f .
1
Mathematically, what we are doing here is constructing a quotient vector space. If we write Z k (U ) for the
vector space of closed k-forms on U , and B k (U ) for the vector space of exact k-forms on U , then the de Rham
cohomology space is constructed as the quotient space H k (U ) = Z k (U )/B k (U ).
CHAPTER 4. k-FORMS 112

2. A one-form Ê on R3 is closed if and only if its associated vector field F is curl-free,

that is
Ò◊F=0
(see Remark 4.4.5).
3. The fact that exact one-forms are closed is the statement that if F is conservative, then
it is curl-free, namely Ò ◊ F = 0. We call this the “screening test for conservative vector
fields”. However, while conservative vector fields are curl-free, curl-free vector fields are
not necessarily conservative.
4. Poincare’s lemma translates into the statement that if F is defined and has continuous
first order partial derivatives on all of R3 (or an open simply connected subset therein),
then F is conservative if and only if it is curl-free.
We can do a similar translation for two-forms. The result is the following four statements.
1. A two-form Ê on R3 is exact if and only if its associated vector field F has a vector
potential, which is a vector field G such that

F = Ò ◊ G.

2. A two-form Ê on R3 is closed if and only if its associated vector field F is divergence-

free, that is
Ò · F = 0.

3. The fact that exact two-forms are closed is the statement that if there exists a vector
potential for F, then F is divergence-free, namely Ò · F = 0. We call this the “screening
test for vector potentials”. However, while vector fields with a vector potential are
divergence-free, divergence-free vector fields do not necessarily have a vector potential.
4. Poincare’s lemma translates into the statement that if F is defined and has continuous
first order partial derivatives on all of R3 (or an open ball therein), then F has a vector
potential if and only if it is divergence-free.
Remark 4.6.8 Suppose that a vector field F on R3 has a vector potential G. This means
that there exists a vector field G such that

F = Ò ◊ G.

Finding G is not always obvious, as one would need in principle to integrate fairly complicated
partial differential equations. Moreover, G is far from unique.
It turns out that there is a nice result that drastically simplifies calculations. One can
show that, if F has a vector potential G, then we can always choose G to have vanishing
z-component function. In other words, if F has a vector potential, then there always exists a
G = (g1 , g2 , 0) such that
F = Ò ◊ G.
This is very helpful when trying to find a vector potential.
In the language of differential forms, this corresponds to the statement that if Ê is an
exact two-form on R3 , then there exists a one-form ÷ = f dx + g dy (with vanishing third
CHAPTER 4. k-FORMS 113

component function) such that Ê = d÷. We prove this statement in Exercise 4.6.3.6.

4.6.3 Exercises
CHAPTER 4. k-FORMS 114

1. Let Ê be the following two-form on R3 :

Ê = (ex + y) dy · dz + (yex + ez ) dz · dx + (ex+y ≠ 2zex ) dx · dy.
Determine whether Ê is exact. If it is, find a one-form ÷ such that d÷ = Ê.
Solution. First, we notice that Ê is defined on all of R3 , so Poincare’s lemma applies.
So we know that it is exact if and only if it is closed. Let us first determine whether it is
closed.
We calculate:
ˆ ˆ
dÊ = (ex + y) dx · dy · dz + (yex + ez ) dy · dz · dx
ˆx ˆy
ˆ x+y
+ (e ≠ 2zex ) dz · dx · dy
ˆz
=(ex + ex ≠ 2ex ) dx · dy · dz
=0.
Thus Ê is closed, and hence, by Poincare’s lemma, it is also exact.
We now want to find a one-form ÷ such that d÷ = Ê. We assume that ÷ takes the
form
÷ = f dx + g dy.
Its exterior derivative is
3 4
ˆg ˆf ˆg ˆf
d÷ = ≠ dy · dz + dz · dx + ≠ dx · dy.
ˆz ˆz ˆx ˆy
So we need to solve the three equations:
ˆg ˆf ˆg ˆf
≠ = ex + y, = yex + ez , ≠ = ex+y ≠ 2zex .
ˆz ˆz ˆx ˆy
Integrating the first one, we get:
g(x, y, z) = ≠zex ≠ yz + –(x, y)
for some function –(x, y). Integrating the second one, we get
f (x, y, z) = yzex + ez + —(x, y)
for some function —(x, y). Substituting these expressions in the third equation, we get:
ˆg ˆf ˆ–(x, y) ˆ—(x, y)
≠ = ≠zex + ≠ zex ≠ = ex+y ≠ 2zex ,
ˆx ˆy ˆx ˆy
so we must have
ˆ–(x, y) ˆ—(x, y)
≠ = ex+y .
ˆx ˆy
We can choose any function – and — that work. So why not choose — = 0, and
–(x, y) = ex+y . Then we get that the one-form
÷ = (yzex + ez ) dx + (≠zex ≠ yz + ex+y ) dy
is such that d÷ = Ê.
2. Let Ê be the following two-form on R3 :

Ê = xy dy · dz + yz dz · dx + zx dx · dy.
CHAPTER 4. k-FORMS 115

Determine whether Ê is exact. If it is, find a one-form ÷ such that d÷ = Ê.

Solution. Ê is defined on all of R3 , so Poincare’s lemma applies. Thus Ê is exact if
and only if it is closed. Let us determine whether it is closed.
We calculate:
ˆ ˆ ˆ
dÊ = (xy) dx · dy · dz + (yz) dy · dz · dx + (zx) dz · dx · dy
ˆx ˆy ˆz
=(y + z + x) dx · dy · dz.

As this is non-zero, Ê is not closed. We then conclude that Ê is not exact.

3. Let F be the following vector field on R3 :

F = (yz, xz, xy).

Determine whether F has a vector potential G. If it does, find such a vector potential.
Solution. Since the component functions of F are smooth on all of R3 , Poincare’s
lemma applies. So F will have a vector potential if and only if it is divergence-free. We
calculate its divergence:
ˆ ˆ ˆ
Ò·F= (yz) + (xz) + (xy)
ˆx ˆy ˆz
=0,

and thus we conclude that there exists a vector potential G such that Ò ◊ G = F.
Now we need to find G. We assume that G = (g1 , g2 , 0). Then the condition that
Ò ◊ G = F is
3 4
ˆg2 ˆg1 ˆg2 ˆg1
Ò◊G= ≠ , , ≠ = (yz, xz, xy).
ˆz ˆz ˆx ˆy
So we get three equations to solve. We integrate the first one (equality of the x-component
functions) to get:
yz 2
g2 (x, y, z) = ≠ + –(x, y).
2
We integrate the second one (equality of the y-component functions) to get:

xz 2
g1 (x, y, z) = + —(x, y).
2
Substituting in the third one (equality of the z-component functions), we get:
ˆg2 ˆg1 ˆ–(x, y) ˆ—(x, y)
≠ = ≠ = xy.
ˆx ˆy ˆx ˆy
We need to find any two functions – and — that satisfy this condition. We choose — = 0,
2
– = x2y . As a result, we have found a vector potential
A B
xz 2 y 2
G(x, y, z) = , (x ≠ z 2 ), 0 .
2 2
CHAPTER 4. k-FORMS 116

4. Consider the two-form

1
Ê= (x dy · dz + y dz · dx + z dx · dy),
(x2 + y2+ z 2 )3/2

which is defined on U = R3 \ {(0, 0, 0)}. Show that Ê is closed.

Solution. We want to show that dÊ = 0. First, we calculate

ˆ x (x2 + y 2 + z 2 )3/2 ≠ 3x2 (x2 + y 2 + z 2 )1/2

=
ˆx (x2 + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )3
(x + y + z ) ≠ 3x2
2 2 2
= .
(x2 + y 2 + z 2 )5/2

Similarly,
ˆ y (x2 + y 2 + z 2 ) ≠ 3y 2
=
ˆy (x2 + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )5/2
and
ˆ z (x2 + y 2 + z 2 ) ≠ 3z 2
= .
ˆz (x2 + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )5/2
Now, we get:
ˆ x ˆ y
dÊ = dx · dy · dz + dy · dz · dx
ˆx (x + y + z )
2 2 2 3/2 ˆy (x + y + z 2 )3/2
2 2

ˆ z
+ dz · dx · dy
ˆz (x2 + y 2 + z 2 )3/2
3(x2 + y 2 + z 2 ) ≠ 3x2 ≠ 3y 2 ≠ 3z 2
= dx · dy · dz
(x2 + y 2 + z 2 )5/2
=0.

We conclude that Ê is closed.

What is interesting with this example is that Ê is closed, but one can show that it is
not exact. This doesn’t contradict Poincare’s lemma, as Ê is not defined on all of R3 (or
an open ball thereof). But it is interesting since Ê is defined on a simply connected subset
of R3 , so it shows that there exists k-forms with k Ø 2 defined on simply connected
subsets that are closed but not exact.
5. Consider the vector field
3 4
y x
F(x, y, z) = ≠ , 2 ,z .
x + y x + y2
2 2

(a) Find the domain of definition of F. Is it path connected? Simply connected?

(b) Determine the divergence of F.

(c) Determine the curl of F.

(d) Does F have a vector potential? Justify your answer. If it does, find such a vector
potential.
CHAPTER 4. k-FORMS 117

(e) Is F conservative? Justify your answer. If it is, find a potential function.

Solution.

(a) F is defined (and in fact, is smooth) wherever the denominator is non-zero. This is
wherever (x, y) ”= (0, 0). So the domain of definition of F is

U = {(x, y, z) œ R3 | (x, y) ”= (0, 0)}.

This is R3 minus the z-axis. It is path connected, since any two points in U can
be connected by a path. It is however not simply connected, since a closed curve
around the z-axis cannot be continuously contracted to a point within U (it would
hit the z-axis, which is not in U ).

(b) We calculate the divergence of F:

3 4 3 4
ˆ y ˆ x ˆ
Ò·F= ≠ 2 + + (z)
ˆx x +y 2 ˆy x + y 2
2 ˆz
2xy 2xy
= 2 ≠ 2 +1
(x + y )
2 2 (x + y 2 )2
=1.

(c) We calculate the curl of F:

3 3 4 3 4
ˆ ˆ x ˆ y ˆ
Ò◊F= (z) ≠ , ≠ 2 ≠ (z),
ˆy ˆz x2 + y 2 ˆz x + y2 ˆx
3 4 3 44
ˆ x ˆ y
≠ ≠ 2
ˆx x2 + y 2 ˆy x + y2
A B
x2 + y 2 ≠ 2x2 x2 + y 2 ≠ 2y 2
= 0, 0, +
(x2 + y 2 )2 (x2 + y 2 )2
=(0, 0, 0).

So F is curl-free.

(d) We found in (b) that the divergence of F is non-zero. This means that F cannot have
a vector potential, since vector fields that have a vector potential are divergence-free.

(e) We found in (c) that F is curl-free, so it passes the screening test for conservative
vector fields. However, since its domain of definition U is not simply connected,
Poincare’s lemma does not apply. We thus cannot conclude whether F is conserva-
tive from the statement that it is curl-free.
In fact, we can show that it is not conservative by showing that its integral along a
closed loop is non-zero, as in Exercise 3.4.3.2. Let Ê be the one-form associated to
F:
y x
Ê=≠ 2 dx + 2 dy + z dz.
x + y2 x + y2
CHAPTER 4. k-FORMS 118

Consider the parametric curve – : [0, 2fi] æ R3 with –(t) = (cos(t), sin(t), 0), which
is the unit circle (counterclockwise) around the origin in the xy-plane. The pullback
of Ê is:
3 4
sin(t) cos(t)
–ú Ê = ≠ 2 (≠ sin(t)) + cos(t) dt
cos (t) + sin (t)
2 cos (t) + sin2 (t)
2

=dt.

The line integral of Ê along – is thus:

⁄ ⁄ 2fi ⁄ 2fi
Ê= – Ê=
ú
dt = 2fi.
– 0 0

Since this is non-zero, by Corollary 3.4.3 (or, in other words, the Fundamental
Theorem for line integrals), Ê cannot be exact, since the line integrals of exact
one-forms along closed curves vanish. Equivalently, the vector field F is not
conservative.
6. Let Ê be an exact two-form on R3 . Show that there exists a one-form ÷ of the form
÷ = f dx + g dy
(i.e. with a vanishing z-component function) such that d÷ = Ê.
In other words, in the language of vector calculus, if a vector field F has a vector
potential G, then G can always be chosen to take the form G = (g1 , g2 , 0).
Solution. Since Ê is exact, we know that there exists a one-form — such that d— = Ê.
Suppose that we find such a —:
— = b1 dx + b2 dy + b3 dz,
for some component functions b1 , b2 , b3 . — is certainly not unique; there are many one-
forms such that their exterior derivative equals Ê. In fact, since d2 = 0, we can add to —
any exact one-form dF for a function F , and we get another one-form whose exterior
derivative is Ê. That is, if we define
—˜ = — ≠ dF
for any function F , then d—˜ = d— ≠ d2 F = Ê.
In particular, since
ˆF ˆF ˆF
dF = dx + dy + dz,
ˆx ˆy ˆz
if we can choose F such that
ˆF
= b3 ,
ˆz
then we see that 3 4 3 4
˜ ˆF ˆF
— = b1 ≠ dx + b2 ≠ dy,
ˆx ˆy
and hence it is of the form that we are looking for (no z-component function). But this
is easy to do; simply pick
⁄
F (x, y, z) = b3 (x, y, z) dz,
CHAPTER 4. k-FORMS 119

i.e. any antiderivative in the z-variable will do. So we conclude that we can always find
a one-form —˜ with no z-component and such that d—˜ = Ê.

4.7 The pullback of a k-form

In Section 2.4 we defined the pullback of a one-form. We now generalize this concept to
k-forms. We first do it using an axiomatic approach as for one-forms, and relate the result
to the concept of Jacobian. We also introduce a more direct definition of pullback using
the algebraic approach to basic k-forms introduced in Subsection 4.1.1, and show that it is
consistent with our axiomatic approach..

Objectives
You should be able to:

• Determine the pullback of a k-form.

• Show that the pullback of a 2-form in R2 and a 3-form in R3 is given by the Jacobian
determinant.

4.7.1 The pullback of a k-form: an axiomatic approach

In Section 2.4 we defined the pullback of a one-form. We used an “axiomatic” approach: we
first defined the properties that we wanted the pullback to satisfy, and showed that there is a
unique construction that satisfies these properties. The properties that we specified were, in
words:
1. The pullback of a sum of one-forms is the sum of the pullbacks.

2. The pullback of a product of a zero-form and a one-form is the product of the pullbacks.

3. The pullback of the exterior derivative of a zero-form is the exterior derivative of the
pullback.
In this section we define the pullback of a k-form using a similar approach. All we need to do
is generalize the first and second properties to k-forms.
More precisely, let „ : V æ U be a smooth function, with V ™ Rm and U ™ Rn open
subsets. We want the pullback „ú to satisfy the following properties:
1. If Ê and ÷ are k-forms on U , then

„ú (Ê + ÷) = „ú Ê + „ú ÷.

2. If Ê is a k-form and ÷ an ¸-form on U , then

„ú (Ê · ÷) = („ú Ê) · („ú ÷).

3. If f is a zero-form (a function) on U , then

„ú (df ) = d(„ú f ).
CHAPTER 4. k-FORMS 120

The third property is unchanged, and the first two are naturally generalized to k-forms.
Imposing these properties gives us a unique definition for the pullback of a k-form.
Lemma 4.7.1 The pullback of a k-form. Let „ : V æ U be a smooth function, with V ™
Rm and U ™ Rn open subsets. We write t = (t1 , . . . , tm ) œ V , and „(t) = (x1 (t), . . . , xn (t)).
Let Ê be a k-form on U , with
ÿ
Ê= fi1 ···ik dxi1 · · · · · dxik ,
1Æi1 <...<ik Æn

for some functions fi1 ···ik : U æ R. Then the pullback „ú Ê is a k-form on V given by:
ÿ 3 4
ˆxi1 ˆxi1
„ Ê=
ú
fi1 ···ik („(t)) dt1 + . . . + dtm
1Æi1 <...<ik Æn
ˆt1 ˆtm
3 4
ˆxik ˆxik
· ··· · dt1 + . . . + dtm .
ˆt1 ˆtm
Proof. Start with a basic k-form
dxi1 · · · · · dxik .
Since we impose (Property 2) that „ú (Ê · ÷) = („ú Ê) · „ú ÷), we know that

„ú (dxi1 · · · · · dxik ) = „ú (dxi1 ) · · · · · „ú (dxik )..

But we already calculated how basic one-forms transform in Lemma 2.4.4; they transform as
differentials would. This calculation followed from Property 3, which we still impose here, so
it is still valid. This gives us the pullback of basic k-forms.
Then we use Property 2 to extended it to a function times a basic k-form, and then
Property 1 to extend it to linear combination of such terms, to get the final result for the
pullback of a generic k-form. ⌅
This formula certainly looks ugly, but it is really not that bad. Concretely, all you need to
do is compose the component functions with „, and transform the basic one-forms one by one
in the wedge product as in Lemma 2.4.4, i.e. they transform as you would expect differentials
transform. That’s it! This will certainly be clearer with examples.
Example 4.7.2 The pullback of a two-form. Consider the following two-form on R3 :
Ê = xy dy · dz + (xz + y) dz · dx + dx · dy,
and the smooth function „ : R2 æ R3 given by
„(u, v) = (uv, u2 , v 2 ) = (x(u, v), y(u, v), z(u, v)).
To calculate „ú Ê, let us start by calculating the pullback of the basic one-forms. We get:
ˆx ˆx
„ú (dx) = du + dv = v du + u dv,
ˆu ˆv
ˆy ˆy
„ú (dy) = du + dv = 2u du,
ˆu ˆv
ˆz ˆz
„ú (dz) = du + dv = 2v dv.
ˆu ˆv
CHAPTER 4. k-FORMS 121

Putting this together, we get:

„ú Ê =(uv)(u2 )„ú (dy) · „ú (dz) + ((uv)(v 2 ) + u2 )„ú (dz) · „ú (dx) + „ú (dx) · „ú (dy)
=u3 v(2u du) · (2v dv) + (uv 3 + u2 )(2v dv) · (v du + u dv) + (v du + u dv) · (2u du)
=4u4 v 2 du · dv + 2uv 2 (v 3 + u) dv · du + 2u2 dv · du
=(4u4 v 2 ≠ 2uv 5 ≠ 2u2 v 2 ≠ 2u2 )du · dv.

Note that we used the fact that du · du = dv · dv = 0, and dv · du = ≠du · dv. ⇤

Example 4.7.3 The pullback of a three-form. Consider the following three-form on R3 :

Ê = ex+y+z dx · dy · dz,

and the smooth function „ : R3>0 æ R3 given by

„(u, v, w) = (ln(uv), ln(vw), ln(wu)).

The pullback of the basic one-forms is:

1 1
„ú (dx) = du + dv,
u v
1 1
„ (dy) = dv +
ú
dw,
v w
1 1
„ú (dz) = dw + du.
w u
We get:
3 4 3 4 3 4
1
ln(uv)+ln(vw)+ln(wu) 1 1 1 1 1
„ Ê =e
ú
du + dv · dv + dw · dw + du
u v v w w u
3 4
1 1
=u2 v 2 w2 du · dv · dw + dv · dw · du
uvw uvw
=2uvw du · dv · dw,

where we used the fact that the basic three-forms vanish whenever one of the factor is repeated,
and dv · dw · du = du · dv · dw since we need to exchange two basic one-forms twice to
relate the two basic three-forms. ⇤
Good, so we are now experts at computing pullbacks! Calculating the pullback of a k-form
is not more difficult than calculating the pullback of a one-form, but the calculation may be
longer, and you need to use the anti-symmetry properties of basic k-forms in Lemma 4.1.6 to
simplify the result at the end of your calculation.
Property 3 in our axiomatic definition states that the pullback commutes with the exterior
derivative for zero-form. It turns out that this property, which is very important, holds in
general for k-forms. Let us prove that.
Lemma 4.7.4 The pullback commutes with the exterior derivative. Let „ : V æ U
be a smooth function, with V ™ Rm and U ™ Rn open subsets. Let Ê be a k-form on U . Then:

„ú (dÊ) = d(„ú Ê).

CHAPTER 4. k-FORMS 122
ÿ
Proof. Recall the definition of the exterior derivative Definition 4.3.1. Let Ê = fi1 ···ik dxi1 ·
1Æi1 <···<ik Æn
· · · · dxik . Then ÿ
dÊ = d(fi1 ···ik ) · dxi1 · · · · · dxik .
1Æi1 <···<ik Æn

Using Properties 1 and 2, we can write

ÿ
„ú (dÊ) = „ú (d(fi1 ···ik )) · „ú (dxi1 ) · · · · · „ú (dxik ).
1Æi1 <···<ik Æn

Now we can use Property 3, which states that

„ú (d(fi1 ···ik )) = d(„ú fi1 ···ik ),

since the fi1 ···ik are just functions (zero-forms). Thus we have:
ÿ
„ú (dÊ) = d(„ú fi1 ···ik ) · „ú (dxi1 ) · · · · · „ú (dxik )
1Æi1 <···<ik Æn
Q R
ÿ
=d a („ú fi1 ···ik )„ú (dxi1 ) · · · · · „ú (dxik )b
1Æi1 <···<ik Æn

=d(„ Ê),ú

where the last line follows from Properties 1 and 2 again. ⌅

4.7.2 The pullback of a top form in Rn and the Jacobian determinant

There is a special case for which the pullback takes a very nice form. This case will play a
role shortly in our theory of integration, as it will be related to the transformation formula
(the generalization of the substitution formula) for multiple integrals.
Let us start by recalling the definition of the Jacobian of a smooth function.
Definition 4.7.5 The Jacobian. Let „ : V æ U be a smooth function with U, V ™ Rn
open subsets. Let us write t = (t1 , . . . , tn ) œ V , and

„(t) = (x1 (t), . . . , xn (t)).

1 ,...,xn )
The Jacobian of „, which we denote by ˆ(x ˆ(t1 ,...,tn ) or J„ or D„ (lots of different notations!),
is the n ◊ n matrix of first partial derivatives:
Q ˆx ˆx1
R
1
···
ˆ(x1 , . . . , xn ) c
ˆt1 ˆtn
D„ = J„ = =c .. .. .. d
. . d
. b
ˆ(t1 , . . . , tn ) a
ˆxn ˆxn
ˆt1 ··· ˆtn

Its determinant is called the Jacobian determinant. ⌃

It turns out that if we pullback an n-form with respect to a „ as in Definition 4.7.5, we
can write „ú Ê in terms of the Jacobian determinant. First, let us introduce the common name
“top form” for an n-form on an open subset U ™ Rn .
CHAPTER 4. k-FORMS 123

Definition 4.7.6 Top form. We call an n-form on an open subset U ™ Rn a top form. ⌃
Such forms are called “top forms” because all forms with k Ø n necessarily vanish on Rn .
Going back to the pullback, we get the nice following result when we pullback a top form:
Lemma 4.7.7 The pullback of a top form in Rn in terms of the Jacobian determinant.
Suppose that Ê is a top form on U ™ Rn , and „ : V æ U a smooth function with V ™ Rn . As
above we write
„(t) = (x1 (t), . . . , xn (t))
with t = (t1 , . . . , tn ) œ V , and
Ê = f dx1 · . . . · dxn
for some function f : U æ R. Then

„ú Ê = f („(t))(det J„ ) dt1 · . . . · dtn ,

where det J„ is the Jacobian determinant.

Proof for R2 and R3 . It is not so easy to write a general proof for Rn using the computational
approach for the pullback that we have used so far. We will be able to write down a general
proof easily in the next subsection after having introduced a more algebraic approach to the
pullback. For the time being, let us prove the statement explicitly for R2 and R3 .
For R2 , our function „ takes the form

„(t1 , t2 ) = (x1 (t1 , t2 ), x2 (t1 , t2 )).

What we need to show is that

„ú (dx1 · dx2 ) = (det J„ )dt1 · dt2 ,

where A B
ˆx1 ˆx1
ˆx1 ˆx2 ˆx2 ˆx1
det J„ = det ˆt1
ˆx2
ˆt2
ˆx2 = ≠ .
ˆt1 ˆt2 ˆt1 ˆt2 ˆt1 ˆt2
But
3 4 3 4
ˆx1 ˆx1 ˆx2 ˆx2
„ú (dx1 · dx2 ) = dt1 + dt2 · dt1 + dt2
ˆt1 ˆt2 ˆt1 ˆt2
3 4
ˆx1 ˆx2 ˆx2 ˆx1
= ≠ dt1 · dt2 ,
ˆt1 ˆt2 ˆt1 ˆt2
and the lemma is proved.
The calculation is similar but more involved for R3 . We have:

„(t1 , t2 , t3 ) = (x1 (t1 , t2 , t3 ), x2 (t1 , t2 , t3 ), x3 (t1 , t2 , t3 )).

What we need to show is that

„ú (dx1 · dx2 · dx3 ) = (det J„ )dt1 · dt2 · dt3 ,

CHAPTER 4. k-FORMS 124

where
Q ˆx1 ˆx1 ˆx1 R
ˆt ˆt2 ˆt3
c 12 ˆx2 d
det J„ = det a ˆx
ˆt1
ˆx2
ˆt2 ˆt3 b
ˆx3 ˆx3 ˆx3
ˆt1 ˆt2 ˆt3
ˆx1 ˆx2 ˆx3 ˆx1 ˆx2 ˆx3 ˆx1 ˆx2 ˆx3
= + +
ˆt1 ˆt2 ˆt3 ˆt2 ˆt3 ˆt1 ˆt3 ˆt1 ˆt2
ˆx1 ˆx2 ˆx3 ˆx1 ˆx2 ˆx3 ˆx1 ˆx2 ˆx3
≠ ≠ ≠ .
ˆt2 ˆt1 ˆt3 ˆt1 ˆt3 ˆt2 ˆt3 ˆt2 ˆt1
But
3 4
ˆx1 ˆx1 ˆx1
„ú (dx1 · dx2 · dx3 ) = dt1 + dt2 + dt3
ˆt1 ˆt2 ˆt3
3 4
ˆx2 ˆx2 ˆx2
· dt1 + dt2 + dt3
ˆt ˆt2 ˆt3
3 1 4
ˆx3 ˆx3 ˆx3
· dt1 + dt2 + dt3
ˆt1 ˆt2 ˆt3
1 ˆx ˆx ˆx ˆx ˆx ˆx ˆx1 ˆx2 ˆx3
1 2 3 1 2 3
= + +
ˆt1 ˆt2 ˆt3 ˆt2 ˆt3 ˆt1 ˆt3 ˆt1 ˆt2
ˆx1 ˆx2 ˆx3 ˆx1 ˆx2 ˆx3 ˆx1 ˆx2 ˆx3 2
≠ ≠ ≠ dt1 · dt2 · dt3 ,
ˆt2 ˆt1 ˆt3 ˆt1 ˆt3 ˆt2 ˆt3 ˆt2 ˆt1

which completes the proof in R3 . Phew, this was painful. ⌅

The appearance of the Jacobian determinant here is quite nice. It will be related to the
transformation formula for multiple integrals, when we integrate a top form over a bounded
region in Rn . We note however that we obtain the Jacobian determinant here, not its absolute
value (in comparison to what you may have seen in previous calculus classes); this will be
related to the fact that our theory of integration is oriented.

4.7.3 The Jacobian matrix and two important theorems in multivariable

calculus (optional)
In the previous section we introduced the Jacobian of a function and its determinant. As you
may already know, the Jacobian determinant plays an important role in multivariable calculus.
In this section we review two important theorems that involve the Jacobian determinant, for
the sake of completeness. This section however is tangential to the rest of the text.
The first theorem is the Inverse Function Theorem, which gives a sufficient condition to
determine whether a function is invertible on an open neighbhorhood of a point in its domain.
Remarkably, invertibility of the function is related to invertibility of the Jacobian matrix;
indeed, if the Jacobian matrix is invertible at a point, i.e., its determinant is non-zero at this
point, then the function is invertible on an open neighbhorhood of this point. Neat!
Theorem 4.7.8 Inverse Function Theorem. Let „ : U æ Rn be a C 1 (continously
differentiable) function, with U ™ Rn an open subset. Let a œ U , and suppose that the
determinant of the Jacobian of „ at a is non-zero, that is, det J„ (a) ”= 0. Then there exists
open neighborhoods A ™ U of a and B ™ Rn of b = „(a) such that „ : A æ B is bijective (that
CHAPTER 4. k-FORMS 125

is invertible).
This is a very powerful theorem, which relates invertibility of a function to invertibility of
its Jacobian matrix.
Another important theorem that involves the Jacobian is the Implicit Function Theorem.
This theorem concerns the following question: suppose that we are given a number of relations
between a set of variables, can we represent these relations as the graph of a function? Let us
be more precise.
Start with single variable calculus. Let f : R2 æ R be a C 1 -function, which we write as
f (x, y). Setting f (x, y) = 0 gives a relation between the variables x and y. Can we think
of this relation as implicitly defining y as a function of x? The answer is yes, under mild
conditions, and only locally.
More precisely, the statement is that, for any point (a, b) in the domain of f such that
f (a, b) = 0, if ˆf
ˆy (a, b) ”= 0, then there exists an open neighborhood U ™ R of a such that
there exists a unique C 1 -function g : U æ R such that g(a) = b and f (x, g(x)) = 0 for all
x œ U . In other words, if ˆf ˆy (a, b) ”= 0, the relation f (x, y) = 0 implicitly and uniquely defines
1
y as a C -function y = g(x) locally near the point (a, b).
This statement can be naturally generalized to multivariable calculus using the Jacobian
determinant.
Theorem 4.7.9 Implicit Function Theorem. Let f : Rn+m æ Rm be a C 1 -function. Let
us denote by (x, y) = (x1 , . . . , xn , y1 , . . . , ym ) coordinates on Rn+m . We can write the function
f in terms of its component functions as

f (x, y) = (f1 (x, y), . . . , fm (x, y)).

Let (a, b) œ Rn+m be a point such that f (a, b) = 0. Suppose that det Jf,y (a, b) ”= 0, where
Jf,y is the Jacobian of f with respect to the variables y:
Q ˆf1 ˆf1 R
ˆy1 ··· ˆym
c . .. .. d
a ..
Jf,y = c . d
. b.
ˆfm ˆfm
ˆy1 ··· ˆym

(In other words, this Jacobian matrix is invertible.) Then there exists an open neighborhood
U ™ Rn of a such that there exists a unique C 1 -function g : U æ Rm such that g(a) = b and
f (x, g(x)) = 0 for all x œ U .
This is the natural multivariable generalization. If the Jacobian matrix of the function
f (x, y) with respect to the y-variables is invertible at a point (a, b), the relation f (x, y) = 0
implicitly and uniquely defines the variables y as C 1 -functions y = (g1 (x), . . . , gm (x)) locally
near (a, b).

4.7.4 The pullback of a k-form: an algebraic approach (optional)

In this section we introduce a direct definition of the pullback using the algebraic approach
to basic k-forms introduced in Subsection 4.1.1. We then show that the three fundamental
properties that we used to define the pullback are satisfied, thus justifying our original
axiomatic approach.
CHAPTER 4. k-FORMS 126

Recall from Definition 4.1.1 (naturally generalized to Rn ) that a basic one-form dxi is a
linear map dxi : Rn æ R which acts as

dxi (u1 , . . . , un ) = ui ,

i.e. it outputs the i’th component of the vector u œ Rn . As we are now thinking of the basic
one-forms dxi as linear maps, we can define their pullback by composition, as we originally
did for functions in Definition 2.4.1.
We first define and study the pullback when „ is a linear map between vector spaces. We
will generalize to the case of a smooth function afterwards.
Definition 4.7.10 The pullback of a basic one-form with respect to a linear map.
Let dxi : Rn æ R be a basic one-form, and let „ : Rm æ Rn be a linear map. We define the
pullback „ú (dxi ) : Rm æ R by composition:

„ú (dxi ) = dxi ¶ „ : Rm æ R.

⌃
Let us now give an explicit formula for the pullback of a basic one-form.
Lemma 4.7.11 An explicit formula for the pullback of a basic one-form. Let
dxi : Rn æ R, i = 1, . . . , n be the basic one-forms on Rn , and dtj : Rm æ R, j = 1, . . . , m
be the basic one-forms on Rm . The linear map „ : Rm æ Rn can be represented by a matrix
A = (aij ). Then we can write
m
ÿ
„ú (dxi ) = aij dtj = ai1 dt1 + . . . + aim dtm .
j=1

Proof. Write v = (v1 , . . . , vm ) œ Rm . Then

„ú (dxi )(v) =dxi („(v))

=dxi (a11 v1 + . . . + a1m vm , · · · , an1 v1 + . . . anm vm )
=ai1 v1 + . . . + aim vm
=ai1 dt1 (v) + . . . + aim dtm (v)
= (ai1 dt1 + . . . + aim dtm ) (v).

⌅
It is then easy to generalize the definition of pullback to the basic k-forms.
Definition 4.7.12 The pullback of a basic k-form with respect to a linear map.
Let dxi1 · . . . · dxik : (Rn )k æ R be a basic k-form, and let „ : Rm æ Rn be a linear map.
Let v1 , . . . , vk œ Rm be vectors. We define the pullback „ú (dxi1 · . . . · dxik ) : (Rm )k æ R by:

„ú (dxi1 · . . . · dxik )(v1 , . . . , vk ) = dxi1 · . . . · dxik („(v1 ), . . . , „(vk )).

⌃
It follows directly from the definition that the pullback commutes with the wedge product:
CHAPTER 4. k-FORMS 127

Lemma 4.7.13 The pullback commutes with the wedge product. Under the setup
above,
„ú (dxi1 · . . . · dxik ) = „ú (dxi1 ) · . . . · „ú (dxik ).
Proof. By definition,

„ú (dxi1 ) · . . . · „ú (dxik )(v1 , . . . , vk ) =dxi1 · . . . · dxik („(v1 ), . . . , „(vk ))

=„ú (dxi1 · . . . · dxik )(v1 , . . . , vk ).

⌅
As a corollary, we obtain an explicit formula to calculate the pullback of a basic k-form.
Corollary 4.7.14 An explicit formula for the pullback of a basic k-form. Let
dxi1 · . . . · dxik : (Rn )k æ R be a basic k-form, and let „ : Rm æ Rn be a linear map. Let
dtj : Rm æ R, j = 1, . . . , m be the basic one-forms on Rm . The linear map „ : Rm æ Rn can
be represented by a matrix A = (aij ). Then we can write

„ú (dxi1 · . . . · dxik ) = (ai1 1 dt1 + . . . + ai1 m dtm ) · · · · · (aik 1 dt1 + . . . + aik m dtm ) .
We can also write down an explicit formula in terms of the determinant when we pullback
a basic n-form from Rn to Rn .
Lemma 4.7.15 The pullback of a basic n-form in Rn . Let dx1 · · · · · dxn : (Rn )n æ R,
and let „ : Rn æ Rn be a linear map, which can be representated by an n ◊ n matrix A = (aij ).
Then
„ú (dx1 · · · · · dxn ) = (det A) dx1 · · · · · dxn .

Proof. Let v1 , . . . , vn œ Rn . Then

„ú (dx1 · · · · · dxn )(v1 , . . . , vn ) =dx1 · · · · · dxn („(v1 ), . . . , „(vn ))

Q R
A11 v11 + . . . + A1n vn1 · · · A11 v1n + . . . + A1n vnn
c
= det c .. .. .. d
a . . .
d
b
An1 v11 + . . . + Ann vn1 · · · An1 v1 + . . . + Ann vn
n n
QQ RQ RR
A11 · · · A1n v11 · · · v1n
cc . .. .. d c .. . .
d c . dd
aa ..
= det c . . .. d
c d
. ba . bb
An1 · · · Ann vn1 · · · vnn
Q R
v 1 · · · v1n
c .1 .
. . ... d
d
a ..
=(det A) det c b
1
vn · · · vn n

=(det A)dx1 · . . . · dxn (v1 , . . . , vn ),

which concludes the proof. ⌅

This is all very nice, but so far we only looked at basic k-forms, and linear maps „ : R æ m

R . How do we generalize this to general differential k-forms on U ™ Rn , and to smooth

n

functions „ : V æ U with V ™ Rm ? The idea is simple. For any point t œ V , the Jacobian
matrix of „ (i.e. the matrix of first partial derivatives), also called the “total derivative of „”,
CHAPTER 4. k-FORMS 128

provides a linear map Rm æ Rn .1

In other words, given a smooth function „ : V æ U , if we write „(t) = (x1 (t), . . . , xn (t)),
we can construct the pullback of k-forms exactly as above, but with the specific linear map
Rm æ Rn given by the Jacobian matrix:
A B
ˆxi
A = (aij ) = .
ˆtj

Then we see that we recover all the formulae that we obtained previously, and that our
three fundamental properties are satisfied! The pullback of a basic one-form becomes
ˆxi ˆxi
„ú (dxi ) = dt1 + . . . + dtn ,
ˆt1 ˆtn
as before. Property 1 is obviously satisfied by definition. Lemma 4.7.13 becomes Property 2,
and Corollary 4.7.14 becomes our general formula for the pullback of k-forms in Lemma 4.7.1.
It is easy to check that Property 3 is satisfied. Finally, we obtain a general proof of Lemma 4.7.7
on Rn , as this is simply Lemma 4.7.15. Neat!

4.7.5 Exercises
1. Consider the basic two-form dx · dy on R2 , and the polar coordinate transformation
– : R2 æ R2 with
–(r, ◊) = (r cos ◊, r sin ◊).
Show by explicit calculation that

–ú (dx · dy) = r dr · d◊ = (det J– )dr · d◊.

Solution. Let us start by calculating the pullback two-form. We get:

–ú (dx · dy) =(cos ◊ dr ≠ r sin ◊ d◊) · (sin ◊ dr + r cos ◊ d◊)

=r cos2 ◊ dr · d◊ ≠ r sin2 ◊ d◊ · dr
=r(cos2 ◊ + sin2 ◊)dr · d◊
=r dr · d◊.

Next, we show that det J– = r. By definition of the Jacobian matrix, we have:

A B
ˆx ˆx
det J– = det ˆr
ˆy
ˆ◊
ˆy
ˆr ˆ◊
A B
cos ◊ ≠r sin ◊
= det
sin ◊ r cos ◊
=r cos2 ◊ + r sin2 ◊
=r.
1
This is called the “differential” or “total derivative” of the smooth function „: in fancier differential
geometry, one would say that it is a linear map from the tangent space of V at t to the tangent space of U at
„(t).
CHAPTER 4. k-FORMS 129

Therefore,
–ú (dx · dy) = r dr · d◊ = (det J– )dr · d◊
as claimed.
2. Let
Ê = (x2 + y 2 + z 2 ) dx · dy · dz
on R3 , and – : R3 æ R3 the spherical transformation

–(r, ◊, „) = (r sin(◊) cos(„), r sin(◊) sin(„), r cos(◊)).

(a) Show by explicit calculation that

–ú (dx · dy · dz) = r2 sin(◊) dr · d◊ · d„ = (det J– )dr · d◊ · d„.

(b) Use this to find –ú Ê.

Solution.
(a) We calculate the pullback:

–ú (dx · dy · dz) =(sin(◊) cos(„)dr + r cos(◊) cos(„)d◊ ≠ r sin(◊) sin(„)d„)

· (sin(◊) sin(„)dr + r cos(◊) sin(„)d◊ + r sin(◊) cos(„)d„)
· (cos(◊)dr ≠ r sin(◊)d◊)
=(r2 sin3 (◊) cos2 („) + r2 sin(◊) cos2 (◊) cos2 („) + r2 sin3 (◊) sin2 („)
+ r2 sin(◊) cos2 (◊) sin2 („))dr · d◊ · d„
=r2 (sin3 (◊) + sin(◊) cos2 (◊))dr · d◊ · d„
=r2 sin(◊)dr · d◊ · d„.

Next we show that this det J– = r2 sin(◊). By definition of the Jacobian matrix,
we get:
Q R
ˆx ˆx ˆx
ˆr
c ˆy ˆ◊ ˆ„ d
det J– = det c
a ˆr
ˆy
ˆ◊
ˆy d
ˆ„ b
ˆz ˆz ˆz
ˆr ˆ◊ ˆ„
Q R
sin(◊) cos(„) r cos(◊) cos(„) ≠r sin(◊) sin(„)
c d
= det a sin(◊) sin(„) r cos(◊) sin(„) r sin(◊) cos(„) b
cos(◊) ≠r sin(◊) 0
=r2 sin(◊) cos2 (◊) cos2 („) + r2 sin3 (◊) sin2 („) + r2 sin3 (◊) cos2 („)
+ r2 sin(◊) cos2 (◊) sin2 („)
=r2 sin(◊).

Therefore

–ú (dx · dy · dz) = r2 sin(◊) dr · d◊ · d„ = (det J– )dr · d◊ · d„

as claimed.
CHAPTER 4. k-FORMS 130

(b) To find –ú Ê we can use the result in (a). We get:

–ú Ê =(r2 sin2 (◊) cos2 („) + r2 sin2 (◊) sin2 („) + r2 cos2 (◊))–ú (dx · dy · dz)
=(r2 sin2 (◊) + r2 cos2 (◊))r2 sin(◊)dr · d◊ · d„
=r4 sin(◊)dr · d◊ · d„.
3. Let
Ê = (x2 + y 2 ) (x dy · dz + y dz · dx + z dx · dy)
on R3 , and – : R3 æ R3 be the cylindrical transformation

–(r, ◊, w) = (r cos ◊, r sin ◊, w).

(a) Find –ú Ê.

(b) Show by explicit calculation that d(–ú Ê) = –ú (dÊ).

Solution.

(a) To simplify the calculation of the pullback, let us first calculate the pullback of the
basic two-forms:

–ú (dy · dz) =(sin(◊)dr + r cos(◊)d◊) · dw

= sin(◊)dr · dw + r cos(◊)d◊ · dw,

–ú (dz · dx) =dw · (cos(◊)dr ≠ r sin(◊)d◊)

= ≠ cos(◊)dr · dw + r sin(◊)d◊ · dw,

and

–ú (dx · dy) =(cos(◊)dr ≠ r sin(◊)d◊) · (sin(◊)dr + r cos(◊)d◊)

=r cos2 (◊)dr · d◊ + r sin2 (◊)dr · d◊
=rdr · d◊.

We also observe that

–ú (x2 + y 2 ) = r2 cos2 (◊) + r2 sin2 (◊) = r2 .

Putting this together, we get:

–ú Ê =–ú (x2 + y 2 )–ú (x dy · dz + y dz · dx + z dx · dy)

=r2 (r cos(◊)(sin(◊)dr · dw + r cos(◊)d◊ · dw)
+ r sin(◊)(≠ cos(◊)dr · dw + r sin(◊)d◊ · dw) + rwdr · d◊)
=r3 (rd◊ · dw + wdr · d◊).
CHAPTER 4. k-FORMS 131

(b) On the one hand, we calculated in (a) the pullback –ú Ê = r3 (rd◊·dw+wdr·d◊).We

can calculate its exterior derivative:

d(–ú Ê) =d(r4 ) · d◊ · dw + d(r3 w) · dr · d◊

=(4r3 + r3 )dr · d◊ · dw.
=5r3 dr · d◊ · dw.

On the other hand, we can calculate first the exterior derivative of Ê. We get:

dÊ =d((x2 + y 2 )x) · dy · dz + d((x2 + y 2 )y) · dz · dx + d((x2 + y 2 )z) · dx · dy

=((3x2 + y 2 ) + (3y 2 + x2 ) + (x2 + y 2 ))dx · dy · dz
=5(x2 + y 2 )dx · dy · dz.

We then calculate its pullback:

–ú (dÊ) =5–ú (x2 + y 2 )–ú (dx · dy · dz)

=5r2 (cos(◊)dr ≠ r sin(◊)d◊) · (sin(◊)dr + r cos(◊)d◊) · dw
=5r2 (r cos2 (◊) + r sin2 (◊))dr · d◊ · dw
=5r3 dr · d◊ · dw.

We conclude that
d(–ú Ê) = –ú (dÊ),
as claimed.
4. Let
Ê = zexy dx · dy
on R3 , and „ : (R”=0 )2 æ R3 with:
3 4
u v
„(u, v) = , , uv .
v u
Find „ú Ê.
Solution. We calculate the pullback:
3 4 3 4
1 u v 1
„ Ê =uve du ≠ 2 dv · ≠ 2 du + dv
u v
ú v u
v v u u
3 4
1 1
=uve ≠ du · dv
uv uv
=0.
5. Show that the pullback of an exact form is always exact.
Solution. Let Ê be an exact k-form on U ™ Rn , and let „ : V æ U be a smooth
function for V ™ Rm . We want to prove that the pullback of Ê is exact.
Since Ê is exact, we know that there exists a (k ≠ 1)-form ÷ on U such that Ê = d÷.
CHAPTER 4. k-FORMS 132

Then
„ú Ê = „ú (d÷) = d(„ú ÷),
where we used the fact that the pullback commutes with the exterior derivative (see
Lemma 4.7.4). Therefore, the k-form „ú Ê on V is the exterior derivative of a (k ≠ 1)-form
„ú ÷ on V , and hence it is exact.
6. Let Ê be a k-form on Rn , and Id : Rn æ Rn be the identity function defined by
Id(x1 , . . . , xn ) = (x1 , . . . , xn ). Show that

Idú Ê = Ê.

Solution. We can write a general k-form on Rn as

ÿ
Ê= fi1 ···ik dxi1 · · · · · dxik
1Æi1 <···<ik Æn

for some smooth functions fi1 ···ik : Rn æ R. We know that Idú (dxi ) = dxi for all
i œ {1, . . . , n}, by definition of the identity map. Similarly, for any function f : Rn æ R,
Idú f = f . As a result, we get:
ÿ
Idú Ê = Idú (fi1 ···ik )Idú (dxi1 ) · · · · · Idú (dxik )
1Æi1 <···<ik Æn
ÿ
= fi1 ···ik dxi1 · · · · · dxik
1Æi1 <···<ik Æn
=Ê.
7. Let Ê be a k-form on U ™ Rn . Let „ : V æ U and – : W æ V be smooth functions,
with V ™ Rm and W ™ R¸ . Show that

(„ ¶ –)ú Ê = –ú („ú Ê).

In other words, it doesn’t matter whether we pullback in one or two steps in the chain of
maps
– „
W æ V æ U.

Solution. We can write a general k-form on U as

ÿ
Ê= fi1 ···ik dxi1 · · · · · dxik
1Æi1 <···<ik Æn

for some smooth functions fi1 ···ik : U æ R. On the one hand, the pullback by „ ¶ – is
ÿ
(„ ¶ –)ú Ê = („ ¶ –)ú (fi1 ···ik )(„ ¶ –)ú (dxi1 ) · · · · · („ ¶ –)ú (dxik ).
1Æi1 <···<ik Æn

On the other hand, pulling back in two steps, we get:

ÿ
–ú („ú Ê) = –ú („ú fi1 ···ik )–ú („ú dxi1 ) · · · · · –ú („ú dxik ).
1Æi1 <···<ik Æn
CHAPTER 4. k-FORMS 133

So all we have to show is that

(„ ¶ –)ú f = –ú („ú f )

for any smooth function f : U æ R, and

(„ ¶ –)ú dxi = –ú („ú dxi )

for any i œ {1, . . . , n}.

First, for any function f : U æ R,

(„ ¶ –)ú f = f ¶ „ ¶ –,

while
–ú („ú f ) = –ú (f ¶ „) = f ¶ „ ¶ –.
Thus
(„ ¶ –)ú f = –ú („ú f ).
As for the basic one-forms, let us introduce further notation for the maps „ : V æ U
and – : W æ V . Let us write z = (z1 , . . . , z¸ ) for coordinates on W ; y = (y1 , . . . , ym )
for coordinates on V ; and x = (x1 , . . . , xn ) for coordinates on U . We write „(y) =
(„1 (y, . . . , „n (y)), and –(z) = (–1 (z), . . . , –m (z)). Then, we have:
m
ÿ ˆ„i
„ú dxi = dya ,
a=1
ˆya

and
m
ÿ̧ ÿ ˆ„i -- ˆ–a
–ú („ú dxi ) = - dzb .
b=1 a=1
ˆya y=–(z) ˆzb
On the other hand, if we pullback by the composition of the maps, we get:

ÿ̧ ˆ(„ ¶ –)i
(„ ¶ –)ú dxi = dzb .
b=1
ˆzb

But the equality

ˆ(„ ¶ –)i ÿm
ˆ„i -- ˆ–a
= -
ˆzb a=1
ˆya y=–(z) ˆzb
is precisely the chain rule for multi-variable functions (written in terms of composition of
functions). Thus we conclude that

–ú („ú dxi ) = („ ¶ –)ú dxi .

Putting all of this together, we conclude that

(„ ¶ –)ú Ê = –ú („ú Ê),

which is the statement of the question.

CHAPTER 4. k-FORMS 134

4.8 Hodge star

In this section we introduce one last operator on differential forms, called the “Hodge star”.
While we will not really use it in this course, it is an integral part of the theory of differential
forms, so it is worth being introduced to it briefly. The Hodge star can be used to recover the
Laplacian operator in the language of vector calculus.

Objectives
You should be able to:

• Define the Hodge star operator in Rn .

• Determine the Hodge star of zero-, one-, two-, and three-forms in R3 .

• Relate to the Laplacian operator in vector calculus.

4.8.1 The Hodge star

The Hodge star is an operator that provides some sort of duality between k-forms and (n ≠ k)-
forms in Rn . It is easiest to define it in terms of the basic k-forms from Definition 4.1.5, and
then extend to general differential forms by applying it to each summand.
Definition 4.8.1 The Hodge star dual of a k-form in Rn . Let Ê = dxi1 · · · · · dxik
be a basic k-form on Rn . Then the Hodge star dual of Ê, which is denoted by ıÊ,1 is the
unique basic (n ≠ k)-form with the property:

Ê · ıÊ = dx1 · · · · · dxn .

To define the Hodge star dual of a general k-form on U ™ Rn , we apply the Hodge star to
each summand. More precisely, if
ÿ
÷= fi1 ···ik dxi1 · · · · · dxik
1Æi1 <...<ik Æn

is a k-form on U , then its Hodge star dual ú÷ is the (n ≠ k)-form given by

ÿ
ı÷ = fi1 ···ik ı (dxi1 · · · · · dxik ).
1Æi1 <...<ik Æn

⌃
To make sense of this definition, let us look at the Hodge star action on the basic k-forms
for low-dimensional space.
Example 4.8.2 The action of the Hodge star in R. There are only two basic k-forms
in R, namely the zero-form 1 and the one-form dx. From the definition, we want 1 · ı1 = dx
and dx · ıdx = dx, from which we conclude that:
ı1 = dx, ıdx = 1.
1
This standard notation should not be confused with the pullback of a differential form; those are very
different things.
CHAPTER 4. k-FORMS 135

The Hodge star thus provides a duality between zero-forms and one-forms in R. ⇤
Example 4.8.3 The action of the Hodge star in R2 . It becomes a little more interesting
in R2 . The basic forms are the zero-form 1, the one-forms dx and dy, and the two-form dx · dy.
From the definition, we want 1 · ı1 = dx · dy, dx · ıdx = dx · dy, dy · ıdy = dx · dy, and
(dx · dy) · ı(dx · dy) = dx · dy. We conclude that

ı1 = dx · dy, ı(dx · dy) = 1,

ıdx = dy, ıdy = ≠dx.

It thus provides a duality between zero-forms and two-forms in R2 , and a “self-duality” for
one-forms. Note that the sign is important here for the action on the basic one-forms. ⇤
Example 4.8.4 The action of the Hodge star in R3 . Things become even more
interesting in R3 . The basic forms are the zero-form 1, the one-forms dx, dy, dz, the two-forms
dy · dz, dz · dx, dx · dy, and the three-form dx · dy · dz. From the definition, we get that:

ı1 = dx · dy · dz, ı(dx · dy · dz) = 1,

ıdx = dy · dz, ıdy = dz · dx, ıdz = dx · dy,
ı(dy · dz) = dx, ı(dz · dx) = dy, ı(dx · dy) = dz.

Thus, in R3 , it provides a duality between zero-forms and three-forms, and between one-forms
and two-forms. ⇤
Example 4.8.5 An example of the Hodge star action in R3 . Consider the two-form
Ê = xyz dy · dz + ex dx · dy. Its Hodge star dual is the one-form:

ıÊ =xyz ı (dy · dz) + ex ı (dx · dy)

=xyz dx + ex dz.

⇤
The action of the Hodge star in R3 naturally justifies our dictionary to translate between
k-forms in R3 and vector calculus objects in Table 4.1.11. Indeed, let Ê = f dx + g dy + h dz
be a one-form on R3 . Then its Hodge dual is the two-form

ıÊ = f dy · dz + g dz · dx + h dx · dy.

This is why we used this particular choice for the basic two-forms in Table 4.1.11; it’s because
this is what one gets through Hodge duality, which identifies one-forms and two-forms in R3 .
In fact, what this means is that we really only needed the first two lines in Table 4.1.11.
Indeed, we can always transform a two-form into a one-form by taking its Hodge dual, and a
three-form into a zero-form. So, in the end, all that we need to establish a dictionary between
differential forms and vector calculus objects is to say that zero-forms are functions, and
one-forms correspond to vector fields.
For instance, if F is the vector field associated to a one-form Ê, we could have defined the
curl Ò ◊ F to be the vector field associated to the one-form ıdÊ. That is,

Ê¡F ı dÊ ¡ Ò ◊ F.
CHAPTER 4. k-FORMS 136

Similarly, we could have defined the divergence Ò · F to be the function given by ıd ı Ê. That
is,
Ê¡F ı d ı Ê ¡ Ò · F.
We could translate all vector calculus identities in Section 4.4 using the Hodge star, but in
the end, as far as we are concerned in this course, this is just a fancier way of saying the same
thing. :-)
Example 4.8.6 Maxwell’s equations using differential forms (optional). Recall
from Example 4.4.7 the statement of Maxwell’s equations, which form the foundations of
electromagnetism. They can be written in terms of the electric vector field E and the magnetic
vector field B on R3 as follows (I am now using units with c = 1 as is standard in modern
physics, and I have rescaled the electric charge fl and the electric current density J to absorb
the factor of 4fi):

Ò · E =fl,
Ò · B =0,
ˆB
Ò◊E+ =0,
ˆt
ˆE
Ò◊B≠ =J.
ˆt
It turns out that there is a very nice way of rewriting Maxwell’s equations using differential
forms, which makes them manifestly relativistic (i.e. consistent with special relativity).
Moreover, this reformulation works in any number of dimensions! It defines the natural
generalization of Maxwell’s equations to higher-dimensional spacetimes, which is useful in
physics theories like string theory.
To write Maxwell’s equations in this form, we need to consider them as living on spacetime,
i.e. R4 , with coordinates (t, x, y, z). We first construct a two-form F on R4 that combines the
electric field E and the magnetic field B as follows:

F = Bx dy · dz + By dz · dx + Bz dx · dy + Ex dx · dt + Ey dy · dt + Ez dz · dt.

We also construct a three-form which combines the electric current J and the electric charge
fl as:
J = fl dx · dy · dz ≠ jx dt · dy · dz ≠ jy dt · dz · dx ≠ jz dt · dx · dy.
Using these definitions, and the definition of the Hodge star on R4 ,2 we can rewrite Maxwell’s
equations neatly as the following two equations (you can check this!):

dF =0,
d ı F =J.

Isn’t that neat? The first equation is simply saying that the two-form F is closed. The second
equation is the “source equation”; if there is no source (i.e. J = 0), it is simply saying that
the two-form ıF is also closed. Moreover, not only is this formulation nice and clean, but it is
manifestly Lorentz invariant (as it is formulated in four-dimensional space time).
Furthermore, this formulation of Maxwell’s equation naturally generalizes to any number
of dimensions. In Rn , F remains a two-form, but J becomes an (n ≠ 1)-form. Then ıF
CHAPTER 4. k-FORMS 137

is an (n ≠ 2)-form, and the two equations make sense in Rn . As mentioned above, this
higher-dimensional generalization is useful in modern physical theories such as string theory.
This is an example of the power of the formalism of differential forms! ⇤

4.8.2 The Hodge star and the Laplacian

We can combine the Hodge star with the exterior derivative to define a new operation on
differential forms, called the “Laplace-Beltrami operator”.
Definition 4.8.7 The codifferential and the Laplace-Beltrami operator. Let Ê be
a k-form on U ™ Rn . We define the codifferential, which is denoted by ”, as the operator
acting on Ê as:
”Ê = (≠1)k ı d ı Ê.
We define the Laplace-Beltrami operator, denoted by , as the operator acting on Ê as:

Ê = ≠(d” + ”d)Ê,

where d is the exterior derivative. ⌃

This looks very fancy, but it is just the natural generalization of the Laplacian of a function
to differential forms, as we now see.
Lemma 4.8.8 The Laplace-Beltrami operator and the Laplacian of a function. Let
f be a zero-form (a function) on U ™ R3 . Then

ˆ2f ˆ2f ˆ2f

f = Ò2 f = Ò · Òf = + + ,
ˆx2 ˆy 2 ˆz 2

where Ò2 f is called the Laplacian of f .

Proof. First, we notice that if f is a function,
”f = ıd ı f = 0,
since ıf is a three-form, and hence its exterior derivative vanishes in R3 . Thus
f = ≠”df = ıd ı df.
We now translate into the language of vector calculus. The vector field associated to the
one-form df is the gradient Òf . Its Hodge dual is a two-form, and taking its exterior derivative
means that we take the divergence of this vector field. So the result is that
f = Ò · Òf.
2
To be precise, we need to modify the definition of the Hodge star operator a little bit here. The reason
is that spacetime is R4 , but with a different definition of length as for standard Euclidean space. It is called
Minkowski spacetime; the definition of length is given by a metric, and in this case the metric is Lorentzian,
which is a bit different from the standard Euclidean metric (it has an extra sign). The reason is that the
fourth dimension of time behaves somewhat differently from the three space dimensions, which is reflected in
this choice of Lorentzian metric. In the end, the correct definition of the Hodge star operator on Minkowski
spacetime R4 is the operator ı that acts on basic k-form as

Ê · ıÊ = ±dt · dx · dy · dz,

with a plus sign whenever Ê does not contain dt, and a minus sign whenever Ê contains dt.
CHAPTER 4. k-FORMS 138

Expanding in coordinates (x, y, z), we get:

ˆ2f ˆ2f ˆ2f

f = Ò · Òf = + + ,
ˆx2 ˆy 2 ˆz 2

which we write as the Laplacian Ò2 f of the function f . ⌅

The Laplace-Beltrami operator applied to a one-form in R3 gives rise to another operation
in vector calculus, called the “Laplacian of a vector field”.
Lemma 4.8.9 The Laplace-Beltrami operator and the Laplacian of a vector field.
Let Ê be a one-form on U ™ R3 , and F be its associated vector field. Then the vector field
associated to the one-form Ê is the Laplacian of the vector field F, which is denoted by
Ò2 F. It acts on the vector field as

ˆ2F ˆ2F ˆ2F

Ò2 F = + + ,
ˆx2 ˆy 2 ˆz 2
and satisfies the identity
Ò2 F = Ò(Ò · F) ≠ Ò ◊ (Ò ◊ F).
Proof. If Ê is a one-form, by definition

Ê = ≠d”Ê ≠ ”dÊ = d ı d ı Ê ≠ ıd ı dÊ,

which is also a one-form. Let us now extract its associated vector field. Let F be the vector
field associated to Ê.
We look at the first term on the right-hand-side. ıÊ is a two-form associated to F. d ı Ê
then takes the divergence Ò · F. ıd ı Ê maps this to a zero-form, and then d ı d ı Ê take
the gradient of the resuling zero-form. The result is that the vector field associated to the
one-form d ı d ı Ê is
Ò(Ò · F).
Let us now look at the second term on the right-hand-side. dÊ takes the curl Ò ◊ F. ıdÊ
then maps it to a one-form associated to the vector field Ò ◊ F. d ı dÊ then takes the curl
again, Ò ◊ (Ò ◊ F), and finaly ıd ı dÊ maps it back to a one-form. The result is that the
vector field associated to the one-form ıd ı dÊ is

Ò ◊ (Ò ◊ F).

We conclude that the vector field associated to the one-form Ê is

Ò(Ò · F) ≠ Ò ◊ (Ò ◊ F),

which we can take as the definition of the Laplacian of the vector field F. To show that it
takes the form
ˆ2F ˆ2F ˆ2F
Ò2 F = + + ,
ˆx2 ˆy 2 ˆz 2
one only needs to do an explicit calculation in R3 , see Exercise 4.8.4.5. ⌅
CHAPTER 4. k-FORMS 139

4.8.3 Two more vector calculus identities

To end this section, let us prove two more vector calculus identities, this time involving the
Laplacian of a function.
Lemma 4.8.10 Vector calculus identities, part 5. Let f and g be smooth functions on
U ™ R3 . Then it satisfies the identities:

1.
Ò2 (f g) = f Ò2 g + 2Òf · Òg + gÒ2 f,

2.
Ò · (f Òg ≠ gÒf ) = f Ò2 g ≠ gÒ2 f.
Proof. These two identities follow from the graded product rule for the exterior derivative.

dF =0,
d ı F =J.

We start with the first one:

Ò2 (f g) = (f g)
= ı d ı d(f g)
= ı d ı (g df + f dg)
= ı d(g ı df + f ı dg)
= ı (dg · ıdf + g d ı df + df · ıdg + f d ı dg)
=Òg · Òf + gÒ2 f + Òf · Òg + f Ò2 g
=f Ò2 g + 2Òf · Òg + gÒ2 f.

In the proof we used the fact that Ò2 f = ıd ı df as in the proof of Lemma 4.8.8.
For the second identity, we get the following:

Ò · (f Òg ≠ gÒf ) = ı d ı (f dg ≠ gdf )
= ı d(f ı dg ≠ g ı df )
= ı (df · ıdg + f d ı dg ≠ dg · ıdf ≠ gd ı df )

To proceed we need to use a result which we haven’t proved. For any two k-forms Ê and
÷ on R3 , there’s a general result that says that

Ê · ı÷ = ıÊ · ÷.

Note that it is important that Ê and ÷ are both k-forms (same k), otherwise it wouldn’t apply.
It it not difficult to prove this statement, but since we do not need it anywhere else, we leave
the proof as an exercise (see Exercise 4.8.4.4).
Now in our previous expression we had the terms df · ıdg and ≠dg · ıdf . Since df and
dg are both one-forms,
df · ıdg = ıdf · dg,
CHAPTER 4. k-FORMS 140

and these two terms cancel out. Thus

Ò · (f Òg ≠ gÒf ) =f (ıd ı dg) ≠ g(ıd ı df )

=f Ò2 g ≠ gÒ2 f.

4.8.4 Exercises
1. Let Ê be the two-form

Ê = xy dy · dz + xyz dz · dx + y dx · dy

on R3 . Find ıÊ.
Solution. We calculate the one-form ıÊ using the action of the Hodge star on basic
two-forms in R3 :

ıÊ =xy ı (dy · dz) + xyz ı (dz · dx) + y ı (dx · dy)

=xy dx + xyz dy + y dz.
2. For any k-form Ê on Rn , show that

ı ı Ê = (≠1)k(n≠k) Ê.

Solution. We only need to prove the statement for basic k-forms as by definition of the
action of the Hodge star operator in Definition 4.8.1 it will then follow for all k-forms.
Let – be a basic k-form on Rn . By definition of the Hodge star, we know that ı– is
the unique basic (n ≠ k)-form such that

– · ı– = dx1 · · · · · dxn .

Now consider the basic (n ≠ k)-form ı–. By definition of the Hodge star, ı ı – will be
the unique basic k-form such that

ı– · ı ı – = dx1 · · · · · dxn .

As the right-hand-side for both equations is the same, we get

– · ı– = ı– · ı ı –.

Using graded commutativity of the wedge product as in Lemma 4.2.6, we can rewrite
the right-hand-side as:
– · ı– = (≠1)k(n≠k) ı ı– · ı–.
Finally, given ı–, we know that the left-hand-side uniquely defines – (by definition of
the Hodge star), while the right-hand-side uniquely deefines ı ı – (again by definition of
the Hodge star), and therefore we must have

– = (≠1)k(n≠k) ı ı–.
CHAPTER 4. k-FORMS 141

3. Let Ê and ÷ be one-forms on Rn , and F and G be the associated vector fields. Show that

ı(Ê · ı÷) = F · G.

Solution. We can write the one-forms Ê and ÷ as:

n
ÿ
Ê= fi dxi ,
i=1
ÿn
÷= gi dxi ,
i=1

for smooth functions fi , gi : Rn æ R. Since ı÷ is an (n ≠ 1)-form, Ê · ı÷ is an n-form on

Rn . It then follows that the only non-vanishing terms in Ê · ı÷ are those of the form
dxi · ıdxi , as all other “cross-terms”, i.e. terms of the form dxi · ıdxj with i ”= j, will
necessarily vanish since ıdxj necessarily contains a dxi . Thus we get:
n
ÿ
Ê · ı÷ = fi gi dxi · ıdxi
i=1
A n B
ÿ
= fi gi dx1 · · · · · dxn ,
i=1

where we used the definition of the Hodge star for basic one-forms. Finally, since
ı(dx1 · · · · dxn ) = 1, we get:
n
ÿ n
ÿ
ı(Ê · ı÷) = fi gi ı (dx1 · · · · · dxn ) = fi gi = F · G,
i=1 i=1

where the last equality is for the associated vector fields F = (f1 , . . . , fn ) and G =
(g1 , . . . , gn ).
4. Let Ê and ÷ be k-forms on Rn . Show that
Ê · ı÷ = (≠1)k(n≠k) ı Ê · ÷.
Note that it is important that Ê and ÷ are both k-forms (same k), otherwise this property
wouldn’t apply.
Solution. The proof is similar in spirit to the solution of the previous problem. Since Ê
and ÷ are both k-forms on Rn , we can write both as linear combinations of basic k-forms
in Rn :
ÿ
Ê= fi1 ···ik dxi1 · · · · · dxik ,
1Æi1 <···<ik Æn
ÿ
÷= gi1 ···ik dxi1 · · · · · dxik ,
1Æi1 <···<ik Æn

for smooth functions fi1 ···ik , gi1 ···ik : Rn æ R. Then

ÿ
ı÷ = gi1 ···ik ı (dxi1 · · · · · dxik ).
1Æi1 <···<ik Æn
CHAPTER 4. k-FORMS 142

But ı(dxi1 · · · · · dxik ) is an (n ≠ k)-form on Rn , and thus Ê · ı÷ is an n-form in Rn . It

follows that the only terms in Ê · ı÷ that will be non-vanishing are those involving the
wedge product of a basic k-form with its own Hodge star dual, as all other “cross-terms”
will necessarily involve the wedge products of repeated dx’s which trivially vanish. So we
get:
ÿ
Ê · ı÷ = fi1 ···ik gi1 ···ik dxi1 · · · · · dxik · ı(dxi1 · · · · · dxik ).
1Æi1 <···<ik Æn

Similarly, we have:
ÿ
ıÊ · ÷ = fi1 ···ik gi1 ···ik ı (dxi1 · · · · · dxik ) · dxi1 · · · · · dxik
1Æi1 <···<ik Æn
ÿ
=(≠1)k(n≠k) fi1 ···ik gi1 ···ik dxi1 · · · · · dxik · ı(dxi1 · · · · · dxik ),
1Æi1 <···<ik Æn

where we used graded commutativity of the wedge product, Lemma 4.2.6. We thus
conclude that
Ê · ı÷ = (≠1)k(n≠k) ı Ê · ÷.
5. Let F = (f1 , f2 , f3 ) be a smooth vector field in R3 . Show that
ˆ2F ˆ2F ˆ2F
Ò(Ò · F) ≠ Ò ◊ (Ò ◊ F) = + + .
ˆx2 ˆy 2 ˆz 2
This is the definition of the Laplacian of the vector field Ò2 F as in Lemma 4.8.9.
Solution. This is just an explicit and rather painful calculation. Let us do it step-by-
step. First,
ˆf1 ˆf2 ˆf3
Ò·F= + + .
ˆx ˆy ˆz
Thus
A B
ˆ 2 f1 ˆ 2 f2 ˆ 2 f3 ˆ 2 f1 ˆ 2 f2 ˆ 2 f3 ˆ 2 f1 ˆ 2 f2 ˆ 2 f3
Ò(Ò · F) = + + , + + , + + .
ˆx2 ˆxˆy ˆxˆz ˆyˆx ˆy 2 ˆyˆz ˆzˆx ˆzˆy ˆz 2
Next, we move on to the curl. First, we have
3 4
ˆf3 ˆf2 ˆf1 ˆf3 ˆf2 ˆf1
Ò◊F= ≠ , ≠ , ≠ .
ˆy ˆz ˆz ˆx ˆx ˆy
Taking the curl again, we get:
A
ˆ 2 f2 ˆ 2 f1 ˆ 2 f1 ˆ 2 f3 ˆ 2 f3 ˆ 2 f2 ˆ 2 f2 ˆ 2 f1
Ò ◊ (Ò ◊ F) = ≠ ≠ + , ≠ ≠ + ,
ˆyˆx ˆy 2 ˆz 2 ˆzˆx ˆzˆy ˆz 2 ˆx2 ˆxˆy
B
ˆ 2 f1 ˆ 2 f3 ˆ 2 f3 ˆ 2 f2
≠ 2
≠ 2
+
ˆxˆz ˆx ˆy ˆyˆz
Putting these two calculations together, and using the fact that partial derivatives
commute by Clairaut’s theorem (since the vector fields are assumed to be smooth), we
get:
Ò(Ò · F) ≠ Ò ◊ (Ò ◊ F)
CHAPTER 4. k-FORMS 143
A B
ˆ 2 f1 ˆ 2 f1 ˆ 2 f1 ˆ 2 f2 ˆ 2 f2 ˆ 2 f2 ˆ 2 f3 ˆ 2 f3 ˆ 2 f3
= + + , + + , + +
ˆx2 ˆy 2 ˆz 2 ˆx2 ˆy 2 ˆz 2 ˆx2 ˆy 2 ˆz 2
ˆ2F ˆ2F ˆ2F
= + + .
ˆx2 ˆy 2 ˆz 2
6. In this problem we prove the statement in Example 4.8.6 about Maxwell’s equations.

(a) Write down the action of the Hodge star operator on basic k-forms in Minkowksi
R4 (see Footnote 4.8.2 ).

(b) Using your result in part (a), show that the two equations

dF =0,
d ı F =J,

with F and J defined in Example 4.8.6, reproduce Maxwell’s equations.

Chapter 5

Integrating two-forms: surface inte-

grals

We study how two-forms can be integrated along surfaces, which leads us to introduce the
concept of surface integrals (also called “flux integrals”).

5.1 Integrating zero-forms and one-forms

Before we move on to two-forms, for completeness we start by defining how zero-forms
can be integrated over oriented points, and review how one-forms can be integrated over
oriented curves. We highlight the main steps of the construction, setting up the stage for the
development of surface integrals.

Objectives
You should be able to:

• Define the integral of a zero-form over oriented points.

• Rephrase the Fundamental Theorem of line integrals in terms of oriented integrals of

one- and zero-forms.

5.1.1 General strategy

Let us start by summarizing step-by-step how we are developing our theory of integration.
Suppose that we want to define integration of k-forms over k-dimensional spaces:

1. We define the orientation of a closed bounded region D µ Rk (such as a closed interval

in R), and the notion of canonical orientation. We define the induced orientation on the
boundary of the region ˆD.

2. We define the integral of a k-form over an oriented region D in terms of standard

multiple integrals from calculus. To define the integral over a bounded region that has
many components, we sum over the integrals on each components.

144
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 145

3. We define the notion of a parametric space, which maps a region D to a k-dimensional

subspace S µ Rn , with n > k. We show that the parametrization induces an orientation
on S.
4. To define the integral of a k-form in Rn on the subspace S µ Rn with a choice of
orientation, we use the parametrization to pullback the k-form to the region D µ Rk ,
and then we integrate as in (2) using standard calculus.
5. We show that the integral is invariant under orientation-preserving reparametrizations
and changes sign under orientation-reversing reparametrizations, thus ensuring that our
theory is oriented and reparametrization-invariant. We conclude that the integral is
defined geometrically in terms of the subspace S and a choice of orientation.
6. As a last step, we study what happens in the case of an exact k-form: this leads to
Stokes’ Theorem, which is the higher dimensional generalization of the Fundamental
Theorem of Calculus and the Fundamental Theorem of line integrals.
This summarizes the conceptual steps to construct our theory of integration. This is pretty
much exactly what we did for line integrals in Chapter 3; we will review these steps below.
But before we do that, let us apply this strategy to construct integrals of zero-forms.

5.1.2 Integrating zero-forms over oriented points

We consider first the very simple and particular case of zero-forms. We would like to integrate
a zero-form over a zero-dimensional space. What is a zero-dimensional space? It is just a
point (or a union of points). But let us construct the theory step-by-step.
STEP 1. We first need to define the orientation of a point.
Definition 5.1.1 Oriented points. Pick a point a œ Rn . The orientation of a point
a œ Rn is given by a choice of + or ≠. We write an oriented point as (a, +) or (a, ≠). The
canonical orientation is +. ⌃
According to our recipe, we should talk about the induced orientation on the boundary,
but a point has no boundary, so this step is meaningless in this case.
STEP 2, 3, 4, 5. The next step is to define the “integral” of a zero-form on an oriented
point in R. Then we would define “parametrizations for points in higher-dimensional spaces”,
but this is all rather trivial here, so we can do steps 2 to 5 all at the same time and simply
define the integral of a zero-form on an oriented point in Rn directly.
Recall that a zero-form is just a function f . In this case, the integral is defined very simply
by just evaluating the function at the point.
Definition 5.1.2 Integral of a zero-form over an oriented point. Let f be a zero-form
on an open subset U ™ Rn and (P, ±) a point in U with a choice of orientation. We define
the integral of f on (P, ±) by: ⁄
f = ±f (P ).
(P,±)

In other words, we just evaluate the function at P œ Rn , and multiply by the sign corresponding
to the chosen orientation.
To define the integral over a set of oriented points, we sum up the integrals over each point
separately. ⌃
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 146

For instance, in this language we can define the integral of a function f over two oriented
points {(P0 , ≠), (P1 , +)} as:
⁄
f = f (P1 ) ≠ f (P0 ).
{(P0 ,≠),(P1 ,+)}

And that’s basically it, as far as zero-forms go. There’s no question of parametrization
here, and the integral is clearly oriented by definition. There’s no step 6, as there is no such
thing as an exact zero-form.
Example 5.1.3 Integral of a zero-form at points. Consider the function f (x, y, z) =
xy + z on R3 . Pick the two points P0 = (0, 0, 0) and P1 = (1, 1, 1). Let’s give P0 a negative
orientation, and P1 a positive orientation. Then
⁄
f = f (P1 ) ≠ f (P0 ) = f (1, 1, 1) ≠ f (0, 0, 0) = 2.
{(P0 ,≠),(P1 ,+)}

5.1.3 Integrating one-forms over oriented curves

We move to the case of one-forms, which was already covered in Chapter 3. Here we review
the construction to highlight how it fits within our general strategy.
STEP 1. We start by considering a closed interval [a, b] œ R. We define its orientation as
in Definition 3.1.3. We also define an induced orientation on the boundary.
Definition 5.1.4 The orientation of an interval. We define the orientation of an
interval [a, b] µ R to be a choice of direction. By [a, b]+ , we mean the interval [a, b] with
the orientation of increasing real numbers (from a to b), and by [a, b]≠ we denote the same
interval but with the orientation of decreasing real numbers (from b to a). We define the
canonical orientation to be the orientation of increasing real numbers.
The boundary of [a, b] is the set of points {a, b}. To the interval [a, b] in canonical
orientation, we define the induced orientation on its boundary to be {(a, ≠), (b, +)}, and
vice-versa for the interval [a, b]≠ in the reverse orientation. ⌃
STEP 2. The next step is to define the integral of a one-form over an interval [a, b] with a
choice of orientation. This is what we did in Definition 3.1.1.
Definition 5.1.5 The integral of a one-form over an oriented interval [a, b]± . Let Ê
be a one-form on U ™ R, with [a, b] µ U . We define the integral of Ê over the oriented
interval [a, b]± as:
⁄ ⁄ b
Ê=± f (x) dx,
[a,b]± a

where on the right-hand-side we use the standard definition of definite integrals from calculus.
⌃
STEP 3. Next step: introduce parametric curves – : [a, b] æ R . This was done in Defini-
n

tion 3.2.1. We do not repeat the definition here, but simply restate that it induces an orientation
on the image curve C = –([a, b]) µ Rn , and on its boundary ˆC = {(–(a), ≠), (–(b), +)} if it
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 147

is not closed. 1 We will write ˆ– to denote the boundary of the parametric curve with its
induced orientation.
STEP 4. We can then define the integral over a parametric curve – using the pullback.
This is Definition 3.3.2.
Definition 5.1.6 (Oriented) line integrals. Let Ê be a one-form on an open subset U of
Rn , and let – : [a, b] æ Rn be a parametric curve whose image C = –([a, b]) µ U . We define
the integral of Ê along – as follows:
⁄ ⁄
Ê= –ú Ê,
– [a,b]

where the integral one the right-hand-side is defined in Definition 5.1.5. ⌃

STEP 5. Then, in Lemma 3.3.5, we showed that the integral above is invariant un-
der orientation-preserving reparametrizations, and changes sign under orientation-reversing
reparametrizations. Thus we can think of the integral as being defined intrinsically in terms
of the image curve C = –([a, b]) and a choice of orientation.
STEP 6. Finally we consider the case of an exact one-form Ê = df . In this case, the
integral simplifies drastically. Let us first look at what happens when we integrate an exact
one-form over an interval [a, b] µ R.
Theorem 5.1.7 The Fundamental Theorem of Calculus. Let f be a zero-form on
U ™ R, and [a, b] µ U with its canonical orientation. Let ˆ([a, b]) = {(a, ≠), (b, +)} be the
boundary of the interval with its induced orientation. Then
⁄ ⁄
df = f,
[a,b] ˆ([a,b])

which is the Fundamental Theorem of Calculus (part II).

It may not seem obvious that this is the Fundamental Theorem of Calculus, but it is! On
the left-hand-side, from our definition of integration, we have:
⁄ ⁄ b
df
df = dx.
[a,b] a dx

On the right-hand-side, from our definition of integration of zero-forms over oriented points
(Definition 5.1.2), we get: ⁄
f = f (b) ≠ f (a).
ˆ([a,b])

So the theorem above is ⁄ b

df
dx = f (b) ≠ f (a),
a dx
which is the Fundamental Theorem of Calculus (part II). Cool!
Now let us move on to integration of exact one-forms over parametric curves.

1
This is clear since the domain of a parametric curve is always the interval [a, b] with its canonical orientation.
Since the canonical orientation induces the orientation {(a, ≠), (b, +)} on the boundary of the interval, it
induces the orientation {(–(a), ≠), (–(b), +)} on the boundary of the parametric curve.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 148

Theorem 5.1.8 The Fundamental Theorem of line integrals. Let f be a zero-form

on U ™ Rn , and – : [a, b] æ Rn be a parametric curve whose image C = –([a, b]) µ U . The
boundary of the parametric curve, with its induced orientation, consists in the two oriented
points ˆ– = {(–(a), ≠), (–(b), +)} if the image curve is not closed; otherwise it is the empty
set. Then: ⁄ ⁄
df = f.
– ˆ–
Again, the statement above may look different from Theorem 3.4.1, but it is really the
same thing. Indeed, the integral on the right-hand-side should be understood as the integral
of a zero-form as in Definition 5.1.2. We can thus write:
⁄ ⁄
f= f = f (–(b)) ≠ f (–(a)),
ˆ– {(–(a),≠),(–(b),+)}

which is the statement in Theorem 3.4.1. In particular, if the curve is closed, ˆ– is the empty
set, and the right-hand-side vanishes, as in Corollary 3.4.3.

5.1.4 Exercises
1. Evaluate the integral of the zero-form f (x, y, z) = sin(x) + cos(xyz) + exy at the set of
oriented points S = {(p, +), (q, ≠), (r, +)} with
3 4
fi 2
p = (0, 0, 5), q= , ,fi , r = (fi, 1, 1).
2 fi

Solution. By definition, the integral is

⁄
f =f (p) ≠ f (q) + f (r)
S
3 4
fi 2
=f (0, 0, 5) ≠ f , , fi + f (fi, 1, 1)
2 fi
=(0 + 1 + 1) ≠ (1 ≠ 1 + e) + (0 ≠ 1 + efi )
=1 + efi ≠ e.
2. Let f be a zero-form on Rn , and suppose that a œ Rn is a zero of the function f . Show
that the integral of f at the point a does not depend on the orientation of the point.
Solution. Let (a, ±) be the point a œ Rn with the positive and negative orientations.
The integral of f at (a, ±) is:
⁄
f = ± f (a)
(a,±)
=0,

since a is a zero of f . Since the result is the same regardless of what orientation the
point a has, we conclude that the integral does not depend on the orientation of the
point. (That’s of course only because the point a is a zero of f , that wouldn’t be true for
an arbitrary point).
3. Suppose that f is a smooth function on Rn , and let p, q œ Rn be two distinct points such
that f (p) = f (q). Show that the line integral of df along any curve starting at p and
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 149

ending at q is zero.
Solution. Let – be any parametric curve whose image starts at p and ends at q. By
the Fundamental Theorem of line integrals, we know that
⁄ ⁄ ⁄
df = f= f = f (q) ≠ f (p) = 0,
– ˆ– {(p,≠),(q,+)}

where the last equality follows since we assume that f (p) = f (q). Therefore, the line
integral of df along any such parametric curve – is zero.
4. Let f be a zero-form on Rn , and S = {(a, +), (≠a, ≠)} for some point a œ Rn (≠a denotes
the point in Rn whose coordinates are minus those of a). Show that
Y
⁄ ]0 if f is even,
f=
S [2f (a) if f is odd.

Solution. The integral of the zero-form is

⁄ ⁄
f= f = f (a) ≠ f (≠a).
S {(a,+),(≠a,≠)}

If f is even, f (≠a) = f (a), and hence

⁄
f = f (a) ≠ f (≠a) = f (a) ≠ f (a) = 0.
S

If f is odd, f (≠a) = ≠f (a), and hence

⁄
f = f (a) ≠ f (≠a) = f (a) ≠ (≠f (a)) = 2f (a).
S
5. Let Ê = 2xy dx + x2 dy be a one-form on R2 . Suppose that
⁄
Ê=5
–

for some parametric curve – : [a, b] æ R2 . Show that the image curve C = –([a, b]) is
not a closed curve, i.e. it must have boundary points.
Solution. First, we notice that Ê is exact. Indeed, let f (x, y) = x2 y. Then

ˆf ˆf
df = dx + dy = 2xy dx + x2 dy = Ê.
ˆx ˆy
But then, by the Fundamental Theorem of line integrals, we know that
⁄ ⁄ ⁄
Ê= df = f.
– – ˆ–

In particular, if the image curve is closed, then the boundary set is empty, i.e. ˆ– = ÿ,
and the right-hand-side is zero. But the question states that it is non-zero; it is equal to
5. Therefore, the image curve cannot be closed.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 150

6. You want to impress your calculus teacher, and you tell her that “integration by parts”
can be rewritten as the “simple” statement that
⁄ ⁄
d(f g) = f g,
[a,b] ˆ([a,b])

i.e. it is just the Fundamental Theorem of Calculus (part II) for a product of functions
(f, g are differentiable functions on R).
Explain why this is equivalent to integration by parts for definite integrals.
Solution. First, using the graded product rule for the exterior derivative, we know
that d(f g) = gdf + f dg. So we can write the left-hand-side as
⁄ ⁄ ⁄ b ⁄ b
d(f g) = (gdf + f dg) = gdf + f dg.
[a,b] [a,b] a a

As for the right-hand-side, the boundary of the interval is ˆ([a, b]) = {(a, ≠), (b, +)}. So
it can be rewritten as ⁄
f g = f (b)g(b) ≠ f (a)g(a).
ˆ([a,b])

Putting this together and rearranging a bit, we get

⁄ b -b ⁄ b
-
f dg = f g - ≠ gdf,
a a a

which is the statement of integration by parts for definite integrals.

5.2 Orientation of a region in R2

We proceed with building our theory of integration for two-forms. Step 1: we define the
orientation of a closed bounded region in R2 , the canonical orientation, and the induced
orientation on its boundary.

Objectives
You should be able to:

• Relate the orientation of Rn to a choice of ordered basis.

• Determine whether two choices of ordered bases on Rn induce the same or opposite
orientations.

• Define the orientation of a closed bounded region in R2 .

• Determine the induced orientation on its boundary.

5.2.1 Orientation of Rn
Recall that a choice of orientation on R is a choice of direction: either that of increasing real
numbers, or of decreasing real numbers. We defined the direction of increasing real numbers
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 151

as being the positive orientation, and called it the canonical orientation. We now generalize
this to Rn .
Definition 5.2.1 Orientation of Rn . An orientation of the vector space Rn is determined
by a choice of ordered basis on Rn . We think of the orientation as a “twirl”, which starts at
the first basis vector, then rotates to the second basis vector, to the third, and so on and so
forth. Ordered bases that generate the same twirl correspond to the same orientation of the
vector space. ⌃
What this definition is saying is that to give an orientation to R , we pick a choice of
n

ordered basis. But not all ordered bases give rise to different orientations; if they generate the
same twirl, they correspond to the same orientation of the vector space. If you think about
it carefully (think about R2 and R3 first), you will see that for any Rn , there are only two
distinct choices of twirl, and hence only two distinct choices of orientation.
This is due to a basic fact in linear algebra. Any two ordered bases of Rn are related
by a linear transformation with non-vanishing determinant. One can see that two ordered
bases that are related by a linear transformation with positive determinant generate the same
twirl, while they generate opposite twirl if they are related by a linear transformation with
negative determinant. So we can group oriented bases into two equivalence classes, depending
on whether they are related by linear transformations with positive or negative determinants.
These equivalence classes of oriented bases correspond to the two choices of orientation on Rn .
Next we define the notion of canonical orientation.
Definition 5.2.2 Canonical orientation of Rn . Let (x1 , . . . , xn ) be coordinates on Rn ,
and {e1 , . . . , en } be the ordered basis

e1 = (1, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), ..., en = (0, . . . , 0, 1),

with ei the unit vector pointing in the positive xi -direction. We call the orientation described
by this ordered basis the canonical orientation of Rn with coordinates (x1 , . . . , xn ), and
denote it by (Rn , +). We denote by (Rn , ≠) the vector space Rn with the opposite choice of
orientation. ⌃
As always, the definition will be clearer with low-dimensional examples. First, let us show
that we recover our previous definition of orientation in R.
Example 5.2.3 Orientation of R and choice of positive or negative direction. If x
is a coordinate on R, the canonical orientation is specified by the basis vector e1 = 1 in the
positive x-direction. Thus the canonical orientation corresponds to the direction of increasing
real numbers, as mentioned before. We denote it by (R, +).
Another choice of basis in R would be f1 = ≠1. It is related to e1 by the linear transfor-
mation f1 = ≠e1 , which has negative determinant. Thus f1 generates the opposite orientation;
indeed, it points in the direction of decreasing real numbers, which correspond to the orientation
(R, ≠). So we recover our previous definition of orientation for R. ⇤
Example 5.2.4 Orientation of R2 and choice of counterclockwise or clockwise
rotation. Let (x, y) be coordinates on R2 . The canonical orientation is specified by the
ordered basis {e1 , e2 }, with e1 = (1, 0) and e2 = (0, 1) being the unit vectors pointing in
the positive x- and y-directions. In two dimensions, the twirl generated by an ordered basis
corresponds to a choice of direction of rotation, from the first basis vector to the second.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 152

So one can think of an orientation of R2 as being a choice of direction of rotation. We

see that the canonical basis corresponds to the choice of counterclockwise rotation. Thus
(R2 , +) corresponds to R2 with the choice of counterclockwise rotation, which is the canonical
orientation.
One could have instead chosen the ordered basis {f1 , f2 } = {e2 , e1 } on R2 (the order of
the vectors is important). Rotating from e2 to e1 corresponds to a clockwise rotation, so
this ordered basis should induce the opposite orientation (R2 , ≠). Indeed, the two bases are
related by the linear transformation (writing basis vectors as column vectors):
A B A B
0 1 0 1
fi = ei , det = ≠1.
1 0 1 0

They therefore generate opposite orientations, as expected. ⇤

Example 5.2.5 Orientation of R3 and choice of right-handed or left-handed twirl.
Let (x, y, z) be coordinates on R3 . The canonical orientation is specified by the ordered basis
{e1 , e2 , e3 } with e1 = (1, 0, 0), e2 = (0, 1, 0) and e3 = (0, 0, 1) being the unit vectors pointing
in the positive x-, y- and z-directions. In three dimensions, the twirl generated by an ordered
basis corresponds to a choice of left-handed or right-handed orientation. We can think of
the first rotation from the first basis vector to the second as being represented by curling
your fingers, and then the rotation from the second basis vector to the third as being in the
direction of your thumb. The canonical basis thus corresponds to the choice of right-handed
twirl, which is the canonical orientation on R3 and denoted by (R3 , +).
Another ordered basis on R3 could have been {f1 , f2 , f3 } = {e2 , e1 , e3 }. Following the twirl
with your fingers and thumb, you will see that this corresponds to a left-handed twirl. So
we expect it to correspond to the opposite orientation (R3 , ≠). Indeed, it is related to the
canonical basis by the linear transformation (writing basis vectors as column vectors):
Q R Q R
0 1 0 0 1 0
c d c d
fi = a1 0 0b ei , det a1 0 0b = ≠1.
0 0 1 0 0 1

The two ordered bases thus generate opposite orientations, as expected. ⇤

5.2.2 Orientation of a closed bounded region in R2

Let us now focus on R2 . Now that we defined the notion of orientation for the vector space
R2 , we can define the orientation of a closed bounded region in R2 , just as we did for closed
intervals in R. But let us first define more carefully what we mean by a closed bounded region
in R2 . We use this to also recall the definition of path connectedness and simple connectedness
(see Subsection 3.6.2), which will be useful later.
Definition 5.2.6 Regions in R2 . Let D µ R2 be a region in R2 .

• A boundary point is a point p œ R2 such that all disks centered at p contain points in
D and also points not in D. We define the boundary of D, which we denote by ˆD, to
be the set of all boundary points of the region D.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 153

• We say that a region D is closed if it contains all its boundary points, that is, ˆD ™ D.
We say that it is open if it contains none of its boundary points, that is, D fl ˆD = ÿ.

• We say that a region D is bounded if it is contained within a finite disk. In other

words, it is bounded if it is finite in extent.

• We say that a region D is path connected (or connected) if any two points in D can
be connected by a path within D. In other words, it is path connected if it has only one
component.

• We say that a region D is simply connected if it is path connected and all simple
closed curves (loops) in D can be continuously contracted to a point within D. In other
words, it is simply connected if it has only one component and no holes.

⌃
Remark 5.2.7 We will often consider closed, simply connected, bounded regions D µ R2 .
Such a region can be constructed by considering a simple closed curve C µ R2 , and letting the
region D be the closed curve and its interior. The boundary of the region is then ˆD = C, i.e.
the closed curve that we started with. With this construction, it is clear that D is bounded,
and it is simply connected since it has one component and no holes.
We note however that the closed curve C does not have to be a parametric curve; it may
have kinks and corners, that is, it could be piecewise parametric. For instance, C could be a
triangle, or a rectangle, etc. It needs to be simple however, i.e. not have self-intersection.
Next we define the orientation of a closed bounded region D µ R2 . Recall that we defined
the orientation of a closed interval in R as being a choice of direction, just as for R; so we
can think of the orientation of an interval as being induced by a choice of orientation on the
surrounding vector space R. We do the same for regions in R2 .
Definition 5.2.8 Orientation of a closed bounded region in R2 . Let D µ R2 be a
closed bounded region in R2 , and choose an orientation on R2 . We define the orientation of
the region D as being the orientation induced by the surrounding vector space R2 . We write
D+ for the region D µ R2 with the canonical (counterclockwise) orientation on R2 , and D≠
for the region with the opposite (clockwise) orientation. When we write D without specifying
the orientation, we always mean the region D with its canonical orientation. ⌃
2
As a last step, given a closed bounded region D µ R with a choice of orientation, we need
to define the induced orientation on its boundary ˆD. In the one-dimensional case, given a
closed interval [a, b] œ R with canonical orientation, we defined the induced orientation on its
boundary to be {(a, ≠), (b, +)}. We do something similar in two dimensions. In this case, the
boundary ˆD is a curve (which may have more than one components), and hence the induced
orientation should be a choice of direction on each component.
Definition 5.2.9 Induced orientation on the boundary of a region in R2 . Let D µ R2
be a closed bounded region with canonical orientation, and ˆD its boundary. Imagine that
the region D is on the floor and that you are walking on its boundary. We define the induced
orientation on each boundary component as being the direction of travel along the boundary
keeping the region on your left. If D has the opposite orientation, the induced orientation on
the boundary is the opposite direction of travel. ⌃
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 154

This will be clearer with some examples.

Example 5.2.10 Closed disk in R2 . Consider the region

D = {(x, y) œ R2 | x2 + y 2 Æ 1},

which is a disk of radius one centered at the origin. The boundary points in D are the points
on the circle x2 + y 2 = 1. Thus ˆD is the circle of radius one centered at the origin. Since it
is included in D, this means that D is closed. It is clearly bounded, as it can be contained
within a disk. It is also simply connected.
We can give D the canonical orientation induced by the standard choice of ordered basis
on R2 , which is the choice of counterclockwise direction of rotation. The induced orientation
on the boundary is the counterclockwise direction of motion along the circle, as this is the
direction that one needs to move along the circle to keep its interior on the left. ⇤
Example 5.2.11 Closed square in R2 . Consider the region

D = {(x, y) œ R2 | ≠ 1 Æ x Æ 1, ≠1 Æ y Æ 1}.

This corresponds to a square and its interior centered at the origin and with side length 2. It
is bounded as it can be contained within a finite disk. Its boundary ˆD is the square itself.
As it is included in D, D is closed. It is also simply connected. If we choose the canonical
orientation on D, the induced orientation on ˆD corresponds to moving counterclockwise
along the square. ⇤
Example 5.2.12 Annulus in R2 . Consider the region

D = {(x, y) œ R2 | 1 Æ x2 + y 2 Æ 2}.

This corresponds to an annulus with inner radius 1 and outer radius 2. D is certainly bounded
as it is contained within a finite disk. Its boundary has two components, ˆD = ˆD1 fi ˆD2 ,
where ˆD1 is the inner circle of radius 1 and ˆD2 is the outer circle of radius 2. While the
region is connected, it is not simply connected, as there is a hole in the middle.
Suppose that we choose the canonical orientation on D. What is the induced orientation
on the boundary? We have to look at the two components separately. First, along the outer
radius ˆD2 , we need to move in the counterclockwise direction to keep the interior of the
annulus on the left. Thus the induced orientation is counterclockwise. However, for the inner
circle ˆD1 , we need to move in the clockwise direction to keep the interior of the annulus on
the left. So the induced orientation on ˆD1 is clockwise. ⇤
Remark 5.2.13 There is another way that one can think of the induced orientation on the
boundary of a region in R2 . It is a little bit more subtle, and while it is not needed at this
stage, it will be useful to generalize to regions in R3 and Rn , so let us mention it here.
Consider first the case where D µ R2 is a closed, simply connected, bounded region.
Then ˆD is a simple closed curve. Since a curve is a one-dimensional subspace of R2 , at a
point p œ ˆD there are two choices of (length one) normal vectors: one that points “inwards”
(towards D), and one that points “outwards” (away from D). For all points p œ ˆD, we pick
the normal vector that points outwards. This defines an orientation on ˆD as follows; start
with the normal vector, and rotate to a tangent vector in a way that reproduces the orientation
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 155

(the twirl) on the ambient space R2 . This defines a choice of tangent vector at all points on
ˆD, which defines the induced direction of motion or orientation on the curve. It is easy to
see in a figure that if the ambient space R2 is equipped with canonical orientation, choosing
the normal vectors pointing outwards corresponds to choosing the direction of motion keeping
the region on the left.
If the region is not simply connected, its boundary may have many components. Do the
same construction for each component, always choosing the normal vectors pointing outwards
(away from D). This will induce the orientation on each component corresponding to the
direction of motion keeping the region on the left.
This construction in terms of normal vectors is more subtle, but it directly generalizes to
closed bounded regions in Rn , which is nice.

5.2.3 Exercises
1. Determine whether the following regions are bounded, closed, connected, and/or simply-
connected.
(a) D = {(x, y) œ R2 | x Ø 2, y Ø 3}.

(b) D = {(x, y) œ R2 | x2 + y 2 < 1}.

(c) D = {(x, y) œ R2 | x œ [0, 1] fi [3, 4], y œ [0, 1]}.

x2 y 2
(d) D = {(x, y) œ R2 | + Æ 1}.
4 9
Solution.

(a) D extends forever in the positive x and y directions, so it is not bounded (it is not
of finite extent). The boundary points are the points in D with x = 2 or y = 3,
since they are at the edge of the region. All those points are included in D, so D is
closed. It is connected, as it has only one component, and it is simply-connected,
as there is no hole.

(b) D is the unit disk of radius one, without its boundary the circle of radius one. It is
bounded, since it can be contained within a disk (it is of finite extent). It is not
closed, since the boundary of D is the circle of radius one, which is not included in
D. It is connected (one component) and simply-connected (no hole).

(c) D consists in two separate square components. It is bounded, since the two square
components can be contained within a disk (finite extent). It is closed, since the
boundary points are the edges of the squares, which are all included in D. It is
however not connected (two components), and hence also not simply-connected.

(d) D is the region bounded by ellipse centered at the origin. It is bounded (finite
extent), closed (the ellipse itself, which is the boundary, is included in D), connected
(one component), and simply-connected (no hole).
2. Let {e1 , e2 , e3 } be the canonical basis on R3 . Show that the ordered basis {e2 , e3 , e1 }
induces the same orientation on R3 as the canonical basis.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 156

Solution. To show that two ordered bases induce the same orientation, we need to show
that they are related by a linear transformation with positive determinant. If we write
{f1 , f2 , f3 } = {e2 , e3 , e1 } for the ordered basis, we see that it is related to the canonical
basis by the linear transformation (thinking of the basis vectors as column vectors):
Q R
0 0 1
c d
fi = a1 0 0b ei .
0 1 0

Indeed,
Q RQ R Q R
0 0 1 1 0
c dc d c d
a1 0 0b a0b = a1b ,
0 1 0 0 0
Q RQ R Q R
0 0 1 0 0
c dc d c d
a1 0 0b a1b = a0b ,
0 1 0 0 1
Q RQ R Q R
0 0 1 0 1
c dc d c d
a1 0 0b a0b = a0b .
0 1 0 1 0

But Q R
0 0 1
c d
det a1 0 0b = 1,
0 1 0
and hence the two ordered bases induce the same orientation on R3 .
3. Let D be the upper half of a disk of radius one, including its boundary. Suppose that D
is given the canonical orientation. Write its boundary, with the induced orientation, as
an oriented parametric curve.
Solution. The upper half of a disk of radius one is the region of R2 defined by:

D = {(x, y) œ R2 | x2 + y 2 Æ 1, y Ø 0}.

Its boundary ˆD is the upper half of the circle, and the x-axis between x = ≠1 and x = 1.
Since D has canonical orientation, the induced orientation corresponds to walking along
the boundary curve keeping the region to the left, which means going counterclockwise
along the boundary.
To realize ˆD as a parametric curve, we need to split it in two, since there are corners
where the upper half disk meets the x-axis. Let C be the upper half disk, and L be the
part of the x-axis in the boundary. We can parametrize the two curves separately. For C
we take –1 : [0, fi] æ R2 with

–1 (t) = (cos(t), sin(t)).

This has the correct orientation, as it goes counterclockwise around the circle. As for the
L, we need to parametrize the line y = 0 from x = ≠1 to x = 1. We take –2 : [≠1, 1] æ R2
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 157

with
–2 (t) = (t, 0).
4. Let A be a closed disk of radius four centered at the origin, and B be the region

B = {(x, y) œ R2 | x œ (≠1, 1), y œ (≠1, 1)}.

Let D be the region that consists of all points in A that are not in B. Is D bounded?
Closed? Connected? Simply-connected? What is the boundary of D? And if D is given
the canonical orientation, what is the induced orientation on the boundary ˆD?
Solution. We note that B is the interior of a square of side length two centered at the
origin. It is enclosed within A, which is a disk of radius four. Therefore, D is certainly
bounded, since it is of finite extent.
The boundary points of D consists of all points on its outer boundary, which is the
circle of radius four, and on its inner boundary, which consists on the points on the square
centered at the origin. Thus, it has two separate components. But all the boundary
points are included in D, and hence D is closed.
It is connected, as it has only one component. However, it is not simply connected,
as any closed loop going around the inner missing square cannot be contracted to a point
within D.
What is the induced orientation on the boundary? Let us denote by C1 the outer
boundary consists of the circle of radius four, and C2 the inner boundary consisting of
the square of side length two. Since D has canonical orientation, the induced orientation
on the boundary will be obtained by walking along the boundary components keeping
the region to the left. Along the outer boundary C1 , we will keep the region to the left if
we walk counterclockwise. However, along the inner boundary C2 , we will keep the region
to the left if we walk clockwise. Therefore, the induced orientation is councterclockwise
on C1 and clockwise on C2 .
5. The parametric curve – : [0, fi] æ R2 with

–(t) = (sin(t), sin(2t))

is shown in the following figure:

CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 158

Figure 5.2.14 The parametric curve – : [0, fi] æ R2 with –(t) = (sin(t), sin(2t)).
Suppose that D is the region consisting of the curve and its interior. What should
the orientation of the region D be so that the induced orientation on its boundary is the
same as the orientation of the parametric curve?
Solution. Let us first find the orientation of the parametric curve. The tangent vector
is
T(t) = (cos(t), 2 cos(2t)).
In particular, at the origin, the tangent vector is T(0) = (1, 2). It points upwards and
in the positive x-direction. Therefore, we see that the parametric curve has clockwise
orientation.
The region D is the region consisting of the curve and its interior. If we walk clockwise
along the curve, the region is on our right. This means that we must give D a clockwise
(or negative) orientation if we want the induced orientation on the boundary to be the
same as the orientation of the parametric curve.

5.3 Integrating a two-form over a region in R2

Step 2: we define the integral of a two-form on U µ R2 along a closed bounded domain
in U with a choice of orientation. The definition is in terms of standard double integrals
from calculus. We show that the resulting integral is invariant under orientation-preserving
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 159

reparametrizations, and changes sign under orientation-reversing reparametrizations. This is

closely connected to the transformation formula for double integrals.

Objectives
You should be able to:

• Define the integral of a two-form over a closed bounded region in R2 .

• Show that invariance under orientation-preserving reparametrizations recovers the

transformation formula for double integrals.

5.3.1 The integral of a two-form over an oriented closed bounded region in

R2
We define the integral of a two-form over an oriented closed bounded region in R2 .
Definition 5.3.1 Integral of a two-form over an oriented closed bounded region in
R2 . Let D µ R2 be a closed bounded region. Let (x, y) be coordinates on R2 . The canonical
(+) orientation on D is described by the ordered basis {e1 , e2 } for R2 , with e1 = (1, 0) and
e2 = (0, 1) the unit vectors pointing in the positive x- and y-directions. Let Ê = f dx · dy be
a two-form on an open subset U ™ R2 such that D µ U . Then, we define the integral of Ê
over D with canonical orientation as:
⁄ ⁄⁄
Ê= f dA,
D+ D

where on the right-hand-side we mean the standard double integral from calculus of the
function f over the region D. If D is given the opposite orientation, we define the integral of
Ê over D≠ as: ⁄ ⁄⁄
Ê=≠ f dA.
D≠ D

⌃
Remark 5.3.2 A bit of notation: we will always use double or triple integral signs to denote
the standard double and triple integrals from calculus, while we will use only one integral sign
when we are integrating a differential form.
It is probably worth recalling here how double integrals are defined, from your previous
calculus course. If D is a rectangular domain, that is,

D = {(x, y) œ R2 | x œ [a, b], y œ [c, d]},

then the double integral is defined as an iterated integral:

⁄⁄ ⁄ d⁄ b
f dA = f dxdy.
D c a

The notation here means that the inner integral is an integral with respect to x (while keeping
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 160

y fixed), while the outer integral is with respect to y.1 We recall Fubini’s theorem, which
states that the order of integration does not matter:
⁄ d⁄ b ⁄ b⁄ d
f dxdy = f dydx,
c a a c

that is, it does not matter whether you integrate in x first and then in y or the other way
around.
To integrate over a more general closed bounded region D, we proceed as follows. Since D
is bounded, we can take it to be inside a rectangular region. We can then extend the function
f to the rectangular region by setting it to zero everywhere outside D. The double integral
over D is then defined to be the integral of the extended function over the rectangular region,
which can be written as an iterated itegral as above.
There are two types of regions that give rise to nice iterated integrals. If D can be written
as follows:
D = {(x, y) œ R2 | x œ [a, b], u(x) Æ y Æ v(x)},
with u, v : R æ R continuous functions, we say that the region is x-supported (or of type I
). In this case, one can show that the double integral can be written as the following iterated
integral:
⁄⁄ ⁄ b ⁄ v(x)
f dA = f dydx,
D a u(x)

where the inner integral is with respect to x (keeping y fixed), while the outer integral is with
respect to y.
If instead D can be written as:

D = {(x, y) œ R2 | y œ [c, d], u(y) Æ x Æ v(y)},

we say that D is y-supported (or of type II). The double integral is then the iterated
integral:
⁄⁄ ⁄ d ⁄ v(y)
f dA = f dxdy,
D c u(y)

with the inner integral being with respect to y (keeping x fixed), and the outer integral with
respect to y.
Note that rectangular regions are particular cases of both x-supported and y-supported
regions. If a region D is either x-supported or y-supported, we say that it is recursively
supported. Most of the regions that we will deal with will be either recursively supported
regions, or regions that can be expressed as unions of recursively supported regions.
Example 5.3.3 Integral of a two-form over a rectangular region with canonical
orientation. Consider the two-form Ê = xy dx · dy on R2 , and the closed bounded region

D = {(x, y) œ R2 | 0 Æ x Æ 2, ≠1 Æ y Æ 3},
1
Note that on the right-hand-side here there is no wedge product: this is not the integral of a two-form, it
is an interated integral in x and y as you have seen in calculus. The inner integral is with respect to x (keeping
y fixed), while the outer integral is with respect to y.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 161

equipped with the canonical orientation. Then

⁄ ⁄⁄
Ê= xy dA
D+ D
⁄ 3 ⁄ 2
= xy dxdy
≠1 0
⁄ 3 C Dx=2
x2
= y dy
≠1 2 x=0
⁄ 3
=2 y dy
≠1
-3
=y 2 -
-
≠1
=8.

To evaluate the double integral, we used the standard procedure for evaluating double integrals
over rectangular regions as iterated integrals, with the inner integral being with respect to x
(keeping y constant), and the outer integral being with respect to y. ⇤
Example 5.3.4 Integral of a two-form over an x-supported (or type I) region with
canonical orientation. Consider the two-form Ê = xey dx · dy, and the region

D = {(x, y) œ R2 | x œ [1, 2], ln(x) Æ y Æ ln(2x)}.

D is an x-supported region. The integral then reads:

⁄ ⁄ 2 ⁄ ln(2x)
Ê= xey dydx
D+ 1 ln(x)
⁄ 2
ln(2x)
= x [ey ]ln(x) dx
1
⁄ 21 2
= xeln(2x) ≠ xeln(x) dx
1
⁄ 21 2
= 2x2 ≠ x2 dx
1
⁄ 2
= x2 dx
1
8 1
= ≠
3 3
7
= .
3
⇤
Remark 5.3.5 It is worth pointing out here that Definition 5.3.1 is actually quite subtle. We
defined the integral of the two-form Ê = f dx · dy over D+ as
⁄ ⁄⁄
f dx · dy = f dA,
D+ D
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 162

where the right-hand-side is the standard double integral of a function in calculus. The subtelty
is that the right-hand-side is not an oriented integral, while the left-hand-side is. Indeed,
suppose for simplicity that D = [a, b] ◊ [c, d] is a rectangular region, as in Example 5.3.3.
Then, we can interpret the right-hand-side as an iterated integral:
⁄⁄ ⁄ d⁄ b
f dA = f dxdy.
D c a

But then, by Fubini’s theorem, we know that we can exchange the order of integration without
issue. That is,
⁄⁄ ⁄ d⁄ b ⁄ b⁄ d
f dA = f dxdy = f dydx.
D c a a c
At first sight, this may appear problematic, as one could be tempted to reinterpret the last
integral as ⁄
f dy · dx,
D+

but since dx · dy = ≠dy · dx, this would be minus the integral we started with! But this is
incorrect. The subtlety is in the choice of orientation.
The key is that in Definition 5.3.1, we started by choosing coordinates (x, y) on R2 , and
then we wrote the one-form Ê = f dx · dy using the basic two-form dx · dy in which the
differentials dx and dy appear in the same order as the coordinates (x, y). This is important,
as dy · dx = ≠dx · dy. So while we can exchange the order of the iterated integrals once we
have written everything in terms of double integrals, when we write the integral in terms of
differential forms, we must use the correct choice of basic two-form dx·dy with the differentials
appearing in the same order as the coordinates of R2 . That’s because integrals of two-forms
are oriented, while double integrals of functions are not.
Note that there is nothing special about the variables x and y. We could name the
coordinates of R2 anything. For instance, if we choose coordinates (u, v) on R2 , then the
integral of a two-form Ê = f du · dv over a region D is equal to the double integral of f over
D (with a positive sign in front) when D is endowed with the canonical orientation described
by the ordered basis {e1 , e2 }, with the basis vectors e1 = (1, 0) and e2 = (0, 1) pointing in the
positive directions of the coordinates u and v respectively.2

5.3.2 Integrals of two-forms over regions in R2 are oriented and reparametrization-

invariant
We already mentioned that integrals of two-forms oriented. Let us now be a little more precise,
and show that, with our definition, integrals of two-forms are invariant under orientation-
preserving reparametrizations, and change sign under orientation-reversing reparametrizations.
Let us first define what we mean by orientation-preserving and orientation-reversing
reparametrizations. We state the definition in R2 , but it naturally generalizes to Rn .
2
In a more careful treatment, we would allow to write the two-form Ê in terms of any basic two-form –. We
would then define the orientation that is compatible with the basic two-form as corresponding to choices of
oriented bases {u, v} such that –(u, v) > 0. Our definition would then state that the integral of a two-form
written in terms of this basic two-form over a region D with the compatible orientation would be given by the
double integral of the function over that region, while the integral over the region with opposite orientation
would be given by minus the double integral.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 163

Definition 5.3.6 Orientation-preserving reparametrizations of regions in R2 . Let

D1 , D2 µ R2 be recursively supported regions,3 and „ : D2 æ D1 a function that can be
extended to a C 1 -function on an open subset U ™ R2 containing D2 . We assume that its
Jacobian determinant det J„ is non-zero on the interior of D2 (recall the definition of the
Jacobian in Definition 4.7.5), which by the Inverse Function Theorem (Theorem 4.7.8) means
that „ is locally bijective (that is invertible) on the interior of D2 . We say that „ is an
orientation-preserving reparametrization if det J„ > 0 for all points in the interior of
D2 , and that it is an orientation-reserving reparametrization if det J„ < 0 for all points
in the interiori of D2 . ⌃
With this definition, we can now show that our theory of integration for two-forms over
regions in R2 is oriented and reparametrization-invariant, our two guiding principles.
Lemma 5.3.7 Integrals of two-forms over regions in R2 are invariant under
orientation-preserving reparametrizations. Let D1 , D2 µ R2 be recursively supported
regions, and „ a reparametrization as in Definition 5.3.6. Let Ê be a two-form on an open
subset U ™ R2 that contains D1 .

• If „ is orientation-preserving, then
⁄ ⁄
„ú Ê = Ê.
D2 D1

• If „ is orientation-reversing, then
⁄ ⁄
„ Ê=≠
ú
Ê.
D2 D1

In other words, the integral is invariant under orientation-preserving reparametrizations, and

changes sign under orientation-reversing reparametrizations.
Proof. The key is to use Lemma 4.7.7 to calculate „ú Ê. Since we are pulling back a two-form
Ê on an open subset U ™ R2 , and that „ : V æ U with V ™ R2 , we can apply the result of
Lemma 4.7.7. We write Ê = f dx · dy, and „(u, v) = (x(u, v), y(u, v)). The jacobian of the
transformation is A B
ˆx ˆx
J„ = ˆu
ˆy
ˆv
ˆy .
ˆu ˆv
Then Lemma 4.7.7 tells us that

„ú Ê = f („(u, v))(det J„ ) du · dv.

This means that ⁄ ⁄

„ Ê=
ú
f („(u, v))(det J„ ) du · dv.
D2 D2
Using our definition of integration in Definition 5.3.1, we know that
⁄ ⁄⁄ ⁄ ⁄⁄
Ê= f (x, y) dxdy, „ú Ê = f („(u, v))(det J„ ) dudv,
D1 D1 D2 D2
3
We could extend the statement more generally to closed bounded regions by taking unions of recursively
supported regions.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 164

where on the right-hand-side of each equation we mean the double integral for the recursively
supported regions D1 in the xy-plane and D2 in the uv-plane.
But recall from your previous calculus course that double integrals satisfy a “transforma-
tion formula”, or “change of variables formula”, which is the natural generalization of the
substitution formula for definite integrals. The transformation formula states that
⁄⁄ ⁄⁄
f (x, y) dxdy = f („(u, v))| det J„ |dudv.
D1 D2

Note that there is now an absolute value around the determinant of the Jacobian. Thus, what
this means is that if our transformation is such that det J„ > 0, then | det J„ | = det J„ , and
⁄ ⁄⁄ ⁄ ⁄
Ê= f (x, y) dxdy = f („(u, v))(det J„ ) du · dv = „ú Ê,
D1 D1 D2 D2

while if det J„ < 0, then | det J„ | = ≠ det J„ , and

⁄ ⁄⁄ ⁄ ⁄
Ê= f (x, y) dxdy = ≠ f („(u, v))(det J„ ) du · dv = ≠ „ú Ê.
D1 D1 D2 D2

This is the statement of the lemma: integrals of two-forms are oriented and reparametrization-
invariant! ⌅
What is particularly nice with the proof of the lemma is that the transformation (or
change of variables) formula for double integrals is simply the statement that
integrals of two-forms over regions in R2 are invariant under orientation-preserving
reparametrizations! Isn’t that cool? It explains why the determinant of the Jacobian
appears; it comes from pulling back the two-form under the change of variables.
The fact that double integrals involve the absolute value of the determinant of the Jacobian,
while our integrals do not (and change signs under orientation-reversing reparametrizations),
is also interesting. As alluded to above, the reason is that our integrals are oriented, while
standard double integrals in calculus are not.
Example 5.3.8 Area of a disk. Consider the basic two-form Ê = dx · dy on R2 . Let us
define the following x-supported domain:
 
D = {(x, y) œ R2 | x œ [≠1, 1], ≠ 1 ≠ x2 Æ y Æ 1 ≠ x2 }.

It is easy to see that D is a closed disk of radius one centered at the origin. The integral of
the basic two-form Ê over D should give us the area of the disk, namely fi. We calculate:
⁄ ⁄⁄
Ê= dA
D D
⁄ Ô
1 ⁄ 1≠x2
= Ô dydx
≠1 ≠ 1≠x2
⁄ 1 1  2
= 1 ≠ x2 + 1 ≠ x2 dx
≠1
⁄ 1 
=2 1 ≠ x2 dx.
≠1
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 165

To evaluate this integral, we do a trigonometric substitution, x = sin(◊):

⁄ 1  ⁄ fi/2 Ò
2 1 ≠ x2 dx =2 1 ≠ sin2 (◊) cos(◊) d◊
≠1 ≠fi/2
⁄ fi/2
=2 cos2 (◊) d◊
≠fi/2
⁄ fi/2
= (1 + cos(2◊)) d◊
≠fi/2
3 4
fi fi 1
= + + (sin(fi) ≠ sin(≠fi))
2 2 2
=fi.

We could have instead use a change of variables to evaluate this integral: polar coordinates.
Define the map „ : R2 æ R2 with

„(r, ◊) = (r cos(◊), r sin(◊)).

Then, if we define the region

D2 = {(r, ◊) œ R2 | r œ [0, 1], ◊ œ [0, 2fi]},

the map „ : D2 æ D is bijective and invertible in the interior of D2 . The determinant of the
Jacobian is A B
cos(◊) ≠r sin(◊)
det J„ = det = r cos2 (◊) + r sin2 (◊) = r,
sin(◊) r cos(◊)
which is positive for all points in the interior of D2 . Thus „ is an orientation-preserving
reparametrization, so the integral of the pullback „ú Ê = r dr · d◊ over D2 should give us fi
again. Indeed, we get:
⁄ ⁄ 2fi ⁄ 1
„ Ê=
ú
(det J„ ) drd◊
D2 0 0
⁄ 2fi ⁄ 1
= r drd◊
0 0
⁄ 2fi
1
= d◊
2 0
=fi.

Notice how easier the integral was! That’s of course because polar coordinates are well suited
for evaluating integrals over regions that have circular symmetry. ⇤

5.3.3 Exercises
1
1. Evaluate the integral of the two-form Ê = (1+x+y)2
dx · dy over the rectangular region

D = {(x, y) œ R2 | x œ [1, 2], y œ [2, 3]}

with canonical orientation.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 166

Solution. By definition of the integral, we have:

⁄ ⁄
1
Ê= dx · dy
D (1 + x + y)
2
D
⁄ 3⁄ 2
1
= dxdy
1 (1 + x + y)
2
2
⁄ 35 6x=2
1
= ≠ dy
2 1 + x + y x=1
⁄ 33 4
1 1
= ≠ dy
2 2+y 3+y
= [ln(2 + y) ≠ ln(3 + y)]y=3
y=2
= ln(5) ≠ ln(6) ≠ ln(4) + ln(5)
3 4
25
= ln .
24
ln(y)
2. Evaluate the integral of the two-form Ê = xy dx · dy over the rectangular region

D = {(x, y) œ R2 | x œ [1, 4], y œ [1, 5]}

with clockwise orientation.

Solution. By definition of the integral, we have (we add a minus sign since we are
evaluating the integral with clockwise orientation):
⁄ ⁄
ln(y)
Ê =≠ dx · dy
D D xy
⁄ 5⁄ 4
ln(y)
=≠ dxdy
1 1 xy
⁄ 5
ln(y)
=≠ [ln(x)]x=4
x=1 dy
1 y
⁄ 5
ln(y)
= ≠ ln(4) dy.
1 y
1
To evaluate the remaining definite integral, we do a substitution u = ln(y), du = y dy,
and y = 1 goes to u = ln(1) = 0, while y = 5 goes to u = ln(5). We get:
⁄ ⁄ ln(5)
Ê = ≠ ln(4) u du
D 0
C Du=ln(5)
u2
= ≠ ln(4)
2 u=0
ln(4)(ln(5))2
=≠ .
2
3. Evaluate the integral of the two-form Ê = cos(y 3 ) dx · dy over the region D bounded by
the parabola x = y 2 and the lines x = 0, y = 0 and y = 1, with canonical orientation.
Solution. We first describe the region explicitly. It is shown in the figure below:
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 167

Figure 5.3.9 The region D is the region bounded by the two curves shown above, the
y-axis, and the x-axis. The green curve is the upper half of the curve x = y 2 , while the
orange line is the line y = 1.
We can describe this region as an x-supported or y-supported region. If we write it
as a y-supported region, we would write 0 Æ y Æ 1, and 0 Æ x Æ y 2 . Thus

D = {(x, y) œ R2 | y œ [0, 1], 0 Æ x Æ y 2 }.

Ô
If we write it as an x-supported region, we would write 0 Æ x Æ 1, and x Æ y Æ 1, that
is, Ô
D = {(x, y) œ R2 | x œ [0, 1], x Æ y Æ 1}.
Which of the two descriptions should we choose? If we use the second description
(x-supported), we first need to integrate in y. But the function that we have to integrate
is cos(y 3 ), which has no elementary antiderivative. So this is problematic. We will not
run into this problem if we use the first description (y-supported); with this description,
we first integrate in x, which is fine. And the next integral in y will then be easy, as we
will see.
So we use the first description of the region as x-supported. The integral then becomes
⁄ ⁄
Ê= cos(y 3 ) dx · dy
D D
⁄ 1 ⁄ y2
= cos(y 3 ) dxdy
0 0
⁄ 1
2
= cos(y 3 ) [x]x=y
x=0 dy
0
⁄ 1
= y 2 cos(y 3 ) dy.
0

To evaluate the remaining definite integral, we do the substitution u = y 3 , du = 3y 2 dy.

Then y = 0 becomes u = 0, and y = 1 becomes u = 1. We get:
⁄ ⁄
1 1
Ê= cos(u) du
D 3 0
1 -u=1
-
= sin(u)-
3 u=0
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 168

1
= sin(1).
3
2 2 2
4. Consider the two-form Ê = e(x +y ) dx · dy, and let D be the disk of radius one with
counterclockwise orientation. Show that
⁄ ⁄ 1
4
Ê = 2fi rer dr.
D 0

Solution. To show this, we change coordinates from Cartesian coordinates to polar

coordinates. We define the function „ : D2 æ D, with

„(r, ◊) = (r cos(◊), r sin(◊)),

and
D2 = {(r, ◊) œ R2 | r œ [0, 1], ◊ œ [0, 2fi]}.
It maps the rectangular region D2 to the unit disk D. The determinant of the Jacobian
of „ is
A B
ˆx ˆx
det J„ = det ˆr
ˆy
ˆ◊
ˆy
ˆr ˆ◊
A B
cos(◊) ≠r sin(◊)
= det
sin(◊) r cos(◊)
=r.

As r œ [0, 1], this is positive on the interior of D2 , and thus we know that the integral
will be invariant under pullback, that is,
⁄ ⁄
„ Ê=ú
Ê.
D2 D

We then calculate the pullback two-form „ú Ê:

4
„ú Ê =er (det J„ )dr · d◊
4
=rer dr · d◊.

Therefore,
⁄ ⁄
4
Ê= rer dr · d◊
D D2
⁄ 1 ⁄ 2fi
4
= rer d◊dr
0 0
⁄ 1
4
= rer [◊]◊=2fi
◊=0 dr
0
⁄ 1
4
=2fi rer dr.
0
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 169

5. Consider the two-form Ê = xy dx · dy, and let D be the region bounded by the four
curves y = x1 , y = x4 , y = x and y = 4x, with canonical orientation. Evaluate the integral
of Ê over D using the change of variables
3 4
u
(x, y) = ,v .
v

Solution. The region D is shown in the figure below:

Figure 5.3.10 The region D is the region bounded by the three curves shown above.
The blue curve is the curve y = 1/x, the orange curve is y = 4/x, the green line is y = x,
and the red line is y = 4x.
We could evaluate the integral of Ê over D by splitting the region D into two sub-
regions, and then realizing these sub-regions as x-supported or y-supported. Or, we can
do a change of variables, as specified in the question.
We consider the change of variables „ : D2 æ D with
3 4
u
„(u, v) = ,v .
v

What is the domain D2 in the (u, v)-plane such that „(D2 ) = D? In the variables u and
v, the four curves y = 1/x, y = 4/x, y = x and y = 4x become
v 4v u u
v= , v= , ,v = , v=4 .
u u v v
Those four bounding equations can be rewritten as
Ô Ô
u = 1, u = 4, v = u, v = 2 u.

Here we used the fact that v is positive, since y = v is positive. So we can describe the
region D2 in the (u, v)-plane as the v-supported region
Ô Ô
D2 = {(u, v) œ R2 | u œ [1, 4], u Æ v Æ 2 u}.

This region is shown in the figure below:

CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 170

Figure 5.3.11 The region D2 in the (u, v)-plane is the region bounded by the four curves
Ô Ô
shown above. The blue curve is the curve v = u, the orange curve is v = 2 u, and the
two grey vertical lines are u = 1 and u = 4.
Next, we calculate the determinant of the Jacobian of „. We get:
A B
ˆx ˆx
det J„ = det ˆu
ˆy
ˆv
ˆy
ˆu ˆv
A B
1
≠ vu2
= det v
0 1
1
= .
v
As v > 0 on D2 , the determinant of the Jacobian is positive on D2 . So we know that the
integral is invariant under pullback:
⁄ ⁄
Ê= „ú Ê.
D D2

The pullback is:

„ú Ê =u(det J„ ) du · dv
u
= du · dv.
v
We then integrate:
⁄ ⁄
u
„ú Ê = du · dv
D2 D2 v
⁄ 4 ⁄ 2Ô u
u
= Ô dvdu
1 u v
⁄ 4 Ô
= u [ln(v)]v=Ôu du
v=2 u
1
⁄ 4
! Ô Ô "
= u ln(2 u) ≠ ln( u) du
1
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 171
⁄ 4
= ln(2) u du
31 4
16 1
= ln(2) ≠
2 2
15 ln(2)
= .
2
For fun, let us show that we would get the same thing by evaluating the integral of
Ê = xy dx · dy directly by splitting the region into two recursively supported sub-regions.
Looking at the graph in Figure 5.3.10, one choice is to split D into the two x-supported
regions D1 and D2 , with

D1 ={(x, y) œ R2 | x œ [1/2, 1], 1/x Æ y Æ 4x},

D2 ={(x, y) œ R2 | x œ [1, 2], x Æ y Æ 4/x}.

We can then evaluate both integrals separately and add them up to get the result. For
D1 , we calculate:
⁄ ⁄ 1 ⁄ 4x
Ê= xy dydx
D1 1/2 1/x
⁄ 1 C Dy=4x
y2
= x dx
1/2 2 y=1/x
⁄ 1 3 4
1
= x 8x2 ≠ dx
1/2 2x2
5 6x=1
1
= 2x4 ≠ ln(x)
2 x=1/2
15 1
= ≠ ln(2).
8 2
As for D2 , we get:
⁄ ⁄ 2 ⁄ 4/x
Ê= xy dydx
D2 1 x
⁄ 1 C Dy=4/x
y2
= x dx
1 2 y=x
⁄ 1 A B
8 x2
= x ≠ dx
1 x2 2
C Dx=2
x4
= 8 ln(x) ≠ ln(x)
8 x=1
15
=8 ln(2) ≠ .
8
Adding those two integrals, we get:
⁄ ⁄ ⁄
Ê= Ê+ Ê
D D1 D2
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 172

15 1 15
= ≠ ln(2) + 8 ln(2) ≠
8 2 8
15 ln(2)
= .
2
This is the same answer that we obtained previously via our change of variables. Great!

5.4 Parametric surfaces in Rn

Step 3: we define parametric surfaces in Rn . This is needed so that we can define integration
of two-forms over surfaces in Rn via pullback. The definition of parametric surfaces is similar
to parametric curves, but there are new subtelties that appear, in particular in relation to the
induced orientation on the surface.

Objectives
You should be able to:

• Define parametric surfaces in Rn .

• Define the tangent plane to a parametric surface in Rn at a point.

• Use different parametrizations for the same surface.

5.4.1 Parametric surfaces in Rn

Definition 5.4.1 Parametric surfaces in Rn . Let D µ R2 be a closed, bounded, and simply
connected region. Let ˆD be its boundary, which is a simple closed curve. A parametric
surface in Rn is a vector-valued function:

– : D æRn
(u, v) ‘æ–(u, v) = (x1 (u, v), . . . , xn (u, v))

such that:
1. – can be extended to a C 1 -function on an open subset U ™ R2 that contains D;

2. The tangent vectors

3 4 3 4
ˆ– ˆx1 ˆxn ˆ– ˆx1 ˆxn
Tu = = ,..., , Tv = = ,...,
ˆu ˆu ˆu ˆv ˆv ˆv

are linearly independent on the interior of D. (If n = 3, this means that Tu ◊ Tv ”= 0

on the interior of D.)

3. If –(u1 , v1 ) = –(u2 , v2 ) for any two distinct (u1 , v1 ), (u2 , v2 ) œ D, then (u1 , v1 ), (u2 , v2 ) œ
ˆD. In other words, – is injective everywhere except possibly on the boundary of D.

The image S = –(D) is a two-dimensional subspace of Rn , which is the surface itself. We say
that the parametric surface is smooth if – can be extended to a smooth function on an open
subset U ™ R2 containing D. ⌃
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 173

This definition is obviously very similar to the definition of parametric curves in Defini-
tion 3.2.1. The three properties play similar roles: together, they ensure that the image surface
S really looks like we expect a surface to look like, that is, some sort of sheet of rubber that may
have been bent, deformed, warped, stretched, what not, but without introducing any puncture
or tear. Property 3 ensures that our parametrization covers the image surface S exactly once
(except possibly for boundary points). Property 2 also ensures that a parametrization induces
a well defined choice of orientation on the image curve S, although this is more subtle than
for parametric curves, as we will see in Section 5.5.
As for parametric curves, we can distinguish between two types of parametric surfaces,
depending on whether the image surface is closed or not. It is a bit more subtle than for
parametric curves, so let us look again at what it means for a parametric curve to be closed.
In Definition 3.2.2, we said that a parametric curve – : [a, b] æ Rn is closed if the image curve
C = –([a, b]) has no endpoints (it is a loop). If it is not closed, then we define the boundary
ˆC = {–(a), –(b)} as containing the endpoints of the image curve.
One way to think about a closed curve is that it is the boundary of a surface. In fact, one
could say that a curve in Rn is closed if and only if it forms the boundary of a surface. If it is
not closed, then it must have endpoints, and those form the boundary of the image curve.
This definition generalizes naturally to surfaces.
Definition 5.4.2 Closed parametric surfaces. Let – : D æ Rn be a parametric surface
with image surface S = –(D) µ Rn . We say that the parametric surface is closed if and only
if the image surface S is the boundary of a solid in Rn . If it is not closed, then it must have
edges: we call the set ˆS consisting of all the points on the edges of S the boundary of the
surface. ⌃
Remark 5.4.3 Given a parametric surface – : D æ Rn , one should not confuse ˆD, the
boundary of the closed bounded region (the domain of –), with ˆS, the boundary of the image
surface. On the one hand, ˆD is never empty, as the domain of – is a closed bounded domain
-- in fact, ˆD is a simple closed curve in R2 as we assume that D is simply connected. On the
other hand, ˆS may or may not be empty, depending on whether the image surface is closed
or not. Thus, generally, –(ˆD) ”= ˆS. However, from the definition of parametric curves it
follows that ˆS ™ –(ˆD); the points on the edges of S can only come from images of points
on the boundary of the domain D.
Example 5.4.4 The graph of a function in R3 . A large number of surfaces in R3 can be
obtained as the graph of a function f (x, y) of two variables:

z = f (x, y).

This defines a surface in R3 . If we choose (x, y) œ D for some closed, bounded, simply
connected domain D, then we can realize the surface as a parametric surface – : D æ R3 with

–(u, v) = (u, v, f (u, v)).

Assuming that f : R2 æ R is a C 1 function, we can check that this satisfies the properties
of a parametric surface. Property one is automatically satisfied, as f is C 1 . The tangent
vectors are 3 4 3 4
ˆf ˆf
Tu = 1, 0, , Tv = 0, 1, .
ˆu ˆv
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 174

Those are clearly linearly independent, regardless of f . Finally, – is certainly injective on the
interior of D; in fact, it is injective everywhere on D.
Since – is injective everywhere on D, including its boundary, this means that the image
surface is not closed. In this case, its boundary ˆS = –(ˆD) is the image of the boundary of
the domain. Basically, the map – takes the domain D and simply deform it continuously in
the z-direction. ⇤
Surfaces realized as the graph of a function, as in the previous example, are very easy to
study parametrically. But many surfaces do not arise in this way, and may instead be given
by an implicit equation in Rn . Finding a parametrization then becomes more difficult.
Example 5.4.5 The sphere. As a second example, we consider the sphere of fixed radius
R centered at the origin in R3 . Its equation is

x2 + y 2 + z 2 = R2 .

We cannot think of this as the graph of a function as in the previous example, because we
cannot solve for z. So we need to think a bit more to realize it as a parametric surface.
As this is a sphere, it is natural that spherical coordinates may be useful. A point on the
sphere radius one can be written as

(x(◊, „), y(◊, „), z(◊, „)) = (R sin(◊) cos(„), R sin(◊) sin(„), R cos(◊)).

It is easy to see that

x(◊, „)2 + y(◊, „)2 + z(◊, „)2 = R2 ,
and thus those points lie on the sphere. Geometrically, ◊ is the inclination angle from the
z direction, and „ is the azimuth angle measured counterclockwise from the x-axis. The
inclination angle runs from 0 to fi, while the azimuth angle runs from 0 to 2fi.
Therefore, a realization of the sphere as a parametric surface is – : D æ R3 , with

D = {(◊, „) œ R2 | ◊ œ [0, fi], „ œ [0, 2fi]},

and
–(◊, „) = (R sin(◊) cos(„), R sin(◊) sin(„), R cos(◊)).
We can check that it satisfies the properties of parametric surfaces. First, – is smooth on
R2 , so Property 1 is fine. Second, it is easy to see geometrically that – is injective everywhere,
except at these points:

• For any two „1 , „2 œ [0, 2fi], –(0, „1 ) = –(0, „2 ) = (0, 0, R). All those points are mapped
to the north pole of the sphere.

• For any two „1 , „2 œ [0, 2fi], –(fi, „1 ) = –(fi, „2 ) = (0, 0, ≠R). All those points are
mapped to the south pole of the sphere.

• For any ◊ œ [0, fi], –(◊, 0) = –(◊, 2fi) = (R sin(◊), 0, R cos(◊)).

The important point is that these points where – is not injective are all on the boundary of
the rectangular region D. Therefore Property 3 is satisfied.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 175

As for Property 2, the tangent vectors are

T◊ =(R cos(◊) cos(„), R cos(◊) sin(„), ≠R sin(◊)),

T„ =(≠R sin(◊) sin(„), R sin(◊) cos(„), 0).

Are those linearly independent? Since the z-coordinate of T„ is zero, the only place where it
could be a multiple of T◊ is when the z-coordinate of T◊ is also zero, that is ≠R sin(◊) = 0.
This will only occur at ◊ = 0, fi, which are on the boundary of D. So we conclude that the
tangent vectors must be linearly independent on the interior of D. Alternatively, we could
have calculated the cross-product T◊ ◊ T„ , and showed that it does not vanish on the interior
of D.
Finally, we note that the image surface, which is the sphere, is closed, since it is the
boundary of a three-dimensional solid (the ball consisting of the interior of the sphere and its
boundary). ⇤
Example 5.4.6 The cylinder. Consider the lateral surface of a cylinder of fixed radius R
extending in the z-direction. Suppose that we look at the part of the cylinder from z = 0 to
z = 2 (we only look at the lateral surface of the cylinder, we do not include the top and the
bottom). How do we realize it as a parametric surface?
The equation of the cylinder is
x2 + y 2 = R2 ,
with z running from 0 to 2. To parametrize it, we introduce polar coordinates. Then a point
on the cylinder can be written as

S(x(◊, w), y(◊, w), z(◊, w)) = (R cos(◊), R sin(◊), w).

If we restrict w from 0 to 2, we get the parametric surface – : D æ R3 with

D = {(◊, w) œ R2 | ◊ œ [0, 2fi], w œ [0, 2]},

and
–(◊, w) = (R cos(◊), R sin(◊), w).
Does it satisfy the properties of a parametric surface? First, – is smooth on R2 , so Property
1 is satisfied. As for Property 2, – is injective except at the points –(0, w) = –(2fi, w) =
(R, 0, w), for all w œ [0, 2]. But those are on the boundary of the rectangular region D, so it is
fine. Finally, the tangent vectors are

T◊ = (≠R sin(◊), R cos(◊), 0), Tw = (0, 0, 1).

Those are clearly linearly independent everywhere, so Property 3 is satisfied.

The image surface here (the cylinder) is not closed; its boundary consists of its two edges,
namely the circle at w = 0 and the circle at w = 2. ⇤

5.4.2 Grid curves

In order to help visualize a parametric surface – : D æ Rn , it is sometimes useful to sketch
the grid curves on the image surface S = –(D). Suppose that D is a closed, simply connected,
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 176

bounded domain in the uv-plane. The idea is to consider the horizontal and vertical lines
in the uv-plane that lie on D; those are the lines u = u0 or v = v0 for constant u0 , v0 . The
image of these lines under the map – will be curves on the image surface S: those are called
the grid curves. They help visualize how the parametrization – maps the region D onto the
image surface S.
Example 5.4.7 Grid curves on the sphere. Consider the sphere of Example 5.4.5, which
is realized as the parametric surface – : D æ R3 with

D = {(◊, „) œ R2 | ◊ œ [0, fi], „ œ [0, 2fi]},

and
–(◊, „) = (R sin(◊) cos(„), R sin(◊) sin(„), R cos(◊)).
The domain D is a rectangular region. Horizontal lines on D correspond to lines with „ = C
for constants C. The image of the horizontal lines would be the grid curves

–(◊, C) = (R cos(C) sin(◊), R sin(C) sin(◊), R cos(◊)).

Since ◊ is the inclination angle, those curves correspond to curves of constant longitude,
starting at the north pole and ending at the south pole.
As for the vertical lines on D, they are given by the lines with ◊ = K for constants K.
The image of the vertical lines would be the grid curves

–(K, „) = (R sin(K) cos(„), R sin(K) sin(„), R cos(K)).

Those correspond to the curves of constant latitude, going all around the sphere. ⇤

5.4.3 The tangent planes

Let – : D æ Rn be a parametric surface, and S = –(D) µ Rn the image curve. In the
definition of parametric surfaces Definition 5.4.1, we introduced the “tangent vectors”
3 4 3 4
ˆ– ˆx1 ˆxn ˆ– ˆx1 ˆxn
Tu = = ,..., , Tv = = ,..., ,
ˆu ˆu ˆu ˆv ˆv ˆv
but we did not really explain their geometric meaning.
At a point P œ S, there is a two-dimensional space of tangent directions to the surface.
This is what is called the “tangent plane” to the surface at P œ S. It gives the best linear
approximation of the surface at that point.
The claim is that the tangent plane at a point p œ S is the two-dimensional vector
space spanned by the vectors Tu and Tv . Indeed, recall that the grid curves are the curves
–(u, C) œ S and –(K, v) œ S for constant C, K. By definition of partial derivatives, it then
follows that the vectors Tu and Tv are tangent vectors at the point p pointing in the direction
of these grid curves. Furthermore, in the definition of parametric surfaces, we assume that Tu
and Tv are linearly independent on the interior of D. Thus, for any point p œ S which is in
the image of the interior of D, we know that the vectors Tu and Tv span a two-dimensional
vector space. We can thus define the tangent planes for a parametric surface as follows:
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 177

Definition 5.4.8 Tangent planes to a parametric surface. Let – : D æ Rn be a

parametric surface, with image surface S = –(D). For any point p œ S which is in the
image of the interior of D, we define the tangent plane Tp S at p to be the two-dimensional
vector space spanned by the tangent vectors Tu and Tv at p. It provides the best linear
approximation of the surface at that point. ⌃
If we now restrict to parametric surfaces in R3 , then there is another way that we can
specify the tangent plane at a point. We can always find the equation of a plane in R3 by
specifying a point on the plane and a normal (perpendicular) vector to the plane. Indeed, if n
is a vector normal to a plane, and (x0 , y0 , z0 ) is a point on the plane, then the equation of the
plane is:
n · (x ≠ x0 , y ≠ y0 , z ≠ z0 ) = 0.
Thus, given a parametric surface – : D æ R3 with image surface S = –(D), at a point p œ S
we can specify the tangent plane by instead specifying the normal vector to the surface at
that point. But what is the normal vector? Well, if Tu and Tv are both tangent vectors,
then we know how to find a new vector that is perpendicular to both vectors: we take the
cross-product! We get:
Definition 5.4.9 Normal vectors to a parametric surface in R3 . Let – : D æ Rn be a
parametric surface, with image surface S = –(D). For any point p œ S which is in the image
of the interior of D, we define the normal vector n at p to be the vector:

n = Tu ◊ Tv .

We note here that this normal vector is not normalized, i.e. it does not have length one. To
get a normalized vector we would divide by its norm. When we talk about the normalized
normal vector later on, we will use the notation
Tu ◊ Tv
n̂ = ,
|Tu ◊ Tv |

to avoid ambiguity. ⌃

5.4.4 Exercises
1. Realize the part of the plane z = x ≠ 2 that lies inside the cylinder x2 + y 2 = 4 as a
parametric surface.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 178

Figure 5.4.10 The surface S is t

2 2
Solution. Let us first sketch a picture of what the surface should look like: x + y = 4 (in orange).
The surface S is the part of the blue plane that lies within the orange cylinder. How
can we realize it as a parametric surface?
We first, we know how to parametrize the plane z = x ≠ 2. The function

–(u, v) = (u, v, u ≠ 2)

is a parametrization of the plane. What we need to determine now is what is the region
D in the (u, v)-plane such that –(D) = S, where S is the part of the plane contained
within the cylinder x2 + y 2 = 4. What we can do is find the boundary curve of the surface
S, which consists in the intersection of the plane z = x ≠ 2 and the cylinder x2 + y 2 = 4.
Using our parametrization above –(u, v), we see that x2 + y 2 = 4 corresponds to the
equation u2 + v 2 = 4. In other words, if we define D to be the disk of radius 2 centered
at the origin, then – maps its boundary (the circle of radius two) to the boundary of the
surface S, and the interior to the interior. So this gives an appropriate choice of domain
D. We can describe D as a u-supported region:
 
D = {(u, v) œ R2 | u œ [≠2, 2], ≠ 4 ≠ u2 Æ v Æ 4 ≠ u2 }.

Then the surface S is realized as the parametric surface – : D æ R3 , with D above and
–(u, v) = (u, v, u ≠ 2).
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 179

Note that there are many other parametrizations that we could have used. For
instance, since the region D is a disk of radius 2, we could have used polar coordinates.
In other words, if we do the change of coordinates u = r cos(◊), v = r sin(◊), the region
becomes
D2 = {(r, ◊) œ R2 | r œ [0, 2], ◊ œ [0, 2fi],
and –2 : D2 æ R3 with

–2 (r, ◊) = (r cos(◊), r sin(◊), r cos(◊) ≠ 2).

That is another parametrization of the same surface S.


2. Realize the part of the sphere x2 + y 2 + z 2 = 4 that lies above the cone z = x2 + y 2 as
a parametric surface.
Solution. Let us first sketch a picture of what the surface should look like:

Figure 5.4.11 The surface S is the part of the sphere with radius 2 (in orange) that lies
above the cone in blue.
How can we realize S as a parametric surface?
We start by parametrizing the sphere of radius 2 centered at the origin. We use
spherical coordinates. A parametrization for the sphere is – : D æ R3 , with

D = {(◊, „) œ R2 | ◊ œ [0, fi], „ œ [0, 2fi]},

and
–(◊, „) = (2 sin(◊) cos(„), 2 sin(◊) sin(„), 2 cos(◊)).
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 180

This is a parametrization of the sphere, but this is not the surface we are
interested in.
We only want to keep the part of the sphere that lies above the cone z = x2 + y 2 . In
other words, we want to restrict the range of the inclination angle ◊ so that it only goes
from 0 to the angle
 where the cone intersects the sphere. What is this angle? The cone
has equation z = x2 + y 2 . Using our parametrization above for points on the sphere,
we see that the points on the sphere that also lie on the cone (i.e. at the intersection of
both surfaces) must satisfy
Ò
2 cos(◊) = (2 sin(◊) cos(„))2 + (2 sin(◊) sin(„))2
=2 sin(◊),

where we used the fact that sin(◊) Ø 0 since ◊ œ [0, fi]. Therefore, we must have

tan(◊) = 1.

The only solution with ◊ œ [0, fi] is ◊ = fi/4.

Therefore, we conclude that the region of the sphere that lies above the cone will be
given by restricting the inclination angle to be between 0 and fi/4. More precisely, a
parametrization of our surface is given by –2 : D2 æ R3 , with

D2 = {(◊, „) œ R2 | ◊ œ [0, fi/4], „ œ [0, 2fi]},

and
–2 (◊, „) = (2 sin(◊) cos(„), 2 sin(◊) sin(„), 2 cos(◊)).
3. Consider the parametric surface – : D æ R3 with D = {(u, v) œ R2 | u œ [0, 2], v œ [0, 3]}
and
–(u, v) = (u, v 3 + 1, u + v).
(a) Find the tangent vectors Tu , Tv , and the normal vector n.

(b) Find an equation for the tangent plane to the image surface –(D) at the point
(1, 2, 2).

Solution. (a) We calculate the tangent vectors by taking partial derivatives:

ˆ– ˆ–
Tu = = (1, 0, 1), Tv = = (0, 3v 2 , 1).
ˆu ˆv
We find the normal vector by taking the cross-product of the tangent vectors:

n =Tu ◊ Tv
Q R
i j k
c d
= det a1 0 1 b
0 3v 2 1
=(≠3v 2 , ≠1, 3v 2 ).

(b) To find an equation of the tangent plane at the point (1, 2, 2), we use the point-
normal form for the equation of a plane. First, we see that –(1, 1) = (1, 2, 2), so the point
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 181

is on the image surface –(D) for the values of the parameters (u, v) = (1, 1). The normal
vector to the plane at that point is n(1, 1) = (≠3, ≠1, 3). Then, by the point-normal
form, we know that the equation of the tangent plane is

n · (x ≠ 1, y ≠ 2, z ≠ 2) = (≠3, ≠1, 3) · (x ≠ 1, y ≠ 2, z ≠ 2) = 0.

Evaluating the dot product, we get the equation of the tangent plane:

≠3x ≠ y + 3z = 1.
4. Consider the curve y = x2 with x œ [0, 2]. Find a parametrization for the surface obtained
by rotating the curve about the y-axis.
Solution. First, we sketch the curve y = x2 in the (x, y)-plane inside R3 :

Figure 5.4.12 The curve y = x2 in the (x, y)-plane inside R3 is shown in blue; the
orange line is the axis of rotation, which is the y-axis.
We rotate the curve about the y-axis, which is the orange line. After rotation, we get
the following surface:
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 182

Figure 5.4.13 The surface obtained by rotating the curve y = x2 about the y-axis.
How do we parametrize this surface? Let’s think about it. First, we can parametrize
the curve y = x2 in the (x, y)-plane within R3 , with x œ [0, 2], by „(t) = (t, t2 , 0) with
t œ [0, 2]. What happens if we rotate the curve about the y-axis? For a fixed value
of t, the point with (x, z)-coordinates (t, 0) gets rotated about the y-axis on a circle
with radius t. We can thus parametrize this circle by (x, z) = (t cos(◊), t sin(◊)), with
◊ œ [0, 2fi]. We do that for all values of t œ [0, 2], and we end up with the surface of
revolution. The resulting parametrization is – : D æ R3 with

D = {(t, ◊) œ R2 | t œ [0, 2], ◊ œ [0, 2fi]}

and
–(t, ◊) = (t cos(◊), t2 , t sin(◊)).

5.5 Orientation of parametric surfaces in R3

Step 3, part II: we show that the parametrization of a surface induces an orientation on
the image surface. However, this is a bit more subtle than for parametric curves, as not all
surfaces are orientable. Thus we first discuss orientability of surfaces, and then show that the
parametrization of an orientable surface induces an orientation on the image surface. We now
focus on surfaces in R3 , which are easier to visualize.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 183

Objectives
You should be able to:

• Define orientable and non-orientable surfaces.

• Determine the orientation of a parametric surface in R3 .

• Relate the orientation of a parametric surface in R3 to the normal vector.

5.5.1 Orientable and non-orientable surfaces

In Section 5.2 we defined the orientation of Rn and of a closed bounded region therein. We
now want to define the orientation of a surface in Rn . We will focus on a surface in R3 in this
section, since it is easier to visualize.
Recall from Definition 5.2.1 that the orientation of the vector space Rn is given by a choice
of “twirl”, specified by a choice of ordered basis on Rn . For R2 , this amounts to specifying
a direction of rotation: either counterclockwise or clockwise. The canonical orientation is
counterclockwise. The orientation of a closed, bounded, and simply connected region D µ Rn
is the orientation induced by a choice of ordered basis on the ambient space Rn .
Now suppose that – : D æ R3 is a parametric surface, with image surface S = –(D) µ R3 .
We would like to define an orientation on S in a way that is similar to what we did for the
plane (or for D µ R2 ) — we want to define a “direction of rotation” on the surface S. But
this is not so obvious anymore, since S is a like a rubber sheet that can be bent, stretched,
deformed, wrapped, what not. So what do we mean by direction of rotation?
The concept of tangent plane introduced in Definition 5.4.8 comes to the rescue. At each
point p œ S (not on its boundary), the tangent plane Tp S is a two-dimensional vector space,
i.e. R2 . So we can naturally define a direction of rotation on the tangent plane at p œ S by
specifying an ordered basis {Tu , Tv }. Then, we can define the orientation of the surface S by
assigning an ordered basis to all tangent planes in a continuous manner.
When the surface is in R3 , there is an equivalent way of defining a direction of rotation on
the tangent planes. Instead of specifying an ordered basis for the tangent planes Tp S, we can
specify a normal vector, as in Definition 5.4.9. If Tu and Tv form an ordered basis for the
tanget plane Tp S at a point p œ S, then we define the normal vector

n = Tu ◊ Tv .

The ordering is important here, since this is what gives the direction of rotation. Changing
from a direction of rotation to the opposite one amounts to exchanging the order of the two
basis vectors, which, in turn, sends n to ≠n. So what matters here is the direction of the
normal vector: it either points in one direction or the opposite direction, and this defines the
two choices of orientation on the tangent plane at this point.
So we can think of an orientation of a surface as an assignment of a normal vector at
all points on S in a continuous manner. Since what matters is the direction of the normal
vector, choosing an assignment of a normal vector at all points on S is basically the same as
choosing a side for the surface S. Which raises the following question: are all surfaces in
R3 orientable? That is, do all surfaces in R3 have two sides?
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 184

Perhaps surprisingly, the answer is no! Not all surfaces are orientable. There exists
surfaces that only have one side! We will come back to this in a second. Let us now define
more carefully the notion of an “orientable surface”.
Definition 5.5.1 Orientable surfaces and orientation. Let S µ R3 be a surface. We
say that S is orientable if there exists a continuous function n : S æ R3 which assigns to all
points p œ S (not on the boundary) the normal vector to the tangent plane Tp S. We say that
it is non-orientable otherwise.
If S is orientable, then an orientation is a choice of continuous function n : S æ R3 . It
assigns to all points on S a normal vector, which basically specifies one side of the surface (and
also determines a direction of rotation on the tangent planes). Saying that the assignment of
the normal vector is continuous amounts to saying that you pick the same side all around the
surface — you do not suddenly jump from one side to the other. ⌃
Most surfaces that we encounter in real life, such as spheres, cylinders, planes, tori, etc.
are orientable. But the prototypical example of a non-orientable surface is the Möbius strip.
This is a strange surface, shown in Figure 5.5.2. (It can be constructed by taking a long
rectangle of paper, giving it one half-twist, and then taping back the ends together.) If you
start at a point on the surface, you can pick a side, which amounts to defining a normal vector.
Then you can move around the surface, always picking the same side. But at some point you
will be back at the point you started with, but on the other side! (Try it!) So the assignment
of normal vectors is not continuous, since you if you stopped just before the point you started
with, the normal vector would suddenly have to jump from one side to the other.
What is going on? The point is that, on the Möbius strip, by doing this assignment of
normal vectors, you end up labeling both sides of the strip. The reason is: the strip really
only has one side! Crazy.

Figure 5.5.2 The Möbius strip (By David Benbennick - Own work, CC BY-SA 3.0, https://
commons.wikimedia.org/w/index.php?curid=50359).
The Möbius strip is very cool, but from now on we will always assume that our surfaces
are orientable.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 185

5.5.2 Orientation of a parametric surface

Now that we know how to define the orientation of a surface in R3 , we can go back to our
parametric surfaces. Are parametric surfaces naturally oriented? The answer is yes, as long
as the image surface is assumed to be orientable (as we could, for instance, realize the Möbius
strip as a parametric surface).
Lemma 5.5.3 Parametric surfaces are oriented. Let

– : D æR3
(u, v) ‘æ–(u, v) = (x(u, v), y(u, v), z(u, v))

be a parametric surface, and assume that the image surface S = –(D) µ R3 is orientable. The
normal vector
n = Tu ◊ Tv ,
with Tu and Tv the two tangent vectors from Definition 5.4.8, naturally induces an orientation
on S.
Note that the order is important here: we must take the cross-product Tu ◊ Tv with the
two vectors in the same order as the coordinates (u, v) on D µ R2 .
Proof. There isn’t much to prove here. We showed in Definition 5.4.8 and Definition 5.4.9
that the normal vector was well defined (and non-zero) for all points that are not in the image
of the boundary of D. So this gives a continuous assignment of a normal vector, and since we
assume that S is orientable, it induces an orientation on S. ⌅
Remark 5.5.4 As for parametric curves, there is another way of thinking about this statement.
We can think of the parametrization – : D æ R3 as not only mapping the region D, but as
also mapping its orientation. When we define parametric surfaces, we always think of the
domain D as being given the canonical orientation (counterclockwise), which is induced by
the canonical basis {e1 , e2 } on R2 , with e1 = (1, 0), e2 = (0, 1). The vector e1 points in the
u-direction, and hence is mapped to Tu , while the vector e2 points in the v-direction and is
mapped to Tv . So the counterclockwise orientation on D induces the orientation given by the
ordered basis {Tu , Tv } on the tangent planes.
Finally, in Definition 5.2.9 we discussed the induced orientation on the boundary curve
ˆD of a closed bounded region D. If D has canonical orientation, the induced orientation of
the boundary is the direction of motion if you walk along the boundary curve keeping the
region on your left. Similarly, a parametric surface induces an orientation on the boundary
ˆS of the image surface S. Since for parametric surfaces we always start with the canonical
orientation on D, we expect the induced orientation on the boundary of the image surface to
be defined similarly, by walking along the boundary curve keeping the surface on your left.
However, things are more subtle here, since the surface is in R3 , so it’s not immediately clear
what it means to walk along the boundary keeping the surface your left... In which direction
should your head be pointing?
For a region D µ R2 , we of course assumed that you were standing up. One way to think
about this is that you had your head in the positive z-direction (if you assume that D is in
the xy-plane embedded in R3 ). We now know that we can think of an orientation as a choice
of normal vector, and the canonical orientation on the xy-plane corresponds to the normal
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 186

vector pointing in the positive z-direction. In other words, we implicitly assumed that your
head was pointing in the direction of the normal vector (or that you were walking on the side
chosen by the orientation). This now generalizes to surfaces in R3 .
Definition 5.5.5 Induced orientation on the boundary of a parametric surface. Let
– : D æ R3 be a parametric surface, with image surface S = –(D) orientable and oriented by
the parametrization. Let ˆS be the boundary (the edges) of the image surface S. The induced
orientation on the boundary curve ˆS is the direction of travel if you keep the region on the
left, with your head in the direction of the normal vector n = Tu ◊ Tv . We denote by ˆ– the
boundary of the image surface with its orientation induced by the parametrization –. ⌃
Example 5.5.6 Upper half-sphere. Let us realize the upper half-sphere of radius R as a
parametric surface, and study its orientation and the induced orientation on its boundary.
The upper half-sphere has equation

x2 + y 2 + z 2 = R2 ,

keeping only the points with z Ø 0. There are many ways that we can parametrize it. For
instance, we can use either Cartesian or spherical coordinates. Let us do it in spherical
coordinates, and leave the Cartesian coordinates parametrization for Exercise 5.5.4.1.
We recall from Example 5.4.5 the parametrization of the sphere of radius R. A parametriza-
tion of the upper half-sphere is obtained in the same way, but restricting the inclination angle
◊ from 0 to fi/2. We get the parametric surface – : D æ R3 , with
5 6
fi
D = {(◊, „) œ R2 | ◊ œ 0, , „ œ [0, 2fi]},
2
and
–(◊, „) = (R sin(◊) cos(„), R sin(◊) sin(„), R cos(◊)).
Now suppose that D has canonical orientation, as always for parametric surfaces. What is
the induced orientation on the upper half-sphere? We calculate the tangent vectors:

T◊ = (R cos(◊) cos(„), R cos(◊) sin(„), ≠R sin(◊))

T„ = (≠R sin(◊) sin(„), R sin(◊) cos(„), 0) .

The normal vector is then:

1 2
n = T◊ ◊ T„ = R2 sin2 (◊) cos(„), sin2 (◊) sin(„), sin(◊) cos(◊) .

The main question is the direction of the normal vector: does it point inwards (towards the
center of the sphere), or outwards? In other words, is the orientation selecting the inner
surface of the upper half-sphere or the outer surface? To find out, we only need to look at
what happens at a given point on the upper half-sphere (we need to pick a point that
!fi "
is not in
the image of the boundary of D). Let’s pick the point with parameters (◊, „) = 4 , 0 . This
Ô
2
is the point 2 R (1, 0, 1). At this point, the normal vector is:
3 4
fi R2
n ,0 = (1, 0, 1) .
4 2
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 187

We thus see that the normal vector points outwards in the radial direction, i.e. away from the
origin. Therefore, the induced orientation on the upper half-sphere it outwards, i.e. the side
of the surface selected is the outer surface.
What about the induced orientation on the boundary? The boundary of the upper half-
sphere is the circle in the xy-plane with radius R (that’s the edge of the upper half-sphere).
! "
In our parametrization, it corresponds to the points with parameters (◊, „) = fi2 , „ for
„ œ [0, 2fi]. The induced orientation should correspond to the direction of motion if we walk
along the circle, with our head in the direction of the normal vector, keeping the surface on
your left. As the normal vector point outwards, we are walking on the circle with our head
pointing away from the origin. If we keep the surface on our left, we end up walking along
the circle counterclockwise in the xy-plane. This is the induced orientation on the boundary
circle. ⇤

5.5.3 Orientation-preserving reparametrizations of a surface

Just as for parametric curves, there are many ways to parametrize a given surface S µ R3 . If
we have a parametric surface – : D æ R3 , with
–(u, v) = (x(u, v), y(u, v), z(u, v)),
and we think of u, v as functions of new variables s, t, then by doing this change of variable we
may end up with a new parametrization of the same image surface. The question is whether
this new parametrization induces the same orientation on the image surface or the opposite
orientation.
Lemma 5.5.7 Orientation-preserving reparametrizations. Let – : D1 æ R3 be a para-
metric surface, with the image surface S = –(C) orientable, and –(u, v) = (x(u, v), y(u, v), z(u, v)).
Let D2 µ R2 be another closed, bounded, and simply-connected region. Let „ : D2 æ D1 be
a bijective and invertible function that can be extended to a C 1 -function on an open subset
U ™ R2 that contains D2 , as in Definition 5.3.6. Then the pullback

„ú – : D2 æ R3
(s, t) ‘æ („ú x(s, t), „ú y(s, t), „ú z(s, t)) = (x(„(s, t)), y(„(s, t)), z(„(s, t)))

is another parametrization of the same image surface S.

Furthermore, if det J„ > 0, the induced orientation on S is the same for both parametriza-
tions – and „ú –, and we say that the reparametrization is orientation-preserving. If
det J„ < 0, the two parametrizations – and „ú – have opposite orientations, and the reparametriza-
tion is orientation-reversing.
Proof. First, it is clear that –(D1 ) = „ú –(D2 ), i.e. the image surfaces are the same, since we
are simply composing maps. But we need to show that „ú – is a parametric surface, according
to Definition 5.4.1.
Property 1 is clearly satisfied for „ú – since „ is assumed to be C 1 . Property 3 is also
satisfied, since „ is bijective. As for Property 2, let us denote the tangent vectors to „ú – by
Vs and Vt . By definition,
3 4
ˆ ˆx(„(s, t)) ˆy(„(s, t)) ˆz(„(s, t))
Vs = „ú –(s, t) = , , ,
ˆs ˆs ˆs ˆs
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 188
3 4
ˆ ˆx(„(s, t)) ˆy(„(s, t)) ˆz(„(s, t))
Vt = „ú –(s, t) = , , .
ˆt ˆt ˆt ˆt

If we denote „(s, t) = (u(s, t), v(s, t)), using the chain rule, we get:
3 4 3 4
ˆu ˆx(u, v) ˆy(u, v) ˆz(u, v) ˆv ˆx(u, v) ˆy(u, v) ˆz(u, v)
Vs = , , + , ,
ˆs ˆu ˆu ˆu ˆs ˆv ˆv ˆv
ˆu ˆv
= Tu + Tv ,
ˆs ˆs
and, similarly,
3 4 3 4
ˆu ˆx(u, v) ˆy(u, v) ˆz(u, v) ˆv ˆx(u, v) ˆy(u, v) ˆz(u, v)
Vt = , , + , ,
ˆt ˆu ˆu ˆu ˆt ˆv ˆv ˆv
ˆu ˆv
= Tu + Tv .
ˆt ˆt
Calculating the cross-product, we get:
3 4
ˆu ˆv ˆv ˆu
Vs ◊ Vt = ≠ Tu ◊ Tv
ˆs ˆt ˆs ˆt
=(det J„ )Tu ◊ Tv .

Therefore, if Tu and Tv are linearly independent, then Tu ◊ Tv ”= 0, and since „ is invertible,

det J„ ”= 0. Therefore Vs ◊ Vt ”= 0, and Property 2 is satisfied.
From this calculation we can also relate the induced orientations of the two parametrization
– and „ú –. If we denote by n the normal vector for –, and m the normal vector for „ú –,
from the calculation above we get:

m = (det J„ )n.

Thus, if det J„ > 0, then the normal vectors of both parametrizations have the same sign (they
pick the same side for the image surface) and the induced orientations are the same, while if
det J„ < 0, they have opposite signs (they pick opposite sides) and the induced orientations
are opposite. ⌅

5.5.4 Exercises
1. Redo the parametrization of the upper half-sphere of radius R, as in Example 5.5.6, but
in Cartesian coordinates. (The region D in this case should be a disk of radius R.) Find
the induced orientation on the image surface by calculating the normal vector.
Solution. The upper half-sphere of radius R is the surface in R3 defined by the equation

x2 + y 2 + z 2 = R2

with z Ø 0. Since z Ø 0, we can solve explicitly for z. We get:

Ò
z= R2 ≠ x2 ≠ y 2 .
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 189

This gives us another way

 of parametrizing the surface directly, since it is the graph
of a function f (x, y) = R2 ≠ x2 ≠ y 2 . We proceed as in Example 5.4.4. We find a
parametrization – : D æ R3 with

–(u, v) = (u, v, R2 ≠ u2 ≠ v 2 ).

However, we need to specify D. The region should be the disk that is at the bottom
of the upper half-sphere, which is the disk of radius R centered at the origin. So the
region D in the (u, v)-plane should be given by u2 + v 2 Æ R2 . We can rewrite this as a
u-supported region:
 
D = {(u, v) œ R2 | u œ [≠R, R], ≠ R2 ≠ u2 Æ v Æ R 2 ≠ u2 .

To find the induced orientation, we need to calculate the normal vector. We first find
the tangent vectors:
3 4
ˆ– u
Tu = = 1, 0, ≠ Ô ,
ˆu R 2 ≠ u2 ≠ v 2
3 4
ˆ– v
Tv = = 0, 1, ≠ Ô .
ˆv R 2 ≠ u2 ≠ v 2
The normal vector is given by the cross-product:

n =Tu ◊ Tv
3 4
u v
= Ô ,Ô ,1 .
2 2
R ≠u ≠v 2 R ≠ u2 ≠ v 2
2

We need to determine in which direction it points (outwards of the sphere, or inwards),

to determine the induced orientation. Let’s pick the point on the sphere with parameters
(u, v) = (0, 0). This is the point –(0, 0) = (0, 0, R), which is the north pole. The normal
vector at this point is n(0, 0) = (0, 0, 1). It points upwards, that is, in the outwards
direction (away from the origin, which is the center of the sphere). Therefore, this
parametrization induces the orientation on the upper half-sphere given by a normal
vector pointing outwards (it picks the outside of the surface).
We note that this is the same induced orientation as in Example 5.5.6. Indeed, we
can relate the two parametrizations by the change of variables

(u, v) = (R sin(◊) cos(„), R sin(◊) sin(„)).

The determinant of the Jacobian of the transformation is

A ˆu B A B
R cos(◊) cos(„) ≠R sin(◊) sin(„)
ˆu
det ˆ◊ ˆ„
= det = R sin(◊) cos(◊),
ˆv
ˆ◊
ˆv
ˆ„ R cos(◊) sin(„) R sin(◊) cos(„)

which is positive for ◊ œ [0, fi/2]. Therefore, the two parametrizations should induce the
same orientation, as we found.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 190

2. Consider the surface S consisting of the part of the plane z + x + y = 2 that lies inside
the cylinder x2 + y 2 = 1. Find two parametrizations of S that have opposite orientation.
Solution. First, we can parametrize the plane by –(u, v) = (u, v, 2 ≠ u ≠ v). Now we
need to find a region D in the (u, v)-plane such that –(D) is the part of the plane that
lies inside the cylinder x2 + y 2 = 1. We see that if we take the region D to be the disk
u2 + v 2 = 1, then the boundary circle of D is mapped to the boundary of the surface S,
that is, the closed curve at the intersection of the plane and the cylinder. So this is the
region D that we are looking for. We can realize it as a u-supported region. The final
parametrization would be – : D æ R3 with
 
D = {(u, v) œ R2 | u œ [≠1, 1], ≠ 1 ≠ u2 Æ v Æ 1 ≠ u2 }

and
–(u, v) = (u, v, 2 ≠ u ≠ v).
What is the induced orientation? The tangent vectors are:

Tu = (1, 0, ≠1), Tv = (0, 1, ≠1).

The normal vector is

n = Tu ◊ Tv = (1, 1, 1),
which defines the induced orientation on the planar region S. (Note that it is a constant
vector since the region S is planar (i.e. part of a plane), and so the normal direction is
the same everywhere on S.)
How can we find a second parametrization that has the opposite orientation? Well,
there are many possibilities. But one easy way to get the normal vector to have opposite
sign is to exchange the order of the two tangent vectors, i.e. exchange the variables u
and v. In other words, we could define the parametrization –2 : D æ R3 with the same
region D but
–2 (u, v) = (v, u, 2 ≠ u ≠ v).
(All we did is interchange u and v -- the image surface is obviously the same.) Then the
tangent vectors are
Tu = (0, 1, ≠1), Tv = (1, 0, ≠1),
and the normal vector is

n = Tu ◊ Tv = (≠1, ≠1, ≠1),

which points in the opposite direction, and hence this parametrization induces the
opposite orientation.
3. We have seen in Example 5.4.4 how we can realize the graph of a function f (x, y) as a
parametric surface in R3 .

(a) Show that the natural parametrization of Example 5.4.4 always induces an orien-
tation on the graph of the function with normal vector pointing in the positive
z-direction.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 191

(b) Find another parametrization of the graph of the function that induces the opposite
orientation.

Solution. (a) Let S µ R3 be the surface given by the graph z = f (x, y) of a function
f , for some region D in the (x, y)-plane. We parametrize it as – : D æ R3 with

–(u, v) = (u, v, f (u, v)),

where D is the same region D but in the (u, v)-plane. To calculate the normal vector,
we calculate the tangent vectors:
3 4 3 4
ˆf ˆf
Tu = 1, 0, , Tv = 0, 1, .
ˆu ˆv
The normal vector is given by the cross-product:

n =Tu ◊ Tv
Q R
i j k
c d
= det a1 0 ˆf ˆu b
0 1 ˆfˆv
3 4
ˆf ˆf
= ≠ ,≠ ,1 .
ˆu ˆv
In particular, we see that it always points in the positive z-direction, regardless of the
details of the function f .
(b) We can do just as in the previous question; we can exchange u and v. We can
write a new parametrization as –2 : D2 æ R3 , with D2 being the same region as D but
with u and v exchanged, and

–(u, v) = (v, u, f (v, u)).

Then the tangent vectors are

3 4 3 4
ˆf (v, u) ˆf (v, u)
Tu = 0, 1, , Tv = 1, 0, ,
ˆu ˆv
and the normal vector is
3 4
ˆf (v, u) ˆf (v, u)
n = Tu ◊ Tv = , , ≠1 ,
ˆv ˆu

which is the same vector (with u and v exchanged) but with the opposite sign. This
parametrization therefore induces the opposite orientation to the parametrization in (a).

4. Consider the surface S consisting of the part of the cone z = x2 + y 2 between z = 1
and z = 5, with normal vector pointing outside the cone. The boundary ˆS of S consists
in two separate oriented closed curves. Realize these two curves as parametric curves,
with orientations that are consistent with the induced orientation on the boundary ˆS.
Solution. The cone is shown in the figure below:
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 192


Figure 5.5.8 The part of the cone z = x2 + y 2 between z = 1 and z = 5.
The boundary has two components: the top boundary circle and the bottom boundary
circle. If we call C1 the top boundary circle, and C2 the bottom one, then ˆS = C1 fi C2 .
These two curves are given by the curves:

C1 ={(x, y, z) œ R3 | x2 + y 2 = 25, z = 5},

C2 ={(x, y, z) œ R3 | x2 + y 2 = 1, z = 1}.

What is the induced orientation on these boundary curves? The surface of the cone is
oriented with an outwards pointing normal vector. To get the induced orientation on
the boundary curves, we should walk along the curves, keeping the region on our left,
with our head pointing in the direction of the normal vector. We see that C1 should be
oriented clockwise, while C2 should be oriented counterclockwise. We now want to find
parametrization of these two curves that are consistent with the induced orientations.
We start with C1 . This is a circle of radius 5 in the plane z = 5, oriented clockwise.
We can parametrize the circle using polar coordinates, but we want to use sine for x and
cosine for y so that the resulting curve is oriented clockwise. We get the parametrization
–1 : [0, 2fi] æ R3 with
–1 (◊) = (5 sin(◊), 5 cos(◊), 5).
As for C2 , it is the circle of radius 1 in the plane z = 1, oriented counterclockwise.
We use polar coordinates again, but cosine for x and sine for y so that it is oriented
counterclockwise. We get –2 : [0, 2fi] æ R3 with:

–2 (◊) = (cos(◊), sin(◊), 1).

5.6 Surface integrals

Step 4: we define integration of a two-form along a parametric surface via pullback. We call
those “surface integrals”. Step 5: we show that the integral is well-defined: it is invariant
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 193

under orientation-preserving reparametrizations, and changes sign under orientation-reversing

reparametrizations. This ensures that the integral of the two-form is intrinsically defined in
terms of the geometry of the image surface with its induced orientation. We also rewrite
surface integrals in terms of vector fields.

Objectives
You should be able to:

• Define the surface integral of a two-form along a parametric surface in R3 , and evaluate
it.

• Rewrite the definition of surface integrals as flux integrals for the associated vector field.

• Show that surface integrals are invariant under orientation-preserving reparametrizations

of the surface.

• Show that surface integrals change sign under reparametrizations of the surface that
reverse its orientation.

5.6.1 The definition of surface integrals

We define the integral of a two-form along a parametric surface: we pull back to the region D
and integrate.
Definition 5.6.1 Surface integrals. Let Ê be a two-form on an open subset U ™ R3 . Let
– : D æ R3 be a parametric surface whose image surface S = –(D) µ U is orientable. We
define the surface integral of Ê along – as:
⁄ ⁄
Ê= –ú Ê,
– D

where the integral on the right-hand-side is defined in Definition 5.3.1, with D given the
canonical orientation (which is consistent with the induced orientation on S). ⌃
What is neat is that by Definition 5.3.1, the integral on the right-hand-side can be written
as a standard double integral from calculus. So using the pullback we reduced the evaluation
of integrals of two-forms over surfaces to standard double integrals!
Example 5.6.2 An example of a surface integral. Consider the two-form Ê =
x dy · dz + z dx · dy. Evaluate its surface integral along the oriented parametric surface
– : D æ R3 with D = {(u, v) œ R2 | u œ [0, 1], v œ [0, 2]} and –(u, v) = (u, uv 2 , v).
We calculate the pullback of the two-form:
1 2 1 2
–ú Ê =u v 2 du + 2uv dv · dv + vdu · v 2 du + 2uv dv
=(uv 2 + 2uv 2 )du · dv
=3uv 2 du · dv.
The surface integral then becomes:
⁄ ⁄
Ê= 3uv 2 du · dv
– D
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 194
⁄ 2⁄ 1
=3 uv 2 dudv
0 0
⁄ 2 C D1
2 u2
=3 v dv
0 2 0
⁄
3 2 2
= v dv
2 0
v 3 --2
= -
2 0
=4.

Note that it is important here that we wrote the two-form –ú Ê in terms of the basic two-form
du · dv (not dv · du) before rewriting as a double integral, as the orientation on D is the
canonical orientation with respect to the coordinates (u, v) on D µ R2 . ⇤

5.6.2 Reparametrization-invariance and orientability of surface integrals

We defined surface integrals in terms of parametric surfaces: the parametrization was key,
as we used it to pullback to D. But in the end, we would like our integral to be defined
solely in terms of the image surface S itself with its orientation. If we use two different
parametrizations that describe the same surface with the same induced orientation, we want
the integral to be the same. The integral should not depend on how we describe the surface,
as long as we preserve the induced orientation.
Lemma 5.6.3 Surface integrals are invariant under orientation-preserving reparametriza-
tions. Let Ê be a two-form on an open subset U ™ R3 . Let – : D1 æ R3 be a parametric
surface whose image surface S = –(D1 ) µ U is orientable. Let „ : D2 æ D1 be as in
Lemma 5.5.7, so that „ú – : D2 æ R3 is another parametrization of the same image surface.

1. If „ is orientation-preserving (that is, det J„ > 0), then

⁄ ⁄
Ê= Ê.
– „ú –

2. If „ is orientation-reversing (that is, det J„ < 0), then

⁄ ⁄
Ê=≠ Ê.
– „ú –

Proof. Let us rewrite the two surface integrals using the pullback. First,
⁄ ⁄
Ê= –ú Ê.
– D1

Second, ⁄ ⁄
Ê= (– ¶ „)ú Ê.
„ú – D2
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 195

From Exercise 4.7.5.7, we know that we can pullback through a chain of maps D2 æ D1 æ R3
„ –

in two different ways, but it gives the same thing: (– ¶ „)ú Ê = „ú (–ú Ê). Therefore,
⁄ ⁄
Ê= „ú (–ú Ê).
„ú – D2

In the end, what we need to compare is two integrals over regions in R2 . But this is precisely
the result of Lemma 5.3.7, for the two-form –ú Ê on D1 . This lemma states that if det J„ > 0,
then ⁄ ⁄
–ú Ê = „ú (–ú Ê),
D1 D2

while if det J„ < 0, ⁄ ⁄

–ú Ê = ≠ „ú (–ú Ê),
D1 D2

We thus conclude that if the reparametrization is orientation-preserving,

⁄ ⁄
Ê= Ê,
– „ú –

while if it is orientation-reversing,
⁄ ⁄
Ê=≠ Ê.
– „ú –

⌅
Surface integrals are oriented and reparametrization-invariant, as we want. Nice! As a
result, while we use a parametrization to define a surface integral, the integral can really be
thought of as being defined intrinsically in terms of the surface S and its orientation.

5.6.3 Surface integrals in terms of vector fields

Let us now translate our definition in surface integrals in terms of vector fields, using the
dictionary between differential forms and vector calculus concepts that we established.
We first need to do a bit of work to rephrase the pullback of a two-form along a parametric
surface in terms of associated vector fields.
Lemma 5.6.4 The pullback of a two-form along a parametric surface in terms of
vector fields. Let Ê = f dy · dz + g dz · dx + h dx · dy be a two-form on U ™ R3 , with
associated vector field F = (f, g, h). Let – : D æ R3 be a parametric surface whose image
surface S = –(D) µ U is orientable, with –(u, v) = (x(u, v), y(u, v), z(u, v)). Let Tu , Tv be
the tangent vectors, and n = Tu ◊ Tv be the normal vector. Then the pullback –ú Ê is:

–ú Ê = (F(–(u, v)) · n) du · dv.

Proof. This is just a calculation:
3 4 3 4
ˆy ˆy ˆz ˆz
– Ê =f (–(u, v))
ú
du + dv · du + dv
ˆu ˆv ˆu ˆv
3 4 3 4
ˆz ˆz ˆx ˆx
+ g(–(u, v)) du + dv · du + dv
ˆu ˆv ˆu ˆv
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 196
3 4 3 4
ˆx ˆx ˆy ˆy
+ h(–(u, v)) du + dv · du + dv
ˆu ˆv ˆu ˆv
3 4
ˆy ˆz ˆy ˆz
=f (–(u, v)) ≠ du · dv
ˆu ˆv ˆv ˆu
3 4
ˆz ˆx ˆz ˆx
+ g(–(u, v)) ≠ du · dv
ˆu ˆv ˆv ˆu
3 4
ˆx ˆy ˆx ˆy
+ h(–(u, v)) ≠ du · dv
ˆu ˆv ˆv ˆu
=F(–(u, v)) · (Tu ◊ Tv ) du · dv.

⌅
It is interesting to remark here that while pulling back a one-form along a parametric
curve picked the tangential component of the vector field along the curve, pulling back a
two-form along a parametric surface picks the normal component of the vector field. This
has an interpretation in physics, as we will see. Just as line integrals were related to the
calculation of work (and hence the tangential component of the force field to the direction of
motion was the relevant one), surface integrals are related to the calculation of flux, for which
the normal component of the force field is the relevant one.
With this result, we can rewrite surface integrals directly as double integrals:
Corollary 5.6.5 Let Ê be a two-form on an open subset U ™ R3 , with associated vector field
F. Let – : D æ R3 be a parametric surface whose image surface S = –(D) µ U is orientable,
with –(u, v) = (x(u, v), y(u, v), z(u, v)). Then
⁄ ⁄⁄
Ê= (F(–(u, v)) · n) dA,
– D

where on the right-hand-side this is a double integral over the region D in the uv-plane.
Proof. This follows directly from Lemma 5.6.4, the definition of surface integrals Definition 5.6.1,
and the definition of integrals over region in R2 Definition 5.3.1. We have:
⁄ ⁄
Ê= –ú Ê
– ⁄D
= (F(–(u, v)) · n) du · dv
D
⁄⁄
= (F(–(u, v)) · n) dA.
D

⌅
This is how surface integrals are generally defined in standard vector calculus textbooks.
Remark 5.6.6 Sometimes, the following shorthand notation is used:
⁄⁄ ⁄⁄
F · dS := (F(–(u, v)) · n) dA.
S D

Such integrals are also called flux integrals, because of the physics interpretation, which
we will study in Section 5.9. The integral calculates the flux of the vector field F across the
surface S in the direction of the normal vector n specified by the orientation.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 197

Example 5.6.7 An example of a surface integral of a vector field. Calculate the

surface integral of the vector field
1
F= (x, y, 0)
x2 + y 2 + z 2
over the upper half-sphere of radius one, with the inwards orientation.
We can use our parametrization from Example 5.5.6 with R = 1, that is, – : D æ R3 , with
5 6
fi
D = {(◊, „) œ R2 | ◊ œ 0, , „ œ [0, 2fi]},
2
and
–(◊, „) = (sin(◊) cos(„), sin(◊) sin(„), cos(◊)).
We showed in that example that the normal vector points outwards, so the orientation on the
upper half-sphere induced by this parametrization is opposite to what is asked in the problem.
Therefore, we will need to take minus the integral over the parametric surface –.
We calculated in Example 5.5.6 the normal vector:
1 2
n = sin2 (◊) cos(„), sin2 (◊) sin(„), sin(◊) cos(◊) .

The integrand is then (noting that x2 + y 2 + z 2 = 1 on the surface):

1 2
F(–(◊, „)) · n = (sin(◊) cos(„), sin(◊) sin(„), 0) · sin2 (◊) cos(„), sin2 (◊) sin(„), sin(◊) cos(◊)
= sin3 (◊) cos2 („) + sin3 (◊) sin2 („)
= sin3 (◊).

The surface integral then becomes (recall that we need to add a minus sign since the surface
integral is with respect to the inwards orientation, while our parametrization induces the
outwards orientation):
⁄⁄ ⁄ 2fi ⁄ fi
2
F · dS = ≠ sin3 (◊)d◊d„
S 0 0
⁄ 2fi 5 6fi
1 2
=≠ ≠ cos(◊) + cos3 (◊) d„
0 3 0
⁄
2 2fi
=≠ d„
3 0
4fi
=≠ .
3
⇤
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 198

5.6.4 Exercises
1. Find the surface integral of the two-form

Ê = y dy · dz ≠ (x + y) dz · dx

along the lower half-sphere x2 + y 2 + z 2 = 4, z Æ 0, with orientation given by an inwards

pointing normal vector.
Solution. We use spherical coordinates to parametrize the lower half sphere of radius
2. That is, – : D æ R3 with
5 6
fi
D = {(◊, „) œ R2 | ◊ œ , fi , „ œ [0, 2fi]},
2
and
–(◊, „) = (2 sin(◊) cos(„), 2 sin(◊) sin(„), 2 cos(◊)).
Here the inclination angle was restricted to be between fi/2 and fi, which amounts to
considering the lower half-sphere with z Æ 0. We have already calculated in Example 5.5.6
the this parametrization induces a normal vector pointing outwards. The question is
asking us to pick the opposite orientation, with normal vector pointing inwards. So we
will have to add an overall minus sign to our surface integral.
To evaluate the surface integral we first need to calculate the pullback –ú Ê. We get:

–ú Ê =2 sin(◊) sin(„)(2 cos(◊) sin(„)d◊ + 2 sin(◊) cos(„)d„)) · (≠2 sin(◊)d◊)

≠ (2 sin(◊) cos(„) + 2 sin(◊) sin(„))(≠2 sin(◊)d◊) · (2 cos(◊) cos(„)d◊ ≠ 2 sin(◊) sin(„)d„)
= ≠ 8 sin3 (◊) sin(„) cos(„)d„ · d◊ ≠ 8 sin3 (◊) sin(„)(cos(„) + sin(„))d◊ · d„
= ≠ 8 sin3 (◊) sin2 („)d◊ · d„.

Then we integrate, recalling that we have to add a minus sign because of the orientation:
⁄ ⁄
≠ Ê =≠ –ú Ê
– D
⁄ fi ⁄ 2fi
=8 sin3 (◊) sin2 („)d„d◊
fi/2 0
⁄ fi
=8fi sin3 (◊)d◊
fi/2
16fi
= ,
3
where I used the fact that
⁄ 2fi ⁄ fi
2
sin2 („)d„ = fi, sin3 (◊)d◊ = ,
0 fi/2 3

which you can check independently using trigonometric identities to evaluate the definite
integrals.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 199

2. Find the flux (the surface integral) of the vector field

F(x, y, z) = (zexy , ≠3zexy , 2xy)

along the parallelogram realized as the parametric surface – : D æ R3 with D = {(u, v) œ

R2 | u œ [0, 2], v œ [0, 1]} and

–(u, v) = (u + v, u ≠ v, 1 + 2u + v).

Solution. Let us solve this problem using the vector field approach. We know the
parametric surface; we need to calculate the normal vector. The tangent vectors are

Tu = (1, 1, 2), Tv = (1, ≠1, 1).

(We note that those are constant vectors, since the image surface is planar.) The normal
vector is then

n =Tu ◊ Tv
Q R
i j k
c d
= det a1 1 2 b
1 ≠1 1
=(3, 1, ≠2).

We then calculate the surface integral:

⁄⁄ ⁄ 2⁄ 1
2 ≠v 2 2 ≠v 2
F · dS = ((1 + 2u + v)eu , ≠3(1 + 2u + v)eu , 2(u2 ≠ v 2 )) · (3, 1, ≠2) dvdu
D 0 0
⁄ 2⁄ 11 2
= ≠4(u2 ≠ v 2 ) dvdu
0 0
⁄ 2C Dv=1
2 v3
=≠4 u v≠ du
0 3 v=0
⁄ 23 4
1
=≠4 u2 ≠ du
0 3
C Du=2
u3 u
=≠4 ≠
3 3 u=0
= ≠ 8.
3. Find the surface integral of the two-form

Ê = x dy · dz + z dx · dy

over the surface of the cube with vertices (±1, ±1, ±1), with orientation given by a
normal vector pointing outwards.
Solution. The cube is shown in the figure below. It has six sides, which we consider as
separate parametric surfaces, Si with i = 1, . . . , 6.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 200

Figure 5.6.8 The surface S is the cube shown in the figure. We consider the six sides as
separate parametric surfaces.
All six sides can be parametrized easily, with the same domain D. We write the
parametrizations as –i : D æ R3 , for i = 1, . . . , 6, with
D = {(u, v) œ R2 | u œ [≠1, 1], v œ [≠1, 1]},
and
–1 (u, v) =(1, u, v)
–2 (u, v) =(≠1, u, v)
–3 (u, v) =(u, 1, v)
–4 (u, v) =(u, ≠1, v)
–5 (u, v) =(u, v, 1)
–6 (u, v) =(u, v, ≠1)
Calculating the normal vectors, we see that the normal vectors for –1 , –4 , –5 point
outwards, while the normal vectors for –2 , –3 , –6 point inwards. So we can write our
desired surface integral as
⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄
Ê= Ê≠ Ê≠ Ê+ Ê+ Ê≠ Ê.
S S1 S2 S3 S4 S5 S6

To calculate these surface integrals, we need the pullbacks. A straightforward calculation

gives:
–1ú Ê = –5ú Ê = du · dv,
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 201

–2ú Ê = –6ú Ê = ≠du · dv,

–3ú Ê = –4ú Ê = 0.

Therefore,
⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄
Ê= Ê≠ Ê≠ Ê+ Ê+ Ê≠ Ê
S S S2 S3 S4 S5 S6
⁄1
=4 du · dv
D
⁄ 1 ⁄ 1
=4 dudv
≠1 ≠1
=16.
4. Let S be the boundary of the region in R3 enclosed by the cylinder x2 + y 2 = 4, the
paraboloid z = x2 + y 2 , and the plane z = 9. Find the surface integral of the vector field

F(x, y, z) = (x2 , y 2 , z 2 )

along S with orientation given by a normal vector pointing inwards.

Solution. We see that the cylinder x2 + y 2 = 4 and the paraboloid z = x2 + y 2 intersect
along the circle x2 + y 2 = 4 in the plane z = 4. Thus the surface S has three components:
S1 is the part the paraboloid z = x2 + y 2 with 0 Æ z Æ 4; S2 is the part of the cylinder
x2 + y 2 = 4 with 4 Æ z Æ 9; and S3 is the disk x2 + y 2 Æ 4 in the plane z = 9. This is
shown in Figure 5.6.9.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 202

Figure 5.6.9 The surface S has three components: S1 is the part of the paraboloid (in
blue) with z Æ 4; S2 is the lateral surface of the cylinder (in orange) for 4 Æ z Æ 9; S3 is
the disk x2 + y 2 Æ 4 in the plane z = 9 (in green).
To evaluate the surface integral of Ê over S, we evaluate it over the three components
separately and then add up the integrals.
Let us consider S1 first, which is the part of the paraboloid z = x2 + y 2 with 0 Æ z Æ 4.
We realize it as the parametric surface –1 : D1 æ R3 with

D1 = {(u, ◊) œ R2 | u œ [0, 2], ◊ œ [0, 2fi]}

and
–1 (u, ◊) = (u cos(◊), u sin(◊), u2 ).
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 203

We need to make sure that the normal vector points inwards. The tangent vectors are

Tu = (cos(◊), sin(◊), 2u), T◊ = (≠u sin(◊), u cos(◊), 0).

The normal vector is

n = Tu ◊ T◊ = (≠2u2 cos(◊), ≠2u2 sin(◊), u).

As u Ø 0, we see that the normal vector points in the positive z-direction, which means
that it points inwards. So we’re good!
We then evaluate the surface integral:
⁄⁄ ⁄ 2 ⁄ 2fi 1 2
F · dS1 = (u2 cos2 (◊), u2 sin2 (◊), u4 ) · (≠2u2 cos(◊), ≠2u2 sin(◊), u) d◊du
S1 0 0
⁄ 2 ⁄ 2fi 1 2
= ≠2u4 (cos3 (◊) + sin3 (◊)) + u5 d◊du
0 0
⁄ 2
=2fi u5 du
0
64fi
= ,
3
s s
where we used the fact that 02fi cos3 (◊)d◊ = 02fi sin3 (◊)d◊ = 0.
Let us now consider S2 , which is the cylinder x2 + y 2 = 4, 4 Æ z Æ 9. We parametrize
it as –2 : D2 æ R3 with

D2 = {(u, ◊) œ R2 | u œ [4, 9], ◊ œ [0, 2fi]}

and
–2 (u, ◊) = (2 cos(◊), 2 sin(◊), u).
We look at the normal vector. The tangent vectors are

Tu = (0, 0, 1), T◊ = (≠2 sin(◊), 2 cos(◊), 0),

and the normal vector is

n = Tu ◊ T◊ = (≠2 cos(◊), ≠2 sin(◊), 0).

We thus see that at a point with coordinates (2 cos(◊), 2 sin(◊), u) on the cylinder, the
normal vector is (≠2 cos(◊), ≠2 sin(◊), 0), which points inwards. Good!
We evaluate the surface integral:
⁄⁄ ⁄ 9 ⁄ 2fi 1 2
F · dS2 = (4 cos2 (◊), 4 sin2 (◊), u2 ) · (≠2 cos(◊), ≠2 sin(◊), 0) d◊du
S2 4 0
⁄ 9 ⁄ 2fi 1 2
= ≠8 cos3 (◊) ≠ 8 sin3 (◊) d◊du
4 0
=0,
s 2fi s 2fi
since, as above, 0 cos3 (◊)d◊ = 0 sin3 (◊)d◊ = 0.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 204

Finally, we consider S3 , which is the disk x2 + y 2 Æ 4 in the plane z = 9. We

parametrize it as –3 : D3 æ R3 with

D3 = {(u, ◊) œ R2 | u œ [0, 2], ◊ œ [0, 2fi]}

and
–3 (u, ◊) = (u sin(◊), u cos(◊), 9).
The tangent vectors are

Tu = (sin(◊), cos(◊), 0), T◊ = (u cos(◊), ≠u sin(◊), 0),

and the normal vector is

n = Tu ◊ T◊ = (0, 0, ≠u).
As u Ø 0, this points in the negative z-direction, that is, it points inwards, as we want.
We evaluate the surface integral:
⁄⁄ ⁄ 2 ⁄ 2fi 1 2
F · dS3 = (u2 sin2 (◊), u2 cos2 (◊), 81) · (0, 0, ≠u) d◊du
S3 0 0
⁄ 2 ⁄ 2fi
= (≠81u)d◊du
0 0
⁄ 2
= ≠ 162fi udu,
0
= ≠ 324fi.

Finally, we compute the surface integral over S by adding up the three surface
integrals aboves. We get:
⁄⁄ ⁄⁄ ⁄⁄ ⁄⁄
F · dS = F · dS1 + F · dS2 + F · dS3
S S1 S2 S3
64fi
= + 0 ≠ 324fi
3
908fi
=≠ .
3
Done! :-)
5. An “inverse square field” is a vector field F that is inversely proportional to the square
of the distance from the origin. It is very important in physics, as it describes many
physical phenomena, such a gravity, electrostatics, etc. It can be written as
1
F(x, y, z) = C (x, y, z),
(x2 + y2+ z 2 )3/2

for some constant C œ R. Show that the flux (the surface integral) of F across a sphere
S centered at the origin (in the outwards direction) is independent of the radius of S.
This is one of the very nice properties of inverse square fields!
Solution. As usual we use spherical coordinates to parametrize the sphere of fixed
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 205

radius R, as – : D æ R3 , with

D = {(◊, „) œ R2 | ◊ œ [0, fi] , „ œ [0, 2fi]},

and
–(◊, „) = (R sin(◊) cos(„), R sin(◊) sin(„), R cos(◊)).
Let use differential forms to calculate the surface integral. To the vector field F we
associated the two-form
1
Ê=C (x dy · dz + y dz · dx + z dx · dy) .
(x2 + y2+ z 2 )3/2

To calculate the surface integral, we calculate the pullback, noting that x(◊, „)2 +y(◊, „)2 +
z(◊, „)2 = R2 :

C
–ú Ê = (R sin(◊) cos(„)(R cos(◊) sin(„)d◊ + R sin(◊) cos(„)d„) · (≠R sin(◊)d◊)
R3
+ R sin(◊) sin(„)(≠R sin(◊)d◊) · (R cos(◊) cos(„)d◊ ≠ R sin(◊) sin(„)d„)
+ R cos(◊)(R cos(◊) cos(„)d◊) ≠ R sin(◊) sin(„)d„) · (R cos(◊) sin(„)d◊ + R sin(◊) cos(„)d„))
CR3
= (sin3 (◊) cos2 („) + sin3 (◊) sin2 („) + cos2 (◊) sin(◊) cos2 („) + cos2 (◊) sin(◊) sin2 („))d◊ · d„
R3
=C sin(◊)d◊ · d„.

In particular, we see that it does not depend on the radius R of the sphere! The surface
integral is then
⁄ ⁄
Ê= –ú Ê
– D
⁄ 2fi ⁄ fi
=C sin(◊)d◊d„
0 0
⁄ 2fi
=2C d„
0
=4fiC,

which is indeed independent of the radius R.

5.7 Green’s theorem

Step 6: we study what happens if we integrate an exact two-form. We start by studying
integrals of exact two-forms over bounded regions in R2 : this gives rise to Green’s theorem.

Objectives
You should be able to:

• State Green’s theorem for integrals of exact two-forms over closed bounded regions in
R2 .
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 206

• Use Green’s theorem to evaluate integrals of exact two-forms over closed bounded regions
in R2 .

• Use Green’s theorem to evaluate line integrals of one-forms along simple closed curves
in R2 .

• Rephrase Green’s theorem in terms of the associated vector fields.

5.7.1 Green’s theorem

Let us start by recalling the corresponding statement for integrals of exact one-forms over
intervals in R. Let f be a zero-form on U µ R. Let [a, b] œ U with the canonical orientation
of increasing numbers. The boundary of the interval, with its induced orientation, is ˆ[a, b] =
{(b, +), (a, ≠)}. Then the integral of the exact one-form Ê = df along [a, b] can be evaluated
as: ⁄ ⁄
df = f = f (b) ≠ f (a).
[a,b] ˆ[a,b]
This is nothing else but the Fundamental Theorem of Calculus, rewritten in a fancy way using
one-forms and zero-forms.
We now want to do something similar for exact two-forms over closed bounded regions in
R2 .
Theorem 5.7.1 Green’s theorem. Let ÷ be a one-form on an open subset U ™ R2 . Let
D µ U be a closed bounded region with canonical orientation, and let ˆD be its boundary with
the induced orientation as in Definition 5.2.9. Then the integral of the exact two-form d÷
along D can be evaluated as: ⁄ ⁄
d÷ = ÷,
D ˆD
where on the right-hand-side this is understood as the integral of the one-form ÷ over the
boundary curve ˆD µ R2 , which is realized as a parametric curve (or union of parametric
curves) compatible with the induced orientation. 1 2
We can be a little more explicit. If we write ÷ = f dx + g dy, then d÷ = ˆx
ˆg
≠ ˆf
ˆy dx · dy.
The statement of Green’s theorem becomes
⁄ 3 4 ⁄
ˆg ˆf
≠ dx · dy = (f dx + g dy) .
D ˆx ˆy ˆD

Proof. We will only prove Green’s theorem for recursively supported regions. (In fact, we
will only write the proof for x-supported regions, but the proof for y-supported regions is
analogous.) For more general closed bounded regions, one can prove Green’s theorem by
rewriting the region as an union of recursively
1 supported
2 regions.
We write ÷ = f dx + g dy and d÷ = ˆx ≠ ˆy dx · dy.
ˆg ˆf

Let D be an x-supported region of the form:

D = {(x, y) œ R2 | x œ [a, b], u(x) Æ y Æ v(x)},

for some continuous functions u, v : R æ R. The boundary ˆD is a closed simple curve, with
counterclockwise orientation. It can be split into four curves:
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 207

1. C1 is the vertical line x = b between y = u(b) and y = v(b);

2. C2 is the curve y = v(x) between x = b and x = a;

3. C3 is the vertical line x = a between y = v(a) and y = u(a);

4. C4 is the curve y = u(x) between x = a and x = b.

In fact, for C2 and C3 , we will pick the opposite orientations (as it simplifies the parametriza-
tions), and add a negative sign in front of the line integrals. Those four curves can be realized
as parametric curves (C2 and C3 have opposite orientations):

1. –1 : [0, 1] æ R2 with –1 (t) = (b, u(b) + (v(b) ≠ u(b))t);

2. –2 : [a, b] æ R2 with –2 (t) = (t, v(t));

3. –3 : [0, 1] æ R2 with –3 (t) = (a, u(a) + (v(a) ≠ u(a))t);

4. –4 : [a, b] æ R2 with –4 (t) = (t, u(t));

Calculating the pullbacks, we evaluate the line integrals:

⁄ ⁄ 1
÷= g(b, u(b) + (v(b) ≠ u(b))t)(v(b) ≠ u(b)) dt,
–1 0
⁄ ⁄ b
! "
÷= f (t, v(t)) + g(t, v(t))v Õ (t) dt,
–2 a
⁄ ⁄ 1
÷= g(a, u(a) + (v(a) ≠ u(a))t)(v(a) ≠ u(a)) dt,
–3 0
⁄ ⁄ b
! "
÷= f (t, u(t)) + g(t, u(t))uÕ (t) dt.
–4 a

Putting those together, remembering that we need to add a negative sign for C2 and C3 , we
get:
⁄ ⁄ ⁄ ⁄ ⁄
÷= ÷≠ ÷≠ ÷+ ÷
ˆD –1 –2 –3 –4
⁄ 1
= (g(b, u(b) + (v(b) ≠ u(b))t)(v(b) ≠ u(b)) ≠ g(a, u(a) + (v(a) ≠ u(a))t)(v(a) ≠ u(a))) dt
0
⁄ b
! "
+ f (t, u(t)) ≠ f (t, v(t)) + g(t, u(t))uÕ (t) ≠ g(t, v(t))v Õ (t) dt.
a
s
Next, we need to evaluate the integral D d÷. Here we will do our favourite trick: the
pullback. Instead of using the x-supported domain D directly, we will pullback to a rectangular
domain. This will allow us to integrate the two-form. Consider the function „ : D2 æ D,
where
D2 = {(s, t) œ R2 | s œ [a, b], t œ [0, 1]},
with
„(s, t) = (s, u(s) + (v(s) ≠ u(s))t).
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 208

We see that „(D2 ) = D, and that the function is bijective. So by Lemma 5.3.7, we know that
⁄ ⁄
„ú (d÷) = d÷,
D2 D1

so we may as well calculate the left-hand-side. In fact, we also know that „ú (d÷) = d(„ú ÷).
So let us calculate this pullback. First,

„ú ÷ = f („(s, t)) ds + g(„(s, t))(uÕ (s) + (v Õ (s) ≠ uÕ (s))t) ds + g(„(s, t))(v(s) ≠ u(s)) dt.

Then
3
ˆf („(s, t)) ˆ ! "
d(„ú ÷) = ≠ ≠ g(„(s, t))(uÕ (s) + (v Õ (s) ≠ uÕ (s))t)
ˆt ˆt
4
ˆ
+ (g(„(s, t))(v(s) ≠ u(s))) ds · dt.
ˆs
Thus
⁄ ⁄ b⁄ 13 4
ˆf („(s, t)) ˆ ! "
„ (d÷) =
ú
≠ ≠ g(„(s, t))(uÕ (s) + (v Õ (s) ≠ uÕ (s))t) dtds
D2 a 0 ˆt ˆt
⁄ 1⁄ b3 ˆ
4
+ (g(„(s, t))(v(s) ≠ u(s))) dsdt,
0 a ˆs
where we used Fubini’s theorem to exchange the order of two integrals in the second line. We
can then evaluate the inner integrals using the Fundamental Theorem of Calculus, since they
are definite integrals of derivatives. We get:
⁄ ⁄ b
! "
„ (d÷) =
ú
f („(s, 0)) ≠ f („(s, 1)) + g(„(s, 0))uÕ (s) ≠ g(„(s, 1))v Õ (s) ds
D2 a
⁄ 1
+ (g(„(b, t))(v(b) ≠ u(b)) ≠ g(„(a, t))(v(a) ≠ u(a))) dt.
0

Substituting back the expression for „(s, t), we get:

⁄ ⁄ b
! "
„ú (d÷) = f (s, u(s)) ≠ f (s, v(s)) + g(s, u(s))uÕ (s) ≠ g(s, v(s))v Õ (s) ds
D2 a
⁄ 1
+ (g(b, u(b) + (v(b) ≠ u(b))t)(v(b) ≠ u(b)) ≠ g(a, u(a) + (v(a) ≠ u(a))t)(v(a) ≠ u(a))) dt.
0

Magic: this
s
is exactly the same as the expression that we found many lines above for the line
integral ˆD ÷! Woot woot! Therefore
⁄ ⁄
d÷ = ÷.
D ˆD

⌅
Remark 5.7.2 If D is simply connected, then ˆD is a simple s
closed curve, and D is the
closed curve with its interior. In this case the line integral ˆD ÷ is simply the line integral of
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 209

the one-form ÷ along the simple closed curve with counterclockwise orientation.
However, in the statement of Green’s theorem, D can be more general. For instance, it
could be an annulus.
s
Then ˆD would have two components: the inner and outer circle. The
line integral ˆD ÷ would then be the sum of the line integrals of the one-form ÷ along the two
circles; the outer circle with counterclockwise orientation, and the inner circle with clockwise
orientation, as this is the induced orientation on the boundary according to Definition 5.2.9.
Remark 5.7.3 Following up on the previous remark, we note that we can read Green’s
theorem in two different ways, which is generally the case for all integral theorems of vector
calculus. We could say:

1. The integral of the exact two-form d÷ over the region D µ R2 with canonical orientation
is equal to the integral of the one-form ÷ over the boundary ˆD with the induced
orientation.

2. The integral of the one-form ÷ over the simple closed curve ˆD µ R2 with canonical
orientation is equal to the integral of its exterior derivative d÷ over the interior D with
canonical orientation.

This is just two different readings of the same statement, depending on whether you start on
the left-hand-side or the right-hand-side. Consequently, we can use Green’s theorem either to
evaluate integrals of exact two-forms via reading 1, or to evaluate line integrals over simple
closed curves via reading 2.
In practice however, Green’s theorem is generally useful mostly to evaluate line integrals
by transforming them into surface integrals.
Example 5.7.4 Using Green’s theorem to calculate line integrals. Find the
line integral of the one-form Ê = xy dx + (x + y) dy over the rectangle with vertices
(≠2, ≠1), (2, ≠1), (2, 0), (≠2, 0), with a clockwise orientation.
We could evaluate the line integral using previous techniques, by rewriting the curve as
four parametric curves for each line segment and then use the definition of line integrals. But
let us instead use Green’s theorem.
Let us denote the rectangular curve by C. It is the boundary of the region D consisting of
the interior of the rectangle with its boundary:

D = {(x, y) œ R2 | x œ [≠2, 2], y œ [≠1, 0]}.

Thus we can use Green’s theorem to rewrite the line integral along C as a surface integral
over D.
We have to be careful with orientation though. If D has canonical orientation, the induced
orientation on the boundary rectangle C will be counterclockwise. But the question is asking
us to evaluate the line integral along C with clockwise orientation. So in Green’s theorem,
if we relate the line integral to the surface integral over D with canonical orientation, we
will need to introduce a minus sign. More precisely, let us denote C≠ to be the rectangle
with clockwise orientation, C+ the rectangle with counterclockwise orientation, and D the
rectangular region with canonical (counterclockwise) orientation. Green’s theorem states:
⁄ ⁄ ⁄
Ê=≠ Ê=≠ dÊ.
C≠ C+ D
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 210
s
So instead of calculating the line integral in the problem, we can calculate ≠ D dÊ.
The two-form dÊ is:

dÊ = xdy · dx + dx · dy = (1 ≠ x)dx · dy.

The integral can be evaluated:

⁄ ⁄ 0 ⁄ 2
dÊ = (1 ≠ x) dxdy
D ≠1 ≠2
⁄ 0 C D2
x2
= x≠ dy
≠1 2 ≠2
⁄ 0
= (2 ≠ (≠2) ≠ 2 + 2) dy
≠1
=4(0 ≠ (≠1))
=4.

Therefore, the line integral of Ê along C≠ is

⁄ ⁄
Ê=≠ dÊ = ≠4.
C≠ D

It is a good exercise to check that this is the correct answer by evaluating the line integral
using the standard approach with parametric curves. ⇤
While Green’s theorem is mostly useful to evaluate line integrals, we can also use Green’s
theorem in the reverse direction, to evaluate integrals of exact two-forms. Here is an example.
Example 5.7.5 Area of an ellipse. Find the area enclosed by the ellipse
x2 y 2
+ 2 = 1.
a2 b
The area is given by integrating the basic two-form Ê = dx · dy over the region D bounded
by the ellipse with canonical orientation. We could calculate this integral directly, as a double
integral. Or we can use Green’s theorem, if Ê is exact. But it certainly is: for instance, we
can write Ê = d÷ with ÷ = 12 (x dy ≠ y dx) (we could choose other one-forms such that d÷ = Ê
as well, but this one gives a nice and easy integral). Then Green’s theorem states:
⁄ ⁄
Ê= ÷,
D C
where on the right-hand-side we are evaluating the line integral of ÷ = 12 (x dy ≠ y dx) along
the ellipse with counterclockwise orientation.
We can parametrize the ellipse as – : [0, 2fi] æ R2 with –(t) = (a cos(t), b sin(t)). The
orientation is counterclockwise as required. The pullback –ú ÷ is:
3 4
ab 2 ab ab
– ÷=
ú
cos (t) ≠ sin(t)(≠ sin(t)) dt = dt.
2 2 2
The line integral is then
⁄ ⁄ 2fi
ab
÷= dt
C 0 2
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 211

=fiab,

which is the area of the ellipse. ⇤

5.7.2 Vector form of Green’s theorem

As always, we can translate our results to vector calculus concepts using our dictionary. It is
a bit artificial though here, since Green’s theorem really lives in two dimensions, while our
vector calculus concepts live in three dimensions. But the idea is to think of the region D as
being the trivial parametric surface in R3 where we embed it trivially in the xy-plane. That is,
– : D æ R3 , with –(x, y) = (x, y, 0). The tangent vectors are Tx = (1, 0, 0) and Ty = (0, 1, 0),
and the normal vector is of course n = (0, 0, 1) = e3 .
To the one-form ÷ = f dx + g dy, we associate the vector field F = (f, g, 0). To the
two-form Ê = d÷ is then associated the vector field Ò ◊ F. Using Corollary 5.6.5, we then see
that we can rewrite the surface integral as
⁄ ⁄⁄
d÷ = ((Ò ◊ F) · e3 ) dA.
D D

In other words, the integrand is the z-component of the curl Ò ◊ F.

As for the other side of Green’s theorem, it is a line integral of the one-form ÷ over the
curve ˆD. Assuming that we parametrize the curve with a position function r, we can rewrite
the line integral as ⁄ ⁄
÷= F · dr.
ˆD ˆD
The result is the vector form of Green’s theorem:
⁄⁄ ⁄
((Ò ◊ F) · e3 ) dA = F · dr.
D ˆD

5.7.3 Exercises
1. Evaluate the line integral of the one-form

Ê = ey dx + ex dy

along the rectangle with vertices (0, 0), (3, 0), (3, 2), (0, 2), counterclockwise, using two
different methods: (a) directly, (b) using Green’s theorem.
Solution. (a) Let us call ˆD the path along the rectangle counterclockwise. It is shown
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 212

in the figure below. Figure 5.7.6 The path ˆD along the rectangle.
To evaluate the line integral directly, we need to parametrize separately the four line
segments. We write the parametrizations as –i : [0, 1] æ R, i = 1, . . . , 4, with

–1 (t) =(3t, 0),

–2 (t) =(3, 2t),
–3 (t) =(3 ≠ 3t, 2),
–4 (t) =(0, 2 ≠ 2t).

We calculate the pullbacks:

–1ú Ê =3 dt,
–2ú Ê =2e3 dt,
–3ú Ê = ≠ 3e2 dt,
–4ú Ê = ≠ 2 dt.

Putting all this together, the line integral becomes

⁄ ⁄ ⁄ ⁄ ⁄
Ê= Ê+ Ê+ Ê+ Ê
ˆD –1 –2 –3 –4
⁄ 11 2
= 3 + 2e3 ≠ 3e2 ≠ 2 dt
0
=1 + 2e3 ≠ 3e2 .

(b) We now use Green’s theorem to calculate the line integral. By Green’s theorem,
we know that ⁄ ⁄
Ê= dÊ,
ˆD D
where D is the interior of the rectangle with its boundary, that is,

D = {(x, y) œ R2 | x œ [0, 3], y œ [0, 2]},

CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 213

with canonical orientation. We calculate the exterior derivative:

dÊ =ey dy · dx + ex dx · dy
=(ex ≠ ey ) dx · dy.

The integral of the two-form is:

⁄ ⁄
dÊ = (ex ≠ ey ) dx · dy
D D
⁄ 2⁄ 3
= (ex ≠ ey )dxdy
0 0
⁄ 21 2
= e3 ≠ 1 ≠ 3ey dy
0
=2e ≠ 2 ≠ 3e2 + 3
3

=1 + 2e3 ≠ 3e2 .

Therefore, ⁄
Ê = 1 + 2e3 ≠ 3e2 ,
ˆD
which is the same answer as in part (a), as it should.
2. Use Green’s theorem to evaluate the line integral of the vector field F(x, y) = (x1/3 +
y, x4/3 ≠ y 1/3 ) along the triangle with vertices (1, 3), (2, 2), (1, 1), clockwise.
Solution. Let us call ˆD the path around the triangle clockwise. It is shown in the
figure below.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 214

Figure 5.7.7 The path ˆD along the triangle.

We associate to the vector field F the one-form Ê = (x1/3 + y) dx + (x4/3 ≠ y 1/3 ) dy.
Since the path ˆD is oriented clockwise (negative orientation), by Green’s theorem we
know that ⁄ ⁄
Ê=≠ dÊ,
ˆD D
where D is the interior of the triangle with its boundary, with canonical orientation. We
can write equations for the three sides of the triangle. The blue line has equation x = 1;
the green line, y = x; the orange line, y = ≠x + 4. Using these equations, we can describe
D as an x-supported region:
D = {(x, y) œ R2 | x œ [1, 2], x Æ y Æ ≠x + 4}.
We calculate the exterior derivative dÊ:
4
dÊ =dy · dx + x1/3 dx · dy
3 34
4 1/3
= x ≠ 1 dx · dy.
3
Its integral along D is:
⁄ ⁄ 2 ⁄ ≠x+4 3 4
4 1/3
dÊ = x ≠1 dydx
D 1 x 3
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 215
⁄ 23 4 Ë È≠x+4
4
= x1/3 ≠ 1
y dx
1 3 x
⁄ 23 4
4 1/3
= x ≠ 1 (≠x + 4 ≠ x) dx
1 3
⁄ 23 4
8 4/3 16 1/3
= ≠ x + 2x + x ≠ 4 dx
1 3 3
3 4x=2
8 7/3 2 4/3
= ≠ x + x + 4x ≠ 4x
7 x=1
32 1/3 8
= ≠ 2 + 4 + 8 · 21/3 ≠ 8 + ≠ 1 ≠ 4 + 4
7 7
31 1/3
2
= 8·2 ≠9 .
7
Therefore, ⁄ ⁄
31 2
Ê=≠ 8 · 21/3 ≠ 9 .
dÊ = ≠
ˆD D 7
3. Use Green’s theorem to find the work done by the force

F(x, y) = (x2 + y 2 , xy)

while moving an object first along the parabola y = x2 from the origin to the point
(2, 4), then along a horizontal line back to the y-axis, and then back to the origin along a
vertical line.
Solution. We denote the closed path by ˆD. It is shown in the figure below.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 216

Figure 5.7.8 The closed path ˆD.

From the description of the path we know that the object is moving counterclockwise
along the path. Therefore, if we associate the one-form Ê = (x2 + y 2 ) dx + xy dy to the
vector field F, by Green’s theorem we know that
⁄ ⁄
Ê= dÊ,
ˆD D

with D being the region enclosed by the path (with its boundary), with canonical
orientation. The region D can be described as an x-supported region:

D = {(x, y) œ R2 | x œ [0, 2], x2 Æ y Æ 4}.

The exterior derivative dÊ is:

dÊ =2y dy · dx + y dx · dy
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 217

= ≠ y dx · dy

Its integral along D is:

⁄ ⁄
dÊ = ≠ y dx · dy
D D
⁄ 2⁄ 4
=≠ y dydx
0 x2
⁄ 2 C 2 Dy=4
y
=≠ dx
0 2 y=x2
⁄ 2A B
x4
=≠ 8≠ dx
0 2
3 4
16
= ≠ 16 ≠
5
64
=≠ .
5
Therefore, by Green’s theorem, the work done by the force F while moving an object
along the path ˆD is: ⁄
64
Ê=≠ .
ˆD 5
4. Suppose that a polygon has vertices (x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ), in counterclockwise
order. The area A of the polygon is given by integrating the basic two-form Ê = dx · dy
over the polygon. Use Green’s theorem to show that
1
A= ((x1 y2 ≠ x2 y1 ) + (x2 y3 ≠ x3 y2 ) + . . . + (xn≠1 yn ≠ xn yn≠1 ) + (xn y1 ≠ x1 yn )) .
2
This is a well known formula for the area of an arbitrary polygon, see for instance
https://en.wikipedia.org/wiki/Polygon#Area.
Solution. We denote by ˆD the closed path that goes around the polygon coun-
terclockwise. It consists of n line segments Li , i = 1, . . . , n, with Li being the line
segment between the points (xi , yi ) and (xi+1 , yi+1 ) (for simplicity of notation, we define
(xn+1 , yn+1 ) := (x1 , y1 )).
Further, we notice that the two Ê = dx · dy is exact. We write it as Ê = d÷ for the
one-form
÷ = x dy.
Note that this is certainly not the only choice of ÷, but any ÷ such that d÷ = Ê will work.
Then, by Green’s theorem, we know that the area A of the polygon is given by
⁄
A= dx · dy
⁄D
= ÷
ˆD
n ⁄
ÿ
= ÷,
i=1 –i
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 218

where in the last line –i is a parametrization of the line segment Li . More explicitly,
since the line segment Li joins the points (xi , yi ) and (xi+1 , yi+1 ), we can parametrize it
as –i : [0, 1] æ R2 with

–i (t) = (xi + t(xi+1 ≠ xi ), yi + t(yi+1 ≠ yi )).

Then
–iú ÷ = (xi + t(xi+1 ≠ xi ))(yi+1 ≠ yi ) dt.
The line integral along –i is then:
⁄ ⁄ 1
÷= (xi + t(xi+1 ≠ xi ))(yi+1 ≠ yi ) dt
–i 0
3 4
1
=(yi+1 ≠ yi ) xi + (xi+1 ≠ xi )
2
1
= (yi+1 ≠ yi )(xi + xi+1 )
2
1
= (xi yi+1 ≠ xi+1 yi + xi+1 yi+1 ≠ xi yi ).
2
Finally, we add up these line integrals to get the area of the polygon. We get:
n ⁄
ÿ
A= ÷
i=1 –i
1 ÿn
= (xi yi+1 ≠ xi+1 yi + xi+1 yi+1 ≠ xi yi )
2 i=1
1
= ((x1 y2 ≠ x2 y1 + x2 y2 ≠ x1 y1 ) + (x2 y3 ≠ x3 y2 + x3 y3 ≠ x2 y2 ) + . . .
2
+ (xn y1 ≠ x1 yn + x1 y1 ≠ xn yn )),

where in the last line we used the fact that xn+1 = x1 and yn+1 = y1 . We see that the
terms x1 y1 , x2 y2 , . . . all cancel out, and we are left with
1
A = ((x1 y2 ≠ x2 y1 ) + (x2 y3 ≠ x3 y2 ) + . . . + (xn y1 ≠ x1 yn )).
2
Bingo!
5. We already studied the one-form
y x
÷=≠ dx + 2 dy,
x2 +y 2 x + y2
defined on U = R2 \ {(0, 0)}, since it is the typical example of a one-form that is closed
but not exact. For instance, we proved in Exercise 3.4.3.2 that it is not exact by showing
that its line integral along a circle of radius one around the origin is non-vanishing. In
this problem we show that the line integral of ÷ is in fact non-vanishing for every simple
closed curve that encloses the origin, and vanishing for every simple closed curve that
does not pass through or enclose the origin.
(a) Consider an arbitrary simple closed curve C0 with canonical orientation that does
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 219

not pass through or enclose the origin. Use Green’s theorem to show that
⁄
÷ = 0.
C0

(b) Let C1 be an arbitrary simple closed curve with canonical orientation that encloses
the origin. Explain why the argument of (a) does not apply here. So we need to do
something else to study the line integral of Ê along C1 .

(c) As in (b), let C1 be an arbitrary simple closed curve with canonical orientation
that encloses the origin. Suppose that K is a circle centered at the origin, with a
radius small enough that K lies completely inside C1 . Give K a counterclockwise
orientation. Use Green’s theorem to show that
⁄ ⁄
÷= ÷.
C1 K

(d) Using part (c), show that it implies that

⁄
÷ = 2fi.
C1

You have showed that the line integral of ÷ along an arbitrary simple closed curve that
encloses the origin is non-vanishing (and is in fact equal to 2fi), while the line integral
along an arbitrary simple closed curve that does not pass through or enclose the origin is
vanishing. Isn’t it neat?
Solution. The key in this problem is to be very careful with the domain of definition
of the one-form ÷. The one-form is define on the open set U = R2 \ {(0, 0)}.
As a starting point, recall that the one-form ÷ is closed, that is, dÊ = 0. Indeed,
A B A B
x2 + y 2 ≠ 2y 2 x2 + y 2 ≠ 2x2
d÷ = ≠ dy · dx + dx · dy
(x2 + y 2 )2 (x2 + y 2 )2
(x2 ≠ y 2 ) + (y 2 ≠ x2 )
= dx · dy
(x2 + y 2 )2
=0.

This will be useful in this problem.

(a) We assume that C0 is a simple closed curve that does not pass through or enclose
the origin, with canonical orientation. Therefore, C0 µ U . Moreover, if we denote by D0
the region consisting of the closed curve C0 and its interior, then D0 µ U as well, since
the origin is not in D0 . Therefore, by Green’s theorem,
⁄ ⁄
÷= d÷
C0 D0
=0,

since d÷ = 0.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 220

(b) If C1 is a simple closed curve that encloses the origin, we cannot apply the
argument in (a). The reason is that, if we denote by D1 the region consistin of the closed
curve C1 and its interior, then D1 is not a subset of U , since D1 contains the origin. In
other words, the one-form ÷ is not defined on D1 , which is one of the assumptions in the
statement of Green’s theorem. Therefore, Green’s theorem does not apply. (Indeed, as
we will see, if Green’s theorem applied, it would imply that the line integral of ÷ along
C1 vanishes, which is not true, as we will see.)
(c) We suppose that C1 is an arbitrary simple closed curve with canonical orientation
that encloses the origin, and that K is a circle centered at the origin that lies completely
inside C1 , also oriented counterclockwise. For instance, C1 could be the orange curve in
the figure below, and K the blue circle.

Figure 5.7.9 An example of curves C1 and K.

We let D1 be the region bounded by the two closed curves C1 and K, and we give
D1 a canonical orientation. By definition, the origin is not in D1 , so D1 µ U and the
one-form is well defined on D1 . We can then apply Green’s theorem. The boundary of
D1 has two components, ˆD1 = C1 fi K, and if D1 is oriented counterclockwise, then
the induced orientation on the boundary is counterclockwise for the outer boundary C1
and clockwise for the inner boundary K. To clarify the notation, we write (C1 )+ for
C1 with counterclockwise orientation, K+ for K with counterclockwise orientation, and
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 221

K≠ for K with clockwise orientation. So we can write the oriented boundary of D1 as

ˆD1 = (C1 )+ fi K≠ . By Green’s theorem, we have:
⁄ ⁄
d÷ = ÷
D1 ˆD1
⁄ ⁄
= ÷+ ÷
(C1 )+ K≠
⁄ ⁄
= ÷≠ ÷,
(C1 )+ K+

where we used the fact that line integrals pick a sign if we change the orientation. But
the left-hand-side of the above equation is necessarily zero, since d÷ = 0. Therefore,
⁄ ⁄
÷= ÷,
(C1 )+ K+

which is the statement that we were trying to prove, with both curves canonically oriented.
(d) The power of what we did in (c) is to replace the evaluation of the line integral of
÷ along an arbitrary simple closed curve C1 that encloses the origin by the evaluation of
the line integral of ÷ along a circle K centered at the origin of a certain non-zero radius.
But we can evaluate the latter explicitly! Indeed, if K has radius R, for some R œ R>0 ,
then we can parametrize K by – : [0, 2fi] æ R2 with

–(◊) = (R cos(◊), R sin(◊)).

The pullback of ÷ is
1
–ú ÷ = (≠R sin(◊)(≠R sin(◊)) + R cos(◊)(R cos(◊))) d◊
R2
=d◊.

Therefore,
⁄ ⁄
÷= –ú ÷
– [0,2fi]
⁄ 2fi
= d◊
0
=2fi.

So the line integral of ÷ along any circle K centered at the origin is always non-zero and
equal to 2fi. By (c), we thus conclude that
⁄
÷ = 2fi
C1

for arbitrary simple closed curves C1 that enclose the origin. Very powerful result!
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 222

5.8 Stokes’ theorem

Step 6, part 2: we continue our study of integration of exact two-forms. But this time we
consider surface integrals, that is integrals of exact two-forms over parametric surfaces. The
result is Stokes’ theorem.

Objectives
You should be able to:

• State Stokes’ theorem for integrals of exact two-forms over parametric surfaces in R3 .

• Rephrase Stokes’ theorem in terms of the associated vector fields.

• Show that Stokes’ theorem implies that the integral of an exact two-form over a closed
surface in R3 vanishes.

• Use Stokes’ theorem to evaluate surface integrals of exact two-forms.

• Use Stokes’ theorem to evaluate line integrals of one-forms over closed curves in R3 .

• Use Stokes’ theorem and its consequences to show that a given two-form cannot be
exact.

5.8.1 Stoke’s theorem

In the previous section we studied integrals of exact two-forms over bounded regions in R2 .
We now look at surface integrals of exact two-forms over parametric surfaces in R3 .
As in the previous section, we start by recalling the corresponding statement for curves,
which we first studied in Theorem 3.4.1, and revisited in Theorem 5.1.8. Without getting into
the details again, the key statement is that the integral of an exact one-form Ê = df over a
parametric curve – satisfies the property
⁄ ⁄
df = f,
– ˆ–

which is the Fundamental Theorem of line integrals rewritten using integrals of zero-forms.
The integral on the left-hand-side is a line integral along the oriented parametric curve –,
while on the right-hand-side we have the integral of the zero-form f over the oriented boundary
ˆ– of the parametric curve –.
Looking back at Theorem 3.4.1, the proof of the Fundamental Theorem of line integrals
basically amounted to pulling back to the interval [a, b] using the definition of line integrals,
and then using the Fundamental Theorem of Calculus. We now do the same thing for surface
integrals: we pull back to the region D and then use Green’s theorem. The result is known as
Stokes’ theorem.
Theorem 5.8.1 Stokes’ theorem. Let ÷ be a one-form on an open subset U ™ R3 , and
– : D æ R3 a parametric surface whose image S = –(D) µ U is orientable and oriented by
the parametrization. Let ˆ– be the boundary of the surface S with its induced orientation, as
in Definition 5.5.5 (if the image surface is closed, ˆ– is the empty set). Then the integral of
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 223

the exact two-form d÷ along the parametric surface – can be evaluated as:
⁄ ⁄
d÷ = ÷.
– ˆ–

In other words, the surface integral of the exact two-form Ê = d÷ along the parametric surface
– is equal to the line integral of the one-form ÷ along its oriented boundary curve.
Proof. Stokes’ theorem pretty much follows from Green’s theorem by pullback. More precisely,
by definition of the surface integral,
⁄ ⁄
d÷ = –ú (d÷).
– D

We know that the exterior derivative commutes with pullback, see Lemma 4.7.4: –ú (d÷) =
d(–ú ÷). So we can rewrite the surface integral as:
⁄ ⁄
d÷ = d(–ú ÷).
– D

This is now the integral of the exact two-form d(–ú ÷) over a closed bounded region D. By
Green’s theorem, Theorem 5.7.1, we know that
⁄ ⁄
d(–ú ÷) = –ú ÷,
D ˆD

where on the right-hand-side we have a line integral of the one-form –ú ÷ along the boundary
of the closed region D µ R2 .
We are almost done, but not quite: the last step is actually quite subtle. We would like to
say that ⁄ ⁄
?
–ú ÷ = ÷.
ˆD ˆ–
If the parametrization – is injective everywhere, then it maps the boundary ˆD one-to-one to
the boundary ˆS, and so this equality follows from the definition of line integrals, thinking
of – restricted to the boundary ˆD as a parametric curve (or a union of parametric curves).
However, in general, it is a little more subtle; as we remarked in Remark 5.4.3, ˆS ™ –(ˆD),
but the two are not necessarily equal.1 The general proof is beyond the scope of these notes.
In any case, the conclusion would be that
⁄ ⁄
d÷ = ÷,
– ˆ–

which is the statement of Stokes’ theorem. ⌅

Just as for the Fundamental Theorem of line integrals, there are two direct consequences
of Stokes’ theorem. First, the integrals of an exact two-form over two surfaces S1 and S2 that
share the same oriented boundary are equal (this statement for surface integrals is analogous
1
In fact, not only can – send boundary points in ˆD to points not on the boundary of S, but it can also
map whole curves in ˆD to points in S, as it does for instance when parametrizing a sphere with spherical
coordinates (two of the boundary lines of the rectangular regions are mapped to the north and south poles of
the sphere).
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 224

to path independence for line integrals of exact one-forms). Second, the integral of an exact
two-form along a closed surface is zero (this statement for surface integrals is analogous to
the statement that the line integral of an exact one-form along a closed curve vanishes). We
state those as the following corollaries.
Corollary 5.8.2 The surface integrals of an exact two-form along two surfaces
that share the same oriented boundary are equal. Let ÷ be a one-form on U ™ R3 .
If – : D1 æ R3 and — : D2 æ R3 are two parametric surfaces whose image surfaces
S1 = –(D1 ) µ U and S2 = —(D2 ) µ U share the same boundary ˆS1 = ˆS2 , with the same
induced orientation, then ⁄ ⁄
d÷ = d÷.
– —

Corollary 5.8.3 The surface integral of an exact two-form along a closed surface
vanishes. Let ÷ be a one-form on U ™ R3 . If – : D æ R3 is a parametric surface whose
image surface S = –(D) µ U is closed (that is, it has no boundary, which means that it is
itself the boundary of a volume), then
⁄
d÷ = 0.
–

Remark 5.8.4 We can read Stokes’ theorem in two ways, just like Green’s theorem.

1. The integral of an exact two-form Ê = d÷ over the parametric surface – : D æ R3

is equal to the line integral of the one-form ÷ over the oriented boundary ˆ– of the
parametric surface.

2. The integral of a one-form ÷ over a closed oriented simple curve ˆS in R3 is equal to the
surface integral of its exterior derivative d÷ over any surface S µ R3 whose boundary is
ˆS, with the appropriate orientation.

These two different readings highlight the two different types of applications of Stokes’ theorem:
either to calculate surface integrals via the first reading, or to calculate line integrals via the
second reading. We note that Corollary 5.8.2 can also be useful computationally, to replace a
complicated surface integral by an easier one.
Example 5.8.5 Using Stokes’ theorem to evaluate a surface integral by transforming
it into a line integral. Use Stokes’ theorem to evaluate the surface integral of the two-form
Ê = 2z dy · dz + dx · dy along the part of the sphere x2 + y 2 + z 2 = 25 that is above the
plane z = 3, with orientation given by an outwards pointing normal vector.
The surface is shown in the following figure:
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 225

Figure 5.8.6 The surface is part of the sphere of radius 5 above the plane z = 3.
To start with, for Stokes’ theorem to apply, the two-form Ê = 2z dy · dz + dx · dy must
be exact. But it easy to see that

÷ = (x ≠ z 2 ) dy ∆ d÷ = Ê,

and thus Ê is exact.

The boundary ˆS of the surface is the intersection of the sphere x2 + y 2 + z 2 = 25 and
the plane z = 3. Setting z = 3 in the equation of the sphere, we get the equation of the circle
x2 + y 2 = 25 ≠ 9 = 16. The boundary is thus the circle x2 + y 2 = 16 in the plane z = 3.
Since we orient the sphere with an outwards pointing normal vector, we see that must walk
counterclockwise along the circle, with our heads pointing outside the sphere, to keep the
region on our left. Therefore, the oriented boundary is the circle x2 + y 2 = 16 in the plane
z = 3 with counterclockwise orientation. Stokes’ theorem then tells us that
⁄ ⁄
d÷ = ÷.
S ˆS

To evaluate the line integral on the right-hand-side we parametrize the oriented boundary
with – : [0, 2fi] æ R3 ,
–(◊) = (4 cos(◊), 4 sin(◊), 3).
We calculate the pullback of the one-form ÷:

–ú ÷ =(4 cos(◊) ≠ 9)(4 cos(◊)) d◊

CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 226

=4(4 cos2 (◊) ≠ 9 cos(◊)) d◊.

The line integral is then:

⁄ ⁄
÷= –ú ÷
– [0,2fi]
⁄ 2fi
=4 (4 cos2 (◊) ≠ 9 cos(◊)) d◊
0
=16fi,
s s
where we used the fact that 02fi cos(◊) = 0 and 02fi cos2 (◊) = fi.
Therefore, by Stokes’ theorem, we conclude that
⁄
d÷ = 16fi.
S

⇤
Example 5.8.7 Using Stokes’ theorem to evaluate a surface integral by using a
simpler surface. Let us consider the same integral as in the previous example, Example 5.8.5.
That is, we want to evaluate the surface integral of the two-form Ê = 2z dy · dz + dx · dy
along the part of the sphere x2 + y 2 + z 2 = 25 that is above the plane z = 3, with orientation
given by an outwards pointing normal vector.
We can use Stokes’ theorem in a different way to evaluate this surface integral. Since we
know that the two-form is exact (as Ê = d÷ with ÷ = (x ≠ z 2 ) du), we know that its surface
integral along any two surfaces S and S Õ that share the same oriented boundary are equal. So
instead of integrating over the part of the sphere above the z = 3 plane, we could replace this
surface by a simpler surface that shares the same boundary, which in this case is the circle
x2 + y 2 = 16 in the plane z = 3, as we saw in Example 5.8.5. In particular, we could take the
surface S Õ to be the disk x2 + y 2 Æ 16 in the plane z = 3, with orientation given by a normal
vector pointing upwards. By Stokes’ theorem, we know that
⁄ ⁄
Ê= Ê,
S SÕ
so we can evaluate the surface integral over the oriented disk instead.
To do so, we parametrize the disk by – : D æ R3 , with
D = {(r, ◊) œ R2 | r œ [0, 4], ◊ œ [0, 2fi]}
and
–(r, ◊) = (r cos(◊), r sin(◊), 3).
It is easy to see that the normal vector n = Tr ◊ T◊ points upwards, as required. We calculate
the pullback two-form:
–ú Ê =2(3)(sin(◊) dr + r cos(◊) d◊) · (0) + (cos(◊) dr ≠ r sin(◊) d◊) · (sin(◊) dr + r cos(◊) d◊)
=r dr · d◊.
The surface integral is then:
⁄ ⁄
Ê= –ú Ê
– D
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 227
⁄ 2fi ⁄ 4
= r drd◊
0 0
⁄ 2fi C 2 Dr=4
r
= d◊
0 2 r=0
⁄ 2fi
=8 d◊
0
=16fi.

Fortunately, this is the same answer that we found in Example 5.8.5, as it should! ⇤
Example 5.8.8 Using Stokes’ theorem to evaluate a line integral by transforming
it into a surface integral. Use Stokes’ theorem to evaluate the line integral of the one-form
Ê = xy dx + yz dy + z 4 dz over the curve C that forms the boundary of the part of the
paraboloid z = 1 ≠ x2 ≠ y 2 with x Ø 0, y Ø 0, z Ø 0. Assume that C is given the orientation
as moving from the x-axis to the y-axis to the z-axis.
It may be hard to visualize the curve and the region at first. A picture is worth a thousand
words. Here is a figure representing the part of the paraboloid z = 1 ≠ x2 ≠ y 2 with x Ø 0,
y Ø 0, z Ø 0.

Figure 5.8.9 The part of the paraboloid z = 1 ≠ x2 ≠ y 2 with x, y, z Ø 0. The curve is the
boundary of the region.
We want to evaluate the line integral of Ê along the boundary C of this region. We could
evaluate the line integral directly. Or we can use Stokes’ theorem, which says that
⁄ ⁄
Ê= dÊ,
C S
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 228

where S is the part of the paraboloid described above. We need to be careful with orientation
here. We assume that C is oriented going from x to y to z. Looking at the figure, we see that
if we walk along the boundary curve in this direction, to keep the surface on the left we must
have our heads pointing away from the origin. So the surface S should be oriented with a
normal vector pointing in that direction.
To evaluate the surface integral of dÊ along S, we first calculate the exterior derivative.
We get:
dÊ = xdy · dx + ydz · dy.
Next, we parametrize the surface S as – : D æ R3 , with

D = {(r, ◊) œ R2 | r œ [0, 1], ◊ œ [0, fi/2]}

and
–(r, ◊) = (r cos(◊), r sin(◊), 1 ≠ r2 ).
The region D was chosen such that x(r, ◊), y(r, ◊), z(r, ◊) Ø 0 on D. Thus it covers the part of
the paraboloid that we are interested in.
The normal vector is

n =Tr ◊ T◊
=(cos(◊), sin(◊), ≠2r) ◊ (≠r sin(◊), r cos(◊), 0)
=(2r2 cos(◊), 2r2 sin(◊), r).

As its z-component is positive, we see that it points away from the origin, as it should. So the
orientation induced by the parametrization is the correct one.
Finally, we calculate the pullback of the two-form dÊ:

–ú (dÊ) =r cos(◊)(sin(◊) dr + r cos(◊) d◊) · (cos(◊) dr

≠ r sin(◊) d◊) + r sin(◊)(≠2r dr) · (sin(◊) dr + r cos(◊) d◊)
=(≠r2 cos(◊) ≠ 2r3 sin(◊) cos(◊))dr · d◊.

We then evaluate the surface integral:

⁄ ⁄
dÊ = –ú (dÊ)
– D
⁄ fi/2 ⁄ 1
= (≠r2 cos(◊) ≠ 2r3 sin(◊) cos(◊)) drd◊
0 0
⁄ fi/2 C 3 Dr=1
r r4
= ≠ cos(◊) ≠ sin(◊) cos(◊) d◊
0 3 2 r=0
⁄ fi/2 3 4
1 1
= ≠ cos(◊) ≠ sin(◊) cos(◊) d◊
0 3 2
1 1
=≠ ≠
3 4
7
=≠ .
12
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 229

Therefore, by Stokes’ theorem, the lintegral of Ê along C is

⁄
7
Ê=≠ .
C 12
It is a good exercise to evaluate the line integral directly along the boundary curve (which
you will need to split into three parametric curves). You should get the same answer ≠7/12.
And, as complicated as the calculation above looked, it is actually simpler than calculating
the line integral directly! ⇤
Remark 5.8.10 We can also use Stokes’ theorem to show that a two-form is not exact, by
showing that its surface integral along a closed surface does not vanish. This is similar to what
we did for one-forms using the Fundamental Theorem of line integrals, see Exercise 3.4.3.2
(and also Exercise 5.7.3.5 ). We use Stokes’ theorem in this way in Exercise 5.8.3.5 for the
two-form in Remark 4.6.6.

5.8.2 Stokes’ theorem in terms of vector fields

As usual we translate Stokes’ theorem in terms of the associated vector fields, as in Corol-
lary 5.6.5.
Theorem 5.8.11 Stokes’ theorem for vector fields. Let F be a smooth vector field on
U ™ R3 . Let – : D æ R3 be a parametric surface whose image surface S = –(D) µ U is
orientable and oriented by parametrization, with oriented boundary ˆ–. Then Stokes’ theorem
is the statement that ⁄⁄ ⁄
(Ò ◊ F) · dS = F · dr,
S ˆS
where on the left-hand-side we used the shorthand notation introduced in Remark 5.6.6 to
denote the surface integral of the vector field Ò ◊ F along the oriented parametric surface S,
and on the right-hand-side we use the shorthand notation from Remark 3.3.8 to denote the
line integral of the vector field F along the oriented boundary curve ˆS parametrized by r.
From Corollary 5.8.2 and Corollary 5.8.3, we conclude that:

• The surface integral of the curl of a vector field is the same for any two surfaces that
share the same oriented boundary.

• The surface integral of the curl of a vector field along a closed surface always vanishes.

As before, we can use Stokes’ theorem in both directions. For instance, if we are interested
in calculating the line integral of a vector field along a closed curve in R3 (say, to calculate
the work done by a force field on an object moving along the closed curve), then we can use
Stokes’ theorem, which tells us that that the line integral is equal to the surface integral of
the curl of the vector field on any surface whose boundary is the closed curve. In particular,
we are free to choose a surface that is simple enough so that the resulting surface integral is
easy to evaluate.
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 230

5.8.3 Exercises

1. Let S be the part of the cone z = 2 x2 + y 2 below the plane x + z = 1, oriented with
an upward 
normal vector. Let S Õ be the part of the plane x + z = 1 contained within the
cone z = 2 x2 + y 2 , also oriented with an upward normal vector. Let ÷ be the one-form
2 3
÷ = ex dx + ey dy + 5(x + z ≠ 1)2 dz.

Use Stokes’ theorem to show that ⁄

d÷ = 0.
S

Solution. First,
 the oriented boundary of S is the closed curve at the intersection of
the cone z = 2 x2 + y 2 and the plane x + z = 1, oriented counterclockwise (looking from
above). We notice that the boundary of S Õ is exactly the same. Therefore, by Stokes’
theorem, we know that ⁄ ⁄
d÷ = d÷.
S SÕ
So we may as well evaluate the integral of d÷ along the surface S Õ , which lies within the
plane x + z = 1.
To do so, we need to evaluate the exterior derivative d÷. We get:
2 3
d÷ =d(ex ) · dx + d(ey ) · dy + 5d((x + z ≠ 1)2 ) · dz
=10(x + z ≠ 1) dx · dz.
s
To evaluate the surface integral S Õ d÷, we would need to parametrize the surface S Õ ,
pullback d÷, and evaluate the integral as a double integral. However, we can show that
the integral will be zero right away without going through the whole calculation. Indeed,
when we pullback d÷ by the parametrization, we need to evaluate its component function
on the surface. But the surface S Õ lies within the plane x + z = 1, and we see right away
that the component function vanishes when x + z ≠ 1. In other words, the two-form
d÷ = 0 when restricted to the plane x + z = 1. So it will certainly vanish on S Õ . Therefore,
we can conclude right away that ⁄
d÷ = 0,
SÕ
and, by Stokes’ theorem, ⁄
d÷ = 0.
S
2. Consider the force field

F(x, y, z) = (≠yx2 + z, xy 2 , exy + z).

Use Stokes’ theorem to find the work done by the force field when moving an object
counterclockwise along the circle x2 + y 2 = 4 in the plane z = 10.
Solution. Let ˆS be the circle x2 + y 2 = 4 in the plane z = 10 with counterclockwise
orientation, and let S be the disk x2 + y 2 Æ 4 within the plane z = 10, with orientation
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 231

given by an upward pointing normal vector. By Stokes’ theorem, we know that

⁄ ⁄⁄
F · dr = (Ò ◊ F) · dS.
ˆS S

So instead of evaluating the line integral of F along ˆS, we can evaluate the surface
integral of the curl of F along S.
First, we calculate the curl of F. We get:

Ò ◊ F = (xexy , 1 ≠ yexy , y 2 + x2 ).

To calculate the surface integral, we need to parametrize the surface S. We use the
following parametrization: – : D æ R3 , with

D = {(r, ◊) œ R2 | r œ [0, 2], ◊ œ [0, 2fi]},

with
–(r, ◊) = (r cos(◊), r sin(◊), 10).
The tangent vectors are

Tr = (cos(◊), sin(◊), 0), T◊ = (≠r sin(◊), r cos(◊), 0).

The normal vector is

n = Tr ◊ T◊ = (0, 0, r),
which points upward as required. Therefore, the surface integral is
⁄⁄ ⁄⁄
(Ò ◊ F) · dS = (Ò ◊ F)(–(r, ◊)) · n dA
S ⁄⁄D
2 sin(◊) cos(◊) 2 sin(◊) cos(◊)
= (r cos(◊)er , 1 ≠ r sin(◊)er , r2 ) · (0, 0, r) dA
⁄⁄D
= r3 dA
D
⁄ 2fi ⁄ 2
= r3 drd◊
0 0
⁄ 2fi
=4 d◊
0
=8fi.

Therefore, by Stokes’ theorem we conclude that the work done by the force field when
moving an object counterclockwise along the circle ˆS is 8fi.
3. Let S be the surface consisting of the top and the four sides (but not the bottom) of the
cube with vertices (±1, ±1, ±1) in R3 , with orientation given by an outward pointing
normal vector, and F the vector field

F = (xy, zx, cos(x2 ) + z 5 ).

CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 232

Use Stokes’ theorem to evaluate the surface integral

⁄⁄
(Ò ◊ F) · dS.
S

Solution. We can use Stokes’ theorem in two different ways here: either to rewrite the
surface integral as a line integral, or to rewrite it as a surface integral over a simpler
surface. For completeness I will do it both ways.
First, we can use Stokes’ theorem to rewrite the surface integral as a line integral.
Let us do this first. We notice that the boundary curve ˆS of the surface S is the square
with vertices (±1, ±1, ≠1), since the bottom side of the cube is not included in S. The
induced orientation on the boundary amounts to moving counterclockwise along the
square, as seen from above. Thus, by Stokes’ theorem, we know that
⁄⁄ ⁄
(Ò ◊ F) · dS = F · dr,
S ˆS

where the right-hand-side is the line integral of the vector field F along the square with
counterclockwise orientation.
To evaluate the line integral, we split the square into four line segments Li , i = 1, . . . , 4.
We parametrize the line segments by –i : [0, 1] æ R3 with

–1 (t) =(≠1 + 2t, ≠1, ≠1),

–2 (t) =(1, ≠1 + 2t, ≠1),
–3 (t) =(1 ≠ 2t, 1, ≠1),
–4 (t) =(≠1, 1 ≠ 2t, ≠1).

We calculate the tangent vectors:

T1 = (2, 0, 0),
T2 = (0, 2, 0),
T3 = (≠2, 0, 0),
T4 = (0, ≠2, 0).

We finally evaluate the line integrals:

⁄ ⁄ 1
F(–1 (t)) · T1 dt =2 (≠1 + 2t)(≠1) dt = 0,
L1 0
⁄ ⁄ 1
F(–2 (t)) · T2 dt =2 (≠1)(1) dt = ≠2,
L2 0
⁄ ⁄ 1
F(–3 (t)) · T3 dt = ≠ 2 (1 ≠ 2t)(1) dt = 0,
L3 0
⁄ ⁄ 1
F(–4 (t)) · T4 dt = ≠ 2 (≠1)(≠1)/dt = ≠2.
L4 0

Adding up these four line integrals, we get ≠4. We conclude, by Stokes’ theorem, that
the surface integral ⁄⁄
(Ò ◊ F) · dS = ≠4.
S
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 233

Another approach would have been to use Stokes’ theorem to rewrite the surface
integral as a simpler surface integral. For instance we could have replaced the surface S
by the missing side of the cube (namely the bottom), as it shares the same boundary
with S. Let us call the bottom side of the cube S Õ . If we give it an upward pointing
normal vector, it induces the same orientation on the boundary as S. Thus, by Stokes’
theorem, ⁄⁄ ⁄⁄
(Ò ◊ F) · dS = (Ò ◊ F) · dSÕ .
S SÕ
To evaluate the integral on the right-hand-side, we first calculate the curl of F. We get:
1 2
Ò ◊ F = ≠x, 2x sin(x2 ), z ≠ x .

Next, we need to parametrize S Õ , which is the square with vertices (±1, ±1, ≠1). We
write – : D æ R3 with

D = {(u, v) œ R2 | u œ [≠1, 1], v œ [≠1, 1]},

and
–(u, v) = (u, v, ≠1).
The tangent vectors are

Tu = (1, 0, 0), Tv = (0, 1, 0),

with normal vector

n = Tu ◊ Tv = (0, 0, 1).
This is the correct orientation. We thus have
⁄⁄ ⁄ 1 ⁄ 1
(Ò ◊ F) · dS =
Õ
(≠u, 2u sin(u2 ), ≠1 ≠ u) · (0, 0, 1) dudv
SÕ ≠1 ≠1
⁄ 1 ⁄ 1
=≠ (u + 1) dudv
≠1 ≠1
= ≠ 4.

The same answer as before!

4. Let C µ R2 be a simple closed curve, and D µ R2 the region consisting of C and its
interior. Let – : D æ R3 be the parametric surface
–(u, v) = (u, v, 1 ≠ u ≠ v),
with S = –(D) and ˆS the boundary curve of S. Let ÷ be the one-form
÷ = ≠z dx ≠ 2x dy + 4y dz.
Use Stokes’ theorem to show that the line integral of ÷ along ˆS is equal to the area of
the region D µ R2 .
Solution. First, by Stokes’ theorem we know that
⁄ ⁄
÷= d÷,
ˆ– –
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 234

with the orientation induced by the parametrization. To evaluate the integral on the
right-hand-side, we calculate the exterior derivative:

d÷ = ≠dz · dx ≠ 2dx · dy + 4dy · dz.

We calculate its pullback:

–ú (d÷) = ≠ (≠du ≠ dv) · du ≠ 2du · dv + 4dv · (≠du ≠ dv)

=du · dv.

Therefore, the surface integral becomes

⁄ ⁄
d÷ = –ú (d÷)
– ⁄D
= du · dv,
D

which is simply the area of the region D. We then conclude, by Stokes’ theorem, that
the line integral of ÷ along ˆS is equal to the area of the region D.
5. Consider the two-form
1
Ê= 2 (x dy · dz + y dz · dx + z dx · dy),
(x + y + z 2 )3/2
2

from Remark 4.6.6. It is defined on U = R3 \ {(0, 0, 0)}.

(a) Show that Ê is closed, that is, dÊ = 0.
(b) Show that Ê is not exact by showing that the surface integral of Ê along the sphere
x2 + y 2 + z 2 = 1 is non-zero.
(c) Does that contradict Poincare’s theorem for two-forms?

Solution. (a) To show that it is closed, we calculate the exterior derivative. We get:
1 (x2 + y 2 + z 2 )3/2 ≠ 3x2 (x2 + y 2 + z 2 )1/2
dÊ =
(x2 + y 2 + z 2 )3
(x2 + y 2 + z 2 )3/2 ≠ 3y 2 (x2 + y 2 + z 2 )1/2
+
(x2 + y 2 + z 2 )3
(x2 + y 2 + z 2 )3/2 ≠ 3z 2 (x2 + y 2 + z 2 )1/2 2
+ dx · dy · dz
(x2 + y 2 + z 2 )3
1 1
2 2 2 2 2 2
2
= 2 3(x + y + z ) ≠ 3(x + y + z ) dx · dy · dz
(x + y 2 + z 2 )5/2
=0.
Therefore, Ê is closed.
(b) In fact, we already did such a surface integral in Exercise 5.6.4.5, but for complete-
ness we redo it here. We parametrize the sphere as always with spherical coordinates,
– : D æ R3 ,
D = {(◊, „) œ R2 | ◊ œ [0, fi], „ œ [0, 2fi]},
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 235

and
–(◊, „) = (sin(◊) cos(„), sin(◊) sin(„), cos(◊)).
The pullback of Ê is:
11
–ú Ê = sin(◊) cos(„)(cos(◊) sin(„)d◊ + sin(◊) cos(„)d„) · (≠ sin(◊)d◊)
1
+ sin(◊) sin(„)(≠ sin(◊)d◊)) · (cos(◊) cos(„)d◊) ≠ sin(◊) sin(„)d„)
2
+ cos(◊)(cos(◊) cos(„)d◊) ≠ sin(◊) sin(„)d„) · (cos(◊) sin(„)d◊ + sin(◊) cos(„)d„)
1 2
= sin3 (◊) cos2 („) + sin3 (◊) sin2 („) + sin(◊) cos2 (◊) d◊ · d„
= sin(◊)d◊ · d„.

We finally calculate the surface integral:

⁄ ⁄
Ê= –ú Ê
– D
⁄ 2fi ⁄ fi
= sin(◊)d◊d„
0 0
⁄ 2fi
=2 d„
0
= 4fi.

The result is certainly non-zero!

As the surface integral of Ê along the sphere is non-zero, we conclude that Ê cannot
be exact, since we know from Stokes’ theorem that the surface of integral of an exact
two-form along a closed surface must vanish.
(c) We have shown that Ê is a closed two-form that is not exact. But it does not
contradict the statement of Poincare’s lemma for two-forms, see Theorem 4.6.4, since
Ê is not defined on all of R3 ; it is not defined at the origin. It also does not contradict
version II of Poincare’s lemma that we saw in Theorem 4.6.5, since Ê is also not defined
on an open ball (the origin is missing).
6. Let F be a smooth vector field on R3 and f a smooth function on R3 . Let – : D æ R3
be a parametric surface. Show that
⁄⁄ ⁄⁄ ⁄
(f Ò ◊ F) · dS = ≠ (Òf ◊ F) · dS + f F · dr.
– – ˆ–

Solution. The key here is recall some of the vector calculus identities that we encoun-
tered previously. From Lemma 4.4.9, we know that
Ò ◊ (f F) = (Òf ) ◊ F + f Ò ◊ F.
From the point of view of differential forms, this identity follows from the graded product
rule for the exterior derivative.
Using this identity, we can write:
⁄⁄ ⁄⁄ ⁄⁄
(f Ò ◊ F) · dS = ≠ (Òf ) ◊ F + Ò ◊ (f F).
– – –
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 236

The second term on the right-hand-side is the surface integral of the curl of the vector
field f F. Using Stokes’ theorem, we can rewrite it as a line integral:
⁄⁄ ⁄
Ò ◊ (f F) = f F · dr.
– ˆ–

Therefore, we conclude that

⁄⁄ ⁄⁄ ⁄
(f Ò ◊ F) · dS = ≠ (Òf ) ◊ F + f F · dr.
– – ˆ–

5.9 Applications of surface integrals

In this section we study some applications of surface integrals, such as calculating the flux of
a vector field across a surface.

Objectives
You should be able to:

• Determine and evaluate appropriate surface integrals and flux integrals in the context
of applications in science.

5.9.1 Surface integrals as flux integrals

The main interpretation of surface integrals consists in calculating the flux of a vector field
across a surface in the direction of the normal vector. The easiest way to understand what
this means is in terms of fluid mechanics.
Suppose that the vector field v(x, y, z) is the velocity field of a fluid. Suppose that the
function fl(x, y, z) is the mass density of the fluid. Thus the vector field flv is the rate of flow
(mass per unit time) per unit area. We would like to calculate the rate of flow (mass per unit
time) of fluid crossing a surface S in the normal direction n. How can we do that?
We use the famous “divide and conquer”, or “slice it till you make it” process of integral
calculus. We divide the surface S into tiny pieces of surface, and calculate the rate of flow
through these tiny pieces of surface. Then we “sum over tiny pieces of surface”, and take the
limit of an infinite number of pieces with infinitesimal size, which turns the calculation into a
double integral.
More precisely, let dS be the area of a tiny piece of surface at (x, y, z) on S. Assuming
that S is a parametric surface, let
Tu ◊ Tv
n̂ =
|Tu ◊ Tv |
be the normalized normal vector at this point. Thus flv · n̂ is the rate of flow per unit area in
the normal direction at the point (x, y, z). Therefore, the rate of flow of fluid through this
tiny piece of surface in the normal direction is

flv · n̂dS.

But... what is the area dS? We will come back to this in Section 7.2. Suppose that the tiny
region of D that is mapped to the tiny piece of surface by the parametrization is a rectangle,
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 237

with sides of lengths du and dv. We can think of these two sides as being vectors in the u and
v directions with length du and dv. We can think of these vectors as being mapped by the
parametrization to the rescaled tangent vectors duTu and dvTv . These two vectors span a
parallelogram in the tangent plane, with area given by |duTu ◊ dvTv | = |Tu ◊ Tv |dudv. The
idea is that the area of this parallelogram is a good approximation of the area dS of the tiny
piece of surface, since the tangent plane is a good approximation of the surface. (And, when
we sum over tiny pieces of surface and take the limit of the Riemann sum, this approximation
will become exact.) As a result, we can write

dS = |Tu ◊ Tv |dA

for the area of the tiny piece of surface, with dA = dudv. Therefore, we get that the rate of
flow of fluid through this tiny piece of surface in the normal direction is:
Tu ◊ Tv
flv · n̂|Tu ◊ Tv |dA =flv · |Tu ◊ Tv |dA
|Tu ◊ Tv |
=flv · ndA,

where n = Tu ◊ Tv (not normalized).

The final step is to sum over tiny pieces of surfaces and take the limit of an infinite number
of pieces of surface of infinitesimal area, which turns the sum into a double integral. The
result is the double integral ⁄⁄
flv · n dA,
D
which we recognize as the surface integral of the vector field flv! Therefore, the surface of
integral of flv calculates the rate of flow of fluid across the surface in the normal direction.
While this was formulated for fluid velocity, one can study similar processes for other
vector fields. The result is called the “flux” of the vector field.
Definition 5.9.1 The flux of a vector field across a surface. Let F(x, y, z) be a vector
field on U ™ R3 , and – : D æ R3 an oriented parametric surface with normal vector n. The
surface integral ⁄⁄ ⁄⁄
F · dS = F(–(u, v)) · n dA
S D
is called the flux of F across S in the normal direction n. ⌃
This is of course just the standard surface integrals that we have studied already. But
it gives it an interpretation as calculating the flux of the vector field, which is why surface
integrals are also known as “flux integrals”.

5.9.2 Flux integrals beyond fluids

The main motivation for introducing the notion of flux is to calculate the rate of flow of a
fluid across a surface. But the concept of flux is also important in other physical applications.
One such example is in electromagnetism. Suppose that E(x, y, z) is an electric field. Then
the surface integral ⁄⁄
E · dS
S
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 238

calculates what is known as the electric flux across the surface S. In fact, one of the
important laws in electromagnetism is Gauss’s law, which relates the electric charge to the
flux of an electric field. More precisely, if S is a closed surface in an electric field E, Gauss’s
law states that the net charge Q enclosed by the surface S is given by
⁄⁄
Q = ‘0 E · dS,
S

i.e. it is the electric flux through S rescaled by a constant ‘0 known as the “permittivity of
free space”. Here the surface S should be given the orientation of an outward normal vector.
Example 5.9.2 The electric flux and net charge of a point source. Let q be a point
charge at the origin. The electric force follows an inverse square law, that is, the magnitude of
the electric field produced by the charge is inversely proportional to the square of the distance
from the charge. More precisely, the electric field produced by the point charge is the vector
field
q
E(x, y, z) = (x, y, z).
4fi‘0 (x + y 2 + z 2 )3/2
2

Now suppose that you want to calculate the electric flux produced by the point charge
across a sphere of radius R centered at the origin. By Gauss’s law, this should calculate the
total charge enclosed by the sphere. Since we only have a point charge at the origin, with
charge q, we expect the electric flux to be given by q. Is that what we get?
The electric flux across the sphere in the outward direction is given by the flux integral
⁄⁄
E · dS.
S

We already calculated such surface integrals in Exercise 5.6.4.5 using spherical coordinates.
The result of this exercise applies here, with C = 4fi‘
q
0
. We thus conclude that
⁄⁄
q q
E · dS = 4fi = .
S 4fi‘0 ‘0
In particular, Gauss’s law states that the total charge enclosed by the sphere is
⁄⁄
q
Q = ‘0 E · dS = ‘0 = q.
S ‘0
Phew!
We note that the flux does not depend on the radius of the sphere, as it should. In fact,
we can go further. An argument very similar to Exercise 5.7.3.5 holds here as well, but using
the divergence theorem (which we will explore in Section 6.2) instead of Green’s theorem.
The conclusion of the argument is that the flux of the electric field of a point charge at the
origin across any closed surface that encloses the origin, not just the sphere, is always equal
to q, as it should by Gauss’s law. Cool! ⇤
The concept of flux is also used for instance in the study of heat flow. Suppose that the
temperature at a point (x, y, z) in a substance is given by the function T (x, y, z). Then the
heat flow is given the gradient of the temperature function, rescaled by a constant. More
precisely, the heat flow is
F = ≠KÒT,
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 239

where K is a constant called the conductivity of the substance. We are then often interested
in calculating the rate of heat flow across a surface S, which is given by the flux of the vector
field F across S: ⁄⁄ ⁄⁄
F · dS = ≠K ÒT · dS.
S S

5.9.3 Exercises
1. Consider a fluid moving with velocity v(x, y, z) = (≠y, x, 0) and constant mass density
fl(x, y, z) = fl0 with fl0 a positive constant. (As we saw in Exercise 2.1.3.2, this type of
fluid motion is a vortex in the (x, y)-plane.) Show that the rate of flow of fluid across
the cylinder x2 + y 2 = R2 , with ≠a Æ z Æ a, for some positive constants a, R, is zero.
Explain why this result makes sense, looking at the fluid motion.
Solution. As we saw in Exercise 2.1.3.2, this type of fluid motion is a vortex in
the (x, y)-plane. Below is, first, a sketch of the velocity vector field of the fluid in
3 dimensions. But since there is no motion in the z-direction, I also present a two-
dimensional plot of the vector field in the (x, y)-plane, which makes the vortex motion more

explicit.Figure 5.9.3 The vector field v(x, y, z) = (≠y, x, 0).

CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 240

Figure 5.9.4 The vector field v(x, y) = (≠y, x), which corresponds to the projection of
the previous vector field in the (x, y)-plane.
Looking at the two-dimensional figure, we see that the motion is moving around in
a circular motion in the (x, y)-plane about the origin. In three dimensions, since there
is no motion in the z-direction, the fluid flow is circular about the z-axis. As such, if
we imagine a cylinder centered on the z-axis in the fluid, we see that there is no fluid
flowing across the surface of the cylinder. Thus we expect that the rate of flow of the
fluid across the cylinder should be zero. Let us show this.
To calculate the rate of flow across the cylinder S, we need to evaluate the surface
integral ⁄⁄
flv · dS.
S

We parametrize the surface of the cylinder as – : D æ R3 with

D = {(u, ◊) œ R2 | u œ [≠a, a], ◊ œ [0, 2fi]}

and
–(u, ◊) = (R cos(◊), R sin(◊), u).
The tangent vectors are

Tu = (0, 0, 1), T◊ = (≠R sin(◊), R cos(◊), 0).

The normal vector is

n = Tu ◊ T◊ = (≠R cos(◊), ≠R sin(◊), 0).

CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 241

It is pointing inward, but it does not matter anyway which orientation we choose as we
will show that the integral is zero. The surface integral is
⁄⁄ ⁄⁄
flv · dS = F(–(u, ◊)) · n dA
S ⁄⁄D
= (≠R sin(◊), R cos(◊), 0) · (≠R cos(◊), ≠R sin(◊), 0) dA
⁄⁄D
= (R2 sin(◊) cos(◊) ≠ R2 sin(◊) cos(◊)) dA
D
=0.

This is just as we expected: there is no rate of flow of the fluid across the cylinder.
2. Find the flux of the vector field

F(x, y, z) = (x, 0, z)

exiting the solid cone

Ò
3
V = {(x, y, z) œ R | 0 Æ z Æ 9 ≠ x2 ≠ y 2 }.

Solution. To find the flux we need to evaluate the surface integral

⁄⁄
F · dS,
S

where S is the surface boundary of the solid cone V .

The cone is shown in the figure below:

Figure 5.9.5 The solid cone V .

The boundary surface of V has two components: the lateral surface of the cone, which
we will call S1 , and the bottom disk, which we will call S2 . We need to evaluate the
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 242

surface integral on both components, and add up the results to calculate the flux of the
vector field exiting the cone.
We parametrize S1 as –1 : D1 æ R3 with

D1 = {(r, ◊) œ R2 | r œ [0, 3], ◊ œ [0, 2fi]},

with 
–(r, ◊) = (r cos(◊), r sin(◊), 9 ≠ r2 ).
The tangent vectors are
3 4
r
Tr = cos(◊), sin(◊), ≠ Ô , T◊ = (≠r sin(◊), r cos(◊), 0).
9 ≠ r2
The normal vector is
A B
r2 r2
n = Tr ◊ T◊ = Ô cos(◊), Ô sin(◊), r .
9 ≠ r2 9 ≠ r2
It points upward in the z-direction, and thus outward of the cone, as we want to calculate
the flux exiting the cone. The surface integral is then
⁄⁄ ⁄⁄ A B
1  2 r2 r2
F · dS1 = r cos(◊), 0, 9 ≠ r2 · Ô cos(◊), Ô sin(◊), r dA
S1 D1 9 ≠ r2 9 ≠ r2
⁄⁄ A B
r3 
= Ô cos2 (◊) + r 9 ≠ r2 dA
D1 9 ≠ r2
⁄ 3 ⁄ 2fi A B
r3 
= Ô cos2 (◊) + r 9 ≠ r2 d◊dr
0 0 9 ≠ r2
⁄ 3A B
r3 
=fi Ô + 2r 9 ≠ r2 dr
0 9 ≠ r2
⁄ 03 4
1u≠9 Ô
=fi Ô ≠ u du
9 2 u
⁄ 03 4
1 9
=fi ≠ u1/2 ≠ u≠1/2 du
9 2 2
5 6u=0
1 3/2 1/2
=fi ≠ u ≠ 9u
3 u=9
=36fi.

As for the bottom disk S2 , it is the disk x2 + y 2 Æ 9 in the z = 0 plane. We can

parametrize it as –2 : D2 æ R3 with

D2 = {(r, ◊) œ R2 | r œ [0, 3], ◊ œ [0, 2fi]},

and
–(r, ◊) = (r sin(◊), r cos(◊), 0).
The tangent vectors are

Tr = (sin(◊), cos(◊), 0), T◊ = (r cos(◊), ≠r sin(◊), 0).

CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 243

The normal vector is

n = Tr ◊ T◊ = (0, 0, ≠r),
which points downward, i.e. outward, which is what we want. The surface integral is
⁄⁄ ⁄⁄
F · dS2 = (r sin(◊), 0, 0) · (0, 0, ≠r) dA
S2 D2
=0.

Thus there is no flux through the bottom of the cone.

We conclude that the flux exiting the cone is given by
⁄⁄
F · dS = 36fi.
S
3. Use Gauss’s law to find the net charge enclosed by the closed surface consisting of
the cylinder x2 + y 2 = 1, ≠2 Æ z Æ 2, with its top and bottom, in the electric field
E(x, y, z) = (x, 0, 0).
Solution. To calculate the charge, by Gauss’s law we need to calculate the eletric flux,
that is, the surface integral of the electric field along the lateral surface of the cylinder as
well as its top and bottom. To do so, we need to split the surface into three components.
However, we can directly conclude that the surface integrals along the top and bottom of
the cylinder will be zero. Why? The surface integrals calculate the flux of the vector
field in the normal direction. That is, in the integrand we take the dot product of the
vector field and the normal vector. For the top and bottom of the cylinder, we know that
the normal vector will point in the z-direction (as the cylinder is centered around the
z-axis). But the z-component of E is zero, and therefore the integrand will vanish.
As a result, we only have to consider the lateral surface S of the cylinder, which we
parametrize as – : D æ R3 with

D = {(u, ◊) œ R2 | u œ [≠2, 2], ◊ œ [0, 2fi]}

and
–(u, ◊) = (cos(◊), sin(◊), u).
The tangent vectors are

Tu = (0, 0, 1), T◊ = (≠ sin(◊), cos(◊), 0),

and the normal vector is

n = Tu ◊ T◊ = (≠ cos(◊), ≠ sin(◊), 0)

This is pointing inward, so we change the sign of the normal vector. The surface integral
is then:
⁄⁄ ⁄⁄
E · dS = (cos(◊), 0, 0) · (cos(◊), sin(◊), 0) dA
– D
⁄ 2 ⁄ 2fi
= cos2 (◊) d◊du
≠2 0
CHAPTER 5. INTEGRATING TWO-FORMS: SURFACE INTEGRALS 244
⁄ 2
=fi du
≠2
=4fi.

We conclude, by Gauss’s law, that the net charge enclosed by the surface is
⁄⁄
Q = ‘0 E · dS = 4fi‘0 .
–
4. Suppose that the temperature distribution in a substance in R3 is given by the function

T (x, y, z) = x2 + xy.

Show that the rate of heat flow along any horizontal plane z = C, where C is a constant,
is zero. Explain why this is consistent with your expectation.
Solution. The heat flow is given by

F = ≠KÒT = ≠K(2x + y, x, 0).

To calculate the rate of heat flow across an horizontal plane z = C, we need to calculate
the surface integral of the heat flow F along the plane. In this process, we take the dot
product of F with the normal vector n to extract the normal component of F. Since the
plane is horizontal, its normal vector will point in the z-direction. But the vector field
F has a vanishing z-component; therefore, F · n = 0, and the surface integral along the
horizontal plane will vanish.
This is consistent with our expectation because the temperature distribution does
not depend on z. So it does not vary as we move in the z-direction; indeed, its gradient
has a vanishing z-component. As a result, there is no heat flowing through horizontal
planes, as we found.
Chapter 6

Beyond one- and two-forms

6.1 Generalized Stokes’ theorem

So far we have seen a number of theorems that take a similar form: the fundamental theorem
of calculus, the fundametal theorem of line integrals, Green’s theorem, and Stokes’ theorem.
In this section we show that they are all special cases of the mother of all integral theorems,
the “generalized Stokes’ theorem”.

Objectives
You should be able to:

• State the generalized Stokes’ Theorem.

• Show that for integration of an exact one-form over a closed interval in R, it reduces to
the Fundamental Theorem of Calculus.

• Show that for integration of an exact one-form over a parametric curve in Rn , it reduces
to the Fundamental Theorem of line integrals.

• Show that for integration of an exact two-form over a closed bounded region in R2 , it
reduces to Green’s theorem.

• Show that for integration of an exact two-form over a parametric surface in R3 , it reduces
to Stokes’ theorem.

6.1.1 The generalized Stokes’ theorem

So we far we have seen four integral theorems, all related to integration of exact one- and
two-forms: the fundamental theorem of calculus, the fundamental theorem of line integrals,
Green’s theorem, and Stokes’ theorem. While at first the theorems look different, you may
have noticed that they all take a similar form. In fact, we could write all four integral theorems
in the following form: ⁄ ⁄
dÊ = Ê.
M ˆM
All that changes is the meaning of Ê and M . More precisely:

245
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 246

1. If Ê is a zero-form (a function) on U ™ R, and M = [a, b] µ U is an oriented interval,

then it becomes the fundamental theorem of calculus, see Theorem 5.1.7.

2. If Ê is a zero-form (a function) on U ™ Rn , and M is a parametric curve – : [a, b] æ Rn

whose image is in U , then it becomes the fundamental theorem of line integrals, see
Theorem 5.1.8.

3. If Ê is a one-form on U ™ R2 , and M µ U is a closed bounded oriented region, then it

becomes Green’s theorem, see Theorem 5.7.1.

4. If Ê is a one-form on U ™ R3 , and M is a parametric surface – : D æ R3 whose image

is in U , then it becomes Stokes’ theorem, see Theorem 5.8.1.

In mathematics, when we see something like this, we dig deeper and try to determine
whether it is a coincidence or not that all these integral theorems pretty much take the same
form. More often than not, such a coincidence is a hint that there is something going on
behind the scenes, that there is a unifying principle at play. This is precisely the case here.
The unifying principle is the mother of all integral theorems, known as the “generalized
Stokes’ theorem”. It states that the relationship above is very general. The precise statement
is the following.
Theorem 6.1.1 The generalized Stokes’ theorem. Let M be a k-dimensional oriented
manifold and ˆM its boundary (which is a (k ≠ 1)-dimensional manifold) with the induced
orientation. Let Ê be a (k ≠ 1)-form on M . Then
⁄ ⁄
dÊ = Ê.
M ˆM
The scope of this theorem is really awe-inspiring, at least in the eye of a mathematician.
We will not prove this theorem as it is beyond the scope of the class, but we can try to make
sense of it.
We know what a k-form is, at least over open subsets in Rn . The key object that we have
not defined and that appears in the statement of the theorem is the notion of a “manifold”,
which is fundamental in differential geometry. So let us say a few words about manifolds.

6.1.2 An informal introduction to manifolds

The concept of manifold is essential in mathematics and physics to do calculations on
complicated geometric spaces. Informally, an n-dimensional manifold M is a space that
“locally looks like Rn ”. What does it mean? It means that for any point p œ M , one can find
an invertible map that sends an open subset of M around p to an open subset of Rn . This
map is called a “coordinate chart”; by mapping the open subset of M to an open subset of
Rn , we are basically defining coordinates on the complicated space M . This is the essence of
a manifold. The description of most manifolds however requires more than one coordinate
charts; we can always map an open subset of M to an open subset of Rn , but we cannot
generally map the whole space M to an open subset of Rn , because the global structure of the
space M may be quite complicated. The different coordinate charts are then “glued” together
in a consistent way, which gives rise to so-called “transition functions”, which are basically
changes of coordinates between different charts. To be able to do calculus on manifolds, we
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 247

usually require that these transition functions, or coordinate changes, are differentiable (or
even smooth).
In the end, the key feature of a manifold is that locally, instead of doing calculations on
the space M itself, you can use the coordinate chart to do calculations on Rn instead. How
do we do this? We use the pullback! Indeed, our coordinate chart is an invertible map, so we
can pullback objects on M via the inverse of the coordinate chart to turn them into objects
on Rn , where we can do calculus. For instance, using pullback with respect to a coordinate
chart, we can define integration of an n-form on a region of an n-dimensional manifold via
integration of an n-form over a region in Rn , which is something that we studied in this class.
We use once again the fundamental principle of reducing something complicated to something
that we already know how to solve, and the pullback is there to help! How neat is this.
Most of the spaces that we encountered in this class, such as parametric curves, parametric
surfaces, etc. are examples of manifolds. But the definition of manifolds is much more general.
A key feature of manifolds is that they are defined “intrinsically”. When we talked about
parametric curves, or parametric surfaces, we introduced complicated geometry, but the way
we did it was by embedding a curve or a surface in a higher-dimensional space Rm . For
manifolds, you do not need to embed them into higher-dimensional spaces to get interesting
geometry; the geometry is intrinsic in the definition of a manifold.
But in the end, you already know many manifolds. Here are a few examples.

• The circle is a one-dimensional manifold, with no boundary.

• The sphere (i.e. the surface of a ball) is a two-dimensional manifold, with no boundary.

• Parametric curves (the way we defined them) are one-dimensional manifolds, possibly
with boundary.

• Parametric surfaces are two-dimensional manifolds, possibly with boundary.

• The graph of a smooth function f : Rn æ R is an n-dimensional manifold.

• Spacetime, where we live, is a manifold!

6.1.3 Back to the generalized Stokes’ theorem

We now understand that an n-dimensional manifold M is basically a complicated looking
space that locally looks like Rn . If the space is orientable (as we saw for surfaces in R3 , this
is not always obvious), we can choose an orientation on M , like we did for parametric curves
and surfaces. The boundary ˆM of M is also a manifold, but one dimension less: it is a
(n ≠ 1)-dimensional manifold. The chosen orientation on M induces an orientation on the
boundary ˆM , just like we did again for parametric curves and surfaces.
Even though we haven’t defined integration of forms over manifolds, as we mentioned
above it can be done via pullback with respect to coordinate charts, and the result is that
integration of forms over manifolds is not much different from what we already did in this class.
So we can, at least informally, understand the truly beautiful statement of the generalized
Stokes’ theorem: ⁄ ⁄
dÊ = Ê.
M ˆM
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 248

To end this section, we summarize in a table how the four integral theorems that we
already saw arise as special cases of the generalized Stokes’ theorem. We add a fifth integral
theorem to the table: the divergence theorem, which is the topic of the next section.
Table 6.1.2 Integral theorems as special cases of the generalized Stokes’ theorem
M Ê Integral theorem
Closed interval in R 0-form Fundamental theorem of calculus
Parametric curve in Rn 0-form Fundamental theorem of line integrals
Closed bounded region in R2 1-form Green’s theorem
Parametric surface in R3 1-form Stokes’ theorem
Closed bounded region in R3 2-form Divergence theorem
Next time someone tells you something about the fundamental theorem of calculus, you
can reply: “oh, I know this theorem, it’s just a special case of the generalized Stokes’ theorem”!
:-)

6.2 Divergence theorem in R3

We show that the generalized Stokes’ theorem for a closed bounded region in R3 reduces to
the divergence theorem of vector calculus.

Objectives
You should be able to:

• Define integration of three-forms over closed bounded regions in R3 .

• For exact three-forms, rephrase the generalized Stokes’ theorem as the divergence
theorem in R3 .

• Summarize all the integral theorems of vector calculus as particular cases of the general-
ized Stokes’ theorem.

6.2.1 Integrating a three-form over a region in R3

Our goal in this section is to study Stokes’ theorem for two-/three-forms. As a first step we
need to define integration of three-forms over regions in R3 , just like we did for two-forms
over regions in R2 in Section 5.2 and Section 5.3.
In this section we concentrate on solid regions E µ R3 that consist of a closed surface
and its interior, such as the regiong bounded by a sphere, or a rectangular box. We saw in
Definition 5.2.1 how to define the orientation of Rn , and in particular in Example 5.2.5 for R3 ,
in which case it is given by a choice of right-handed or left-handed twirl, with right-handed
twirl being the canonical orientation. We define the orientation of a region in R3 as being
induced from the orientation of the ambient space.
Definition 6.2.1 Orientation of a region in R3 and induced orientation on the
boundary. Let E µ R3 be a solid region that consists of a closed surface and its interior. Its
boundary ˆE is the closed surface. Choose an orientation on R3 . We define the orientation
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 249

of E as being the orientation induced by the ambient vector space R3 . We write E+ for the
region with the canonical (right-handed twirl) orientation, and E≠ for the opposite (left-handed
twirl) orientation.
If E is canonically oriented, we define the induced orientation on the boundary ˆE
as corresponding to an outward pointing normal vector. ⌃
Example 6.2.2 Solid region bounded by a sphere in R3 . Let E µ R3 be the solid
region in R3 bounded by the sphere x2 + y 2 + z 2 = R2 . Its boundary ˆE is the sphere itself,
that is, the surface x2 + y 2 + z 2 = R2 . If we give E the canonical orientation given by a choice
of ordered basis on R3 corresponding to a right-handed twirl, then the induced orientation on
the sphere is that of a normal vector pointing outward. ⇤
With this definition of orientation, we can define the integral of a three-form over a region
in R3 .
Definition 6.2.3 Integral of a three-form over a closed bounded region in R3 . Let
E µ R3 be a solid region that consists of a closed surface and its interior. Let Ê = f dx·dy ·dz
be a three-form on an open subset U ™ R3 that contains E. If E has canonical orientation,
we define the integral of Ê over E as:
⁄ ⁄⁄⁄
Ê= f dV,
E E

where on the right-hand-side we mean the standard triple integral from calculus. If E has
opposite orientation, we define ⁄ ⁄⁄⁄
Ê=≠ f dV.
E E
Note that, as for the integral of two-forms over regions, the choice of basic three-form dx·dy·dz
(which is consistent with the ordering of the canonical orientation) when expressing Ê in
terms of the function f is important here, as integrals of three-forms are oriented, while triple
integrals are not. (See Remark 5.3.5.) ⌃
The definition reduces the evaluation of integrals of three-forms to triple integrals, which
you have encountered already in your previous calculus course. We will focus here on regions
that are recursively supported, as we did for regions in R2 (more general regions could be
expressed as unions of recursively supported regions). We say that:
1. A region E µ R3 is xy-supported (also called type 1) if there exists a region D in the
(x, y)-plane and two continuous functions z1 (x, y), z2 (x, y) such that

E = {(x, y, z) œ R3 | (x, y) œ D, z1 (x, y) Æ z Æ z2 (x, y)}.

2. A region E µ R3 is yz-supported (also called type 2) if there exists a region D in the

(y, z)-plane and two continuous functions x1 (y, z), x2 (y, z) such that

E = {(x, y, z) œ R3 | (y, z) œ D, x1 (x, z) Æ x Æ x2 (y, z)}.

3. A region E µ R3 is xz-supported (also called type 3) if there exists a region D in the

(x, z)-plane and two continuous functions y1 (x, z), y2 (x, z) such that

E = {(x, y, z) œ R3 | (x, z) œ D, y1 (x, z) Æ y Æ y2 (x, z)}.

CHAPTER 6. BEYOND ONE- AND TWO-FORMS 250

If the two-dimensional region D µ R2 is also supported on at least one of the two coordinates,
we say that the solid region E is recursively supported. In this case the triple integral can
be evaluated as an interated integral.
We note that in the case of rectangular regions, Fubini’s theorem still applies, and the
order of integration for the iterated integrals does not matter.
Example 6.2.4 Integral of a three-form over a recursively supported region. Let
E µ R3 be the solid region bounded by the planes x = 0, x = 2, y = 2, z = 0 and z = y. Find
the integral of the three-form
Ê = xyz dx · dy · dz
over E with canonical orientation.
By definition, we know that
⁄ ⁄⁄⁄
Ê= xyz dV.
E E

We want to evaluate the triple integral on the right-hand-side.

We can express the region E as an xy-supported region. Indeed, if D µ R3 is the
rectangular region D = [0, 2] ◊ [0, 2] in the (x, y)-plane, then

E = {(x, y, z) œ R3 | (x, y) œ D, 0 Æ z Æ y}.

In fact, using the definition of the rectangular region D, we could write

E = {(x, y, z) œ R3 | x œ [0, 2], y œ [0, 2], 0 Æ z Æ y},

which shows that E is recursively supported. We can then write the triple integral as an
iterated integral, and evaluate:
⁄⁄⁄ ⁄ 2⁄ 2⁄ y
xyz dV = xyz dzdydx
E 0 0 0
⁄ 2⁄ 2 C Dz=y
z2
= xy dydx
0 0 2 z=0
⁄ 2⁄ 2
1
= xy 3 dydx
2 0 0
⁄ 2 C Dy=2
1 y4
= x dx
2 0 4 y=0
⁄ 2
=2 x dx
0
=4.

⇤
To conclude this section, we should show that our integration theory is invariant under
orientation-preserving reparametrizations, just as we did for two-forms in Subsection 5.3.2.
We will be brief here and simply state the result. Let E1 , E2 µ R3 be recursively supported
regions, and let „ : E2 æ E1 be a bijective and invertible function (that can be extended to a
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 251

C 1 function on an open subset U ™ R3 containing E2 ).

Lemma 6.2.5 Integrals of three-forms over regions in R3 are oriented and
reparametrization-invariant. Let Ê be a three-form on an open subset U ™ R3 that
contains E1 , and „ : E2 æ E1 as above. Then:

• If det J„ > 0 in the interior of E2 ,

⁄ ⁄
„ Ê=ú
Ê.
E2 E1

• If det J„ < 0 in the interior of E2 ,

⁄ ⁄
„ Ê=≠
ú
Ê.
E2 E1
We will not write the proof here, as it is almost identical to the proof of the corresponding
statement for two-forms in Lemma 5.3.7. The key is that the pullback brings forth the
determinant of the Jacobian of the transformation, and invariance reduces to the transformation
(or change of variables) formula for triple integrals. As for two-forms, one can say that the
transformation formula for triple integrals is simply the statement the integrals
of three-forms over regions in R3 are invariant under orientation-preserving
reparametrizations.
More precisely, just as for two-forms, the transformation formula for triple integrals
involves the absolute value of the determinant of the Jacobian, while invariance under pullback
for integrals of three-forms involves the determinant of the Jacobian directly. This is because
integrals of three-forms are oriented, while triple integrals are not.

6.2.2 The divergence theorem in R3

Now that we understand integration of three-forms over regions in R3 , we can go back to the
generalized Stokes’ theorem Theorem 6.1.1 and see what it becomes when M is taken to be a
solid region E µ R3 that consists of a closed surface and its interior.
Theorem 6.2.6 The divergence theorem in R3 . Let Ê be a two-form on U ™ R3 . Let
E µ U be a solid region that consists of a closed surface and its interior, and ˆE its surface
boundary. Give E the canonical orientation, and ˆE the induced orientation corresponding to
an outward pointing normal vector. Then
⁄ ⁄
dÊ = Ê,
E ˆE

where the integral on the right-hand-side is a surface integral of Ê over the boundary ˆE
realized as a parametric surface.
In vector calculus language, if F is the vector field associated to the two-form Ê, then
⁄⁄⁄ ⁄⁄
(Ò · F) dV = F · dS,
E ˆE

where the integral on the right-hand-side is the surface integral of the vector field F over the
boundary surface ˆE with normal vector pointing outward.
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 252

We will not prove the divergence theorem here. We note that it follows directly from the
generalized Stokes’ theorem Theorem 6.1.1, just like our four other integral theorems. The
vector calculus translation follows directly from our dictionary between differential forms and
vector calculus concepts.
Remark 6.2.7 Just as for Green’s theorem, we can read the divergence theorem in two
different ways, starting from the left-hand-side or the right-hand-side. This results in two
potential applications: either to evaluate the volume integral of the divergence of a vector
field (or an exact three-form), or to evaluate the surface integral of a vector field (a two-form)
over a closed surface. However, as was the case for Green’s theorem, the divergence theorem
is mostly useful to evaluate surface integrals over closed surfaces by transforming them into
volume integrals over the interior of the region.
Example 6.2.8 Using the divergence theorem to evaluate the flux of a vector field
over a closed surface in R3 . Find the flux of the vector field

F(x, y, z) = (xy, sin(z 2 ) + y + cos(x3 ), exy )

in the outward direction over the surface of the solid region E that lies above the (x, y)-plane
and below the surface z = 2 ≠ x ≠ y 3 , x œ [≠1, 1], y œ [≠1, 1].
We could try to evaluate the surface integral directly, but given how complicated the
vector field is, this would probably be a nightmare. Or we can use the divergence theorem,
which tells us that ⁄⁄ ⁄⁄⁄
F · dS = (Ò · F)dV.
ˆE E
The divergence of the vector field is

Ò · F = y + 1,

which is of course much simpler, so using the divergence theorem looks like a good strategy.
To evaluate the volume integral we need to write the solid region E as a recursively
supported region. Let us first look at what the solid region looks like. The surface Z = 2≠x≠y 3
is shown below.
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 253

Figure 6.2.9 The surface z = 2 ≠ x ≠ y 3 over the rectangle [≠1, 1] ◊ [≠1, 1] in the (x, y)-plane.
The solid region E consists of the region bounded by the surface shown above, the four
sides of the box in the figure, and the bottom of the box. It can be written as an xy-supported
region:
E = {(x, y, z) œ R3 | x œ [≠1, 1], y œ [≠1, 1], 0 Æ z Æ 2 ≠ x ≠ y 3 }.
The volume integral can then be evaluated:
⁄⁄⁄ ⁄ 1 ⁄ 1 ⁄ 2≠x≠y3
(Ò · F)dV = (y + 1) dzdydx
E ≠1 ≠1 0
⁄ 1 ⁄ 1
3
= (y + 1) [z]z=2≠x≠y
z=0 dydx
≠1 ≠1
⁄ 1 ⁄ 1
= (2y ≠ xy ≠ y 4 + 2 ≠ x ≠ y 3 ) dydx
≠1 ≠1
⁄ 1 C Dy=1
2 xy 2 y 5 y4
= y ≠ ≠ + 2y ≠ xy ≠ dx
≠1 2 5 4 y=≠1
⁄ 1 3 4
18
= ≠ 2x dx
≠1 5
36
= .
5
Therefore, by the divergence theorem, the flux of F over the surface ˆE is equal to 36/5. ⇤

6.2.3 Exercises
1. Use the divergence theorem to find the surface integral of the two-form
Ê = 3xy 2 dy · dz + (ex z + yz 2 ) dz · dx + xy dx · dy
over the surface of the solid bounded by the cylinder y 2 + z 2 = 1 and the planes x = ≠1,
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 254

x = ≠2, with orientation given by an outward pointing normal vector.

Solution. Let E be the solid region described in the problem, and ˆE its boundary
surface with outward pointing normal vector. The divergence theorem tells us that
⁄ ⁄
Ê= dÊ,
ˆE E

where the integral on the right-hand-side is over the solid region E with canonical
orientation. Thus instead of evaluating the surface integral, we can evaluate the volume
integral of the three-form dÊ over the solid region E.
We calculate the exterior derivative:

dÊ = (3y 2 + z 2 ) dx · dy · dz.

To evaluate the integral of dÊ over E, we describe the solid region E as a recursively

supported region. In fact, it is easiest to work in cylindrical coordinates (to be clear, you
could do the whole calculation in Cartesian coordinates as well, and you would get the
same answer, but I find it easier to work with cylindrical coordinates). So we do the
change of coordinates „ : E Õ æ E with

„(u, r, ◊) = (u, r cos(◊), r sin(◊)).

The pullback of dÊ is easily calculated to be:

„ú (dÊ) =(3r2 cos2 (◊) + r2 sin2 (◊))r du · dr · d◊

=r3 (1 + 2 cos2 (◊)) du · dr · d◊.

We note that the determinant of the Jacobian is r, which is positive, and hence the
integral is invariant under the change of coordinates (pullback). Thus we can rewrite the
integral as
⁄ ⁄
dÊ = „ú (dÊ)
E Õ
⁄E
= r3 (1 + 2 cos2 (◊)) du · dr · d◊.
EÕ

The region E Õ is easily described as a rectangular region in (u, r, ◊):

E Õ = {(u, r, ◊) œ R3 | u œ [≠2, ≠1], r œ [0, 1], ◊ œ [0, 2fi]}.

The volume integral can finally be evaluated:

⁄ ⁄ ≠1 ⁄ 1 ⁄ 2fi
3 2
r (1 + 2 cos (◊)) du · dr · d◊ = r3 (1 + 2 cos2 (◊))d◊drdu
EÕ ≠2 0 0
⁄ ≠1 ⁄ 1
=4fi r3 drdu
≠2 0
⁄ ≠1
=fi du
≠2
=fi.

Therefore, the integral of Ê over the surface ˆE specified in the question is equal to fi.
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 255

2. Use the divergence theorem to find the flux of the vector field

F(x, y, z) = (z + sin(y), y, ex )

across (in the outward direction) the sphere x2 + y 2 + z 2 = 16.

Solution. Let E be the sphere and its interior, and ˆE be the sphere with outward
pointing normal vector. The divergence theorem tells us that
⁄⁄ ⁄⁄⁄
F · dS = Ò · F dV,
ˆE E

where the integral on the right-hand-side is with canonical orientation.

We first calculate the divergence of F:

Ò · F = 0 + 1 + 0 = 1.

Thus the integral that we are interested in is

⁄⁄⁄ ⁄⁄⁄
Ò · F dV = dV,
E E

which is simply the volume of the solid region E. As E is the interior of the sphere of
radius 4, we know right away that its volume is
4 256fi
fi(43 ) = ,
3 3
so we could conclude right away that the flux of the vector field across the sphere ˆE is
equal to 256fi
3 .
3. Use the divergence theorem to evaluate the surface integral of the two-form

Ê = (3x3 + yz) dy · dz + (y 3 + xz) dz · dx + (3zy 2 + x7 ) dx · dy

over the surface of the solid bounded by the paraboloid z = 1≠x2 ≠y 2 and the (x, y)-plane,
with orientation given by an inward pointing normal vector.
Solution. First, we note that the problem is asking to evaluate the surface integral
with a normal vector pointing inward. Thus, if we denote the solid region by E and its
boundary surface by ˆE the divergence theorem tells us that
⁄ ⁄
Ê=≠ dÊ,
ˆE≠ E

where on the left-hand-side we mean the surface integral with normal vector point-
ing inward, while on the right-hand-side we mean the volume integral with canonical
orientation.
We calculate the exterior derivative:

dÊ = (9x2 + 3y 2 + 3y 2 ) dx · dy · dz = 3(3x2 + 2y 2 ) dx · dy · dz.

We can either work in Cartesian or cylindrical coordinates here. But cylindrical coor-
dinates will make the calculation much easier (believe me: I first did it in Cartesian
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 256

coordinates, and it is no fun! See below. :-). So we do the change of coordinates

„ : E Õ æ E with
„(r, ◊, u) = (r cos(◊), r sin(◊), u).
The pullback of dÊ is easily calculated to be:

„ú (dÊ) =3(3r2 cos2 (◊) + 2r2 sin2 (◊))r dr · d◊ · du.

=3r3 (2 + cos2 (◊)) dr · d◊ · du.

We note that the determinant of the Jacobian is r, which is positive, and hence the
integral is invariant under the change of coordinates (pullback). Thus we can rewrite the
integral as
⁄ ⁄
dÊ = „ú (dÊ)
E Õ
⁄E
= 3r3 (2 + cos2 (◊)) dr · d◊ · du.
EÕ

The region E Õ is easily described as a recursively supported region in (r, ◊, u):

E Õ = {(r, ◊, u) œ R3 | r œ [0, 1], ◊ œ [0, 2fi], 0 Æ u Æ 1 ≠ r2 }.

The volume integral becomes

⁄ ⁄ 1 ⁄ 2fi ⁄ 1≠r2
„ (dÊ) =3
ú
r3 (2 + cos2 (◊)) dud◊dr
EÕ 0 0 0
⁄ 1 ⁄ 2fi
=3 r3 (2 + cos2 (◊))(1 ≠ r2 ) d◊dr
0 0
⁄ 1
=15fi (r3 ≠ r5 ) dr
0
3 4
1 1
=15fi ≠
4 6
5fi
= .
4
Therefore, by the divergence theorem the surface integral of the two-form that we are
asked to evaluate is equal to minus this result (the surface integral is with respect to a
normal vector pointing inward, as mentioned above), and thus is equal to ≠ 5fi 4 .
For completeness, let me do the calculation in Cartesian coordinates as well -- you
will see how much uglier it is. The solid is given by the region enclosed by the parabola
z = 1 ≠ x2 ≠ y 2 above the disk x2 + y 2 = 1 in the (x, y)-plane. We describe the solid
region as a recursively supported region:
 
E = {(x, y, z) œ R3 | x œ [≠1, 1], ≠ 1 ≠ x2 Æ y Æ 1 ≠ x2 , 0 Æ z Æ 1 ≠ x2 ≠ y 2 }.

The volume integral becomes:

⁄ ⁄ 1 ⁄ Ô1≠x2 ⁄ 1≠x2 ≠y2
dÊ =3 Ô (3x2 + 2y 2 ) dzdydx
E ≠1 ≠ 1≠x2 0
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 257

⁄ 1 ⁄ Ô1≠x2
=3 Ô (3x2 + 2y 2 )(1 ≠ x2 ≠ y 2 ) dydx
≠1 ≠ 1≠x2
⁄ 1 ⁄ Ô1≠x2
=3 Ô (3x2 ≠ 3x4 + (≠5x2 + 2)y 2 ≠ 2y 4 ) dydx
≠1 ≠ 1≠x 2
Ô
⁄ 1 5 6
y= 1≠x 2
1 2
=3 y(3x2 ≠ 3x4 + (≠5x2 + 2)y 2 ≠ y 4 ) dx
≠1 3 5 Ô
y=≠ 1≠x2
⁄ 1  3 4
2 4 1 2 2 2 2 2
=6 1 ≠ x 3x ≠ 3x + (≠5x + 2)(1 ≠ x ) ≠ (1 ≠ x )
2 dx
≠1 3 5
⁄ 1  3 4
26 22 4
=6 1 ≠ x2 ≠ x4 + x2 + dx
≠1 15 15 15
⁄ fi/2 3 4
26 22 4
=6 cos2 (◊) ≠ sin4 (◊) + sin2 (◊) + d◊
≠fi/2 15 15 15
5fi
= .
4
Here I used the trigonometric substitution x = sin(◊), and to evaluate the last trigono-
metric integral one needs to use a whole bunch of trigonometric identities (or a computer
algebra system :-). We fortunately get the same result as before for the volume integral,
and we conclude as before that the surface integral is equal to minus this result, that is,
≠ 5fi
4 .
4. Show that the volume V of a solid region E µ R3 bounded by a closed surface ˆE can
be written as ⁄
V = x dy · dz,
ˆE
where the surface integral is evaluated with the orientation given by an outward pointing
normal vector.
Solution. We know that the volume of the region E is given by the integral of the
basic three-form Ê = dx · dy · dz over E with canonical orientation:
⁄
V = dx · dy · dz.
E

But Ê = dx·dy ·dz is exact, as it can be written as Ê = d÷ for the two-form ÷ = x dy ·dz.
Therefore, by the divergence the theorem, we know that
⁄ ⁄
dx · dy · dz = x dy · dz,
E ˆE

where the integral on the right-hand-side is the surface integral with orientation given by
an upward pointing normal vector.
5. We studied the two-form
1
Ê= 2 (x dy · dz + y dz · dx + z dx · dy)
(x + y 2 + z 2 )3/2
in Remark 4.6.6 and Exercise 5.8.3.5. Ê is defined on U = R3 \ {(0, 0, 0)}. We proved
in Exercise 5.8.3.5 that Ê is closed, but that it is not exact, by showing that its surface
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 258

integral along the sphere x2 + y 2 + z 2 = 1 is equal to 4fi. In this problem we show that
the surface integral of Ê is non-vanishing for all closed surface that contain the origin
(and always equal to 4fi), while it vanishes for all closed surfaces that do not pass through
or enclose the origin.

(a) Consider an arbitrary closed surface S0 with outward pointing normal vector that
does not contain or pass through the origin. Use the divergence theorem to show
that ⁄
Ê = 0.
S0

(b) Let S1 be an arbitrary closed surface with outward pointing normal vector that
contains the origin. Explain why the argument of (a) does not apply here.

(c) Let S1 be an arbitrary closed surface with outward pointing normal vector that
contains the origin. Let K be a sphere centered at the origin, with a radius small
enough that it is contained completely inside S1 . Give K the orientation of a
normal vector pointing outward (outward of the sphere K). Use the divergence
theorem to show that ⁄ ⁄
Ê= Ê.
S1 K

(d) Using part (c), show that it implies that

⁄
Ê = 4fi.
S1

You have shown that the surface integral of Ê along any closed surface that contains
the origin is 4fi, while the surface integral of Ê along any closed surface that does not
enclose or pass through the origin is zero. Note that this is the argument that is needed
to show that, using Gauss’s law, the total charge contained within any closed surface
that encloses a point charge q at the origin is always equal to q -- see Example 5.9.2.
Solution. (a) We know that the two-form Ê is closed, that is, dÊ = 0. Furthermore,
Ê is defined on U = R3 \ {(0, 0, 0)}. If the surface S0 does not contain or pass through
the origin, then the surface S0 and the solid region E0 bounded by S0 lie within U , the
domain of definition of Ê. Therefore, by the divergence theorem,
⁄ ⁄
Ê= dÊ = 0.
S0 E0

(b) If S1 contains the origin in its interior, we cannot apply the divergence theorem
as in (a), since the solid region E1 bounded by S1 does not lie within U , the domain of
definition of Ê.
(c) We stated the divergence theorem only for solid regions that consisted of a single
closed surface and its interior, but it in fact applies to any closed bounded region. One
simply needs to make sure that each bounding surface is oriented with an outward
pointing normal vector, where “outward” means away from the solid region bounded by
the surfaces.
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 259

In particular, we can consider the solid region E that is inside S1 , but outside the
sphere K. It is bounded by the two closed surfaces S1 and K. As the origin is inside
K, it is not within the solid region E. Therefore, E µ U , and the divergence theorem
applies to this solid region. In this case, the divergence theorem says that the sum of the
surface integral over the outer boundary S1 with outward pointing normal vector and
the inner boundary K with vector pointing away from the solid region (which means
that it is pointing inside the sphere K) is equal to the volume integral over E, which is
zero since dÊ = 0: ⁄ ⁄ ⁄
Ê+ Ê= dÊ = 0.
S1 ,out K,in E

We conclude that ⁄ ⁄
Ê=≠ Ê.
S1 ,out K,in

To get rid of the minus sign, we can reverse the orientation on K, and consider a normal
vector that points outward of the sphere K. We get:
⁄ ⁄
Ê= Ê,
S1 ,out K,out

as stated in the question.

(d) We already calculated the surface integral of Ê along the sphere x2 + y 2 + z 2 = 1
in Exercise 5.8.3.5 and obtained the result 4fi. In fact, the same calculation for a sphere
of arbitrary radius x2 + y 2 + z 2 = R2 would still give 4fi (in fact we already did this
calculation in Exercise 5.6.4.5). Therefore, by (c), we conclude that the surface integral
of Ê along an arbitrary closed surface S1 that contains the origin, with outward pointing
normal vector, is: ⁄
Ê = 4fi.
S1

6.3 Divergence theorem in Rn

We show that the divergence theorem holds in Rn , not just in R3 . It follows again from the
generalized Stokes’ theorem, but we need to rewrite it a little bit to see this.

Objectives
You should be able to:

• Use the Hodge star operator to rewrite the generalized Stokes’ theorem in Rn , which
can be rewritten as the divergence theorem in Rn .

• Formulate and use the divergence theorem in Rn to calculate integrals.

6.3.1 A divergence theorem in Rn ?

In the previous section, we showed that the generalized Stokes’ theorem, in the particular
case where Ê is a two-form on R3 and M is a solid region E µ R3 , reduces to the divergence
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 260

theorem in R3 , which reads

⁄⁄⁄ ⁄⁄
(Ò · F) dV = F · dS.
E ˆE

Contrary to Green’s and Stokes’ theorem, the divergence theorem involves the divergence
of the vector field, not the curl. While the notion of curl of a vector field is not so easy to
generalize to Rn , the divergence can be generalized easily.
More precisely, let F(x1 , . . . , xn ) = (f1 , . . . , fn ) be a smooth vector field on U ™ Rn , with
f1 , . . . , fn : U æ R smooth functions. We can define the divergence of F naturally as
n
ÿ ˆfi ˆf1 ˆf2 ˆfn
Ò·F= = + + ... + .
i=1
ˆxi ˆx1 ˆx2 ˆxn

Now suppose that E µ Rn is a closed bounded region that consists of a closed (n ≠ 1)-
dimensional space ˆE and its interior. The integral of Ò · F over E is defined naturally in
calculus as a multiple (“n-tuple”) integral, which can be rewritten as an iterated integral if E
is recursively supported. The “surface” integral over ˆE can also be generalized; since ˆE is
a (n ≠ 1)-dimensional subspace in Rn , it has a well defined normal vector. If E is canonically
oriented (choose the canonical ordered basis on Rn ), we say that the induced orientation on
ˆE corresponds to an outward pointing normal vector, as for R3 . A natural question then
arise: does the divergence theorem generalize to any dimension? That is, is it true that
⁄ ⁄ ⁄ ⁄
··· (Ò · F) dVn = ··· (F · n)dVn≠1
E ˆE
¸ ˚˙ ˝ ¸ ˚˙ ˝
n times (n ≠ 1) times

where the integral on the left-hand-side is an n-tuple integral over the region E µ Rn , and
the right-hand-side is an integral of the vector field F over the parametrized surface ˆE with
normal vector pointing outward?
The answer is yes, and it again follows from the generalized Stokes’ theorem. But we need
to rewrite the generalized Stokes’ theorem a little bit to see this.

6.3.2 Rewriting the generalized Stokes’ theorem

Let us recall the generalized Stokes’ theorem from Theorem 6.1.1:
⁄ ⁄
dÊ = Ê,
M ˆM

where M is an oriented n-dimensional manifold, ˆM its boundary, and Ê a (n ≠ 1)-form.

Let us focus on a case similar to the previous section, where we take ˆE to be a closed
(n ≠ 1)-dimensional space in Rn (such as a closed surface in R3 in the previous section), and
E to be the n-dimensional region of Rn consisting of ˆE and its interior. We assign to E the
canonical orientation, and to ˆE the induced orientation corresponding to a normal vector
pointing outwards.
There is then a natural way of constructing a (n ≠ 1)-form, using the Hodge star operator
from Section 4.8. Let ÷ be a one-form on U ™ Rn . Then ıÊ is a (n ≠ 1)-form on U ™ Rn ,
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 261

by definition of the Hodge star. So we can rewrite the generalized Stokes’ theorem for the
one-form ÷ as follows: ⁄ ⁄
d(ı÷) = ı÷.
E ˆE
This is really the same generalized Stokes’ theorem, but instead of writing it in terms of a
(n ≠ 1)-form Ê, we write it in terms of a one-form ÷.
Why would that be of any use? The advantage is that we can easily translate to vector
field concepts for all Rn , since we can establish a direct translation between one-forms and
vector fields regardless of the dimension.

6.3.3 The divergence theorem in Rn

There is a natural dictionary between one-forms and vector fields in Rn . Let ÷ be a one-form
on U ™ Rn . We can write:
n
ÿ
÷= fi dxi ,
i=1
where the fi : U æ R, for i = 1, . . . , n, are smooth functions. We can associate to this
one-form the smooth vector field
F = (f1 , f2 , . . . , fn )
on U ™ Rn .
We would like to rewrite our variant of the generalized Stokes’ theorem as an integral
theorem for the vector field F. Let us first prove a lemma that will enable us to rewrite the
left-hand-side of the generalized Stokes’ theorem.
qn
Lemma 6.3.1 Rewriting the left-hand-side. Let ÷ = i=1 fi dxi be a one-form on Rn
with associated vector field F = (f1 , . . . , fn ). Then

d(ı÷) = (Ò · F) dx1 · . . . · dxn ,

and hence we can write ⁄ ⁄ ⁄

d(ı÷) = ··· (Ò · F) dVn ,
E E
¸ ˚˙ ˝
n times

which is a multiple (“n-tuple”) integral over the closed bounded region E µ Rn .

Proof. By definition of the Hodge star, we have:
n
ÿ
ı÷ = (≠1)i+1 fi dx1 · · · · dx
‰i · · · · · dxn ,
i=1

where the hat notation means that we take the wedge product of all dxj ’s except the dxi .
Calculating the exterior derivative, we get:
n
ÿ ˆfi
d(ı÷) = (≠1)i≠1 ‰i · · · · · dxn
dxi · dx1 · · · · dx
i=1
ˆxi
n
ÿ ˆfi
= dx1 · · · · · dxn
i=1
ˆxi
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 262

=(Ò · F)dx1 · · · · · dxn .

⌅
As for
s
the right-hand-side of the generalized Stokes’ theorem, we need to rewrite the
integral ˆE ı÷ in terms of vector calculus objects. ˆE is a closed (n ≠ 1)-dimensional space
in Rn . We can think of it as a parametric space – : D æ Rn for some closed bounded region
D µ Rn≠1 , like we did for parametric curves in R2 and parametric surfaces in R3 . In this we
case, we can define the integral by pulling back using the parametrization. We claim that the
following lemma holds:
Lemma 6.3.2 Rewriting the right-hand-side. If the boundary space ˆE is realized as a
parametric space – : D æ Rn ,
⁄ ⁄ ⁄
ı÷ = ··· (F · n)dVn≠1 ,
ˆE
¸ ˚˙ D˝
(n ≠ 1) times

which is a multiple (“(n ≠ 1)-tuple”) integral over the closed bounded region D µ Rn≠1 . Here,
n is the normal vector to ˆE pointing outward.1
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 263

Proof. We will not prove this statement in general; we will only prove it for parametric curves
and surfaces. In fact, for parametric surfaces, this is basically the statement that was already
proven in Corollary 5.6.5; indeed, what we have in this case is a surface integral in R3 , and
because by Table 4.1.11 we know that the vector field associated to the two-form ı÷ is the
same as the vector field associated to the one-form ÷, the result of Corollary 5.6.5 still holds
here.
Let us then show that it holds for parametric curves in R2 . In this case, ÷ = f dx + g dy,
with associated vector field F = (f, g), and ı÷ = f dy ≠ g dx. Let – : [a, b] æ R2 be a
parametric curve representing the boundary curve ˆE. Thus we have:
⁄ ⁄ ⁄
ı÷ = ı÷ = –ú (ı÷).
ˆE – [a,b]

If we write –(t) = (x(t), y(t)), the pullback is

! "
–ú (ı÷) = f (–(t))y Õ (t) ≠ g(–(t))xÕ (t) dt.
Now, the tangent vector to the parametric curve is
T(t) = (xÕ (t), y Õ (t)).
The outward pointing normal vector is then
n(t) = (y Õ (t), ≠xÕ (t)),
as the two vectors must be orthogonal, and the overall sign of the normal vector is fixed by
requiring the it points outwards. We thus see that we can write
–ú (ı÷) = (F · n) dt,
and ⁄ ⁄
ı÷ = (F · n)dt.
ˆE [a,b]
⌅
Putting this together, we see that our variant of the generalized Stokes’ theorem gives rise
to a generalization of the divergence theorem of the previous section that now holds in any
dimension.
Theorem 6.3.3 Divergence theorem in Rn . Let F be a vector field on U ™ Rn . Let
– : D æ Rn be a parametric (n ≠ 1)-dimensional space, whose image ˆE = –(D) µ U is
closed. Let E µ Rn be the region consisting of the closed surface ˆE and its interior. Let n be
the normal vector to ˆE pointing outwards. Then
⁄ ⁄ ⁄ ⁄
··· (Ò · F)dVn = ··· (F · n)dVn≠1 ,
¸ ˚˙ E˝ ¸ ˚˙ D˝
n times (n ≠ 1) times

where both sides should be understood as multiple integrals over the corresponding regions.
To be precise, we need to specify what normal vector n we are using here. In R3 , we take
the normal vector n = Tu ◊ Tv induced by a parametrization of the surface ˆE (with the right
1
To be precise, we would need to specify what normal vector we are considering here, since it is not of unit
length. As we will see, in R3 the normal vector is the usual one n = Tu ◊ Tv induced by the parametrization
of the surface, while in R2 it is the normal vector that has the same norm as the tangent vector T to the
parametric curve.
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 264

orientation); in R2 , we take the normal vector n = (y Õ (t), ≠xÕ (t)) in terms of a parmaetrization
–(t) = (x(t), y(t)) of the curve (with the right orientation),  which has the same norm as the
tangent vector to the parametric curve, that is, |n| = |T| = (xÕ (t))2 + (y Õ (t))2 .
Remark 6.3.4 In some textbooks, the divergence theorem in R2 is simply called “another
form of Green’s theorem”. The reason is that it actually follows directly from Green’s theorem.
Recall that, given a vector field F = (f, g) in R2 , Green’s theorem states that
⁄⁄ ⁄
(Ò ◊ F) · e3 dA = (F · T) dt.
D ˆD

The left-hand-side can be rewritten explicitly as

⁄⁄ ⁄⁄ 3 4
ˆg ˆf
(Ò ◊ F) · e3 dA = ≠ dA,
D D ˆx ˆy
while the right-hand-side can be rewriten as
⁄ ⁄
! "
(F · T) dt = f (–(t))xÕ (t) + g(–(t))y Õ (t) dt,
ˆD ˆD

where –(t) = (x(t), y(t)) is a parametrization of the curve ˆD.

Now if we consider a new vector field G = (≠g, f ), Green’s theorem applied to G is the
statement that ⁄⁄ ⁄
(Ò ◊ G) · e3 dA = (G · T) dt,
D ˆD
which becomes, once written out explicitly,
⁄⁄ 3 4 ⁄
ˆf ˆg ! "
+ dA = f (–(t))y Õ (t) ≠ g(–(t))xÕ (t) dt.
D ˆx ˆy ˆD

But if we rewrite this expression in terms of the original vector field F = (f, g), we get
⁄⁄ ⁄
(Ò · F) dA = (F · n) dt,
D ˆD

which is the divergence theorem in R2 for F!

So Green’s theorem and the divergence theorem in R2 are really equivalent. But we prefer
to call the later the divergence theorem in R2 as it is the special case of the general divergence
theorem in Rn .
Remark 6.3.5 Comparing Green’s theorem and the divergence theorem in R2 , it is interesting
to note that the curl is related the tangential component of the vector field, while the divergence
is related to the normal component. This is not a coincidence; if you recall from Section 4.5,
the curl and divergence of vector fields are given a physical interpretation in terms of a moving
fluid. The curl concerns whether a small sphere immersed in the fluid will rotate due to the
fluid motion -- the rotation will be induced by the tangential component of the velocity field
of the fluid on the surface of the sphere. The divergence concerns whether there is more fluid
exiting than entering a small sphere immersed in the fluid -- this is mostly influenced by the
normal component of the velocity field on the surface of the sphere. In fact, we can make
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 265

this physical interpretation of the curl and div precise by applying the Stokes’ and divergence
theorem (respectively) to the small sphere, and take a limit where the volume of the sphere
goes to zero. See for instance Section 4.4.1 in CLP 4 for this detailed calculation.

6.4 Applications of the divergence theorem

In this section we study a few applications of the divergence theorem in Rn .

Objectives
You should be able to:

• Use the divergence theorem in the context of applications in science.

6.4.1 The divergence theorem in R3 and the heat equation

Our first application concerns heat flow in R3 . First, we recall the divergence theorem in R3 .
Let F be a smooth vector field, ˆE a closed surface with normal vector pointing outward, and
E the solid region consisting of ˆE and its interior with canonical orientation. The divergence
theorem in R3 is the statement that
⁄⁄⁄ ⁄⁄
(Ò · F) dV = (F · n)dA.
E ˆE

We consider the case where the vector field F is the heat flow. Recall the context from
Subsection 5.9.2. Suppose that the temperature at a point (x, y, z) in an object (or substance)
is given by the function T (x, y, z). The heat flow is given the gradient of the temperature
function, rescaled by a constant K known as the conductivity of the substance:

F = ≠KÒT.

Now consider any closed surface ˆE, with the surface ˆE and its interior within the object.
The amount of heat flowing across the surface ˆE is given by the flux of the vector field F
across ˆE: ⁄⁄ ⁄⁄
(F · n)dA = ≠K (ÒT · n) dA.
ˆE ˆE
If we are interested in the amount of heat entering the solid region E (instead of flowing
across the surface in the outward direction), we change the sign of the flux integral. Then, by
the divergence theorem, the amount of heat entering the solid region E can be rewritten as a
volume integral:
⁄⁄ ⁄⁄⁄
K (ÒT · n) dA =K (Ò · ÒT ) dV
ˆE ⁄⁄⁄E
=K Ò2 T dV,
E

where Ò2 T is the Laplacian of the temperature function T .

Now we want to consider the situation where the temperature function T is also changing
in time. So we think of T as a function of four variables T = T (x, y, z, t). But t is just a
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 266

“spectator variable” here; we still consider the operator Ò2 in R3 , in terms of the variables
(x, y, z). So we can go through all the steps above, and we obtain that the amount of heat
entering the solid region E during a small (infinitesimal) amount of time dt is
⁄⁄⁄
Kdt Ò2 T dV.
E
2 2 2
Here the Laplacian is only in the variables (x, y, z), that is, Ò2 = ˆx ˆ
2 + ˆy 2 + ˆz 2 .
ˆ ˆ

To proceed further, we need a little bit of physics. It is known in physics that the amount
of heat energy required to raise the temperature of an object by T is given by CM T ,
where M is the mass of the object and C is constant known as the “specific heat” of the
material. Now consider the object consisting of the solid region E. In a small (infinitesimal)
amount of time dt, the temperature changes by ˆT (x,y,z,t)
ˆt dt. If we consider an infinitesimal
volume element dV in E, and fl(x, y, z) is the mass density of the solid region, then the mass
of the volume element is fldV . Thus the heat energy required to change the temperature of
the object in the time interval dt is
ˆT
Cfl dV dt.
ˆt
We then sum over all volume elements, i.e. integrate over E, to get that the total heat energy
required to change the temperature during the time interval dt is
⁄⁄⁄
ˆT
Cdt fl dV.
E ˆt
Assuming that the object is not creating heat energy itself, this heat energy should be
equal to the amount of heat entering the solid region E through the boundary surface ˆE
during the time interval dt, which is what we calculated previously. We thus obtain the
equality: ⁄⁄⁄ ⁄⁄⁄
ˆT
Cdt fl dV = Kdt Ò2 T dV.
E ˆt E
We can cancel the time interval dt on both sides. Rewriting both terms on the same side of
the equality, we get: ⁄⁄⁄ 3 4
2 ˆT
KÒ T ≠ Cfl dV = 0.
E ˆt
But this must be true for all solid regions E within the object, and for all times t. From this
we can conclude that the integrand must be identically zero:
ˆT
KÒ2 T = Cfl .
ˆt
This equation is generally rewritten as
ˆT (x, y, z, t)
= –Ò2 T (x, y, z, t),
ˆt
2 2 2
where – = Cfl
K
is called the “thermal diffusivity”, and Ò2 = ˆx
ˆ
2 + ˆy 2 + ˆz 2 .
ˆ ˆ

This equation is very famous: it is known as the heat equation. As mentioned in the
Wikipedia page on “Heat equation”,
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 267

As the prototypical parabolic partial differential equation, the heat equation is

among the most widely studied topics in pure mathematics, and its analysis is
regarded as fundamental to the broader field of partial differential equations.

The importance of the heat equation goes beyond physics and heat flow. It has a wide range
of applications, from the physics of heat flow of course, to probability theory, to financial
mathematics, to quantum mechanics, to image analysis in computer science. A generalization
of the heat equation is also behind the famous proof of the Poincare conjecture by Pereleman
in 2003 (the only Millenium Prize Problem that has been solved so far). I encourage you to
have a look at the wikipedia page on the heat equation!

6.4.2 The divergence theorem in Rn and Green’s first and second identities
We now consider the divergence theorem in Rn . Let F be a vector field, ˆE a closed (n ≠ 1)-
dimensional subspace with normal vector pointing outward, and E the region of Rn consisting
of ˆE and its interior with canonical orientation. The divergence theorem in Rn is the
statement that ⁄ ⁄ ⁄ ⁄
· · · (Ò · F)dVn = ··· (F · n)dVn≠1 .
¸ ˚˙ E˝ ¸ ˚˙ ˆE˝
n times (n ≠ 1) times

Using this theorem, we can prove the following two identities, known as Green’s first and
second identities.
Lemma 6.4.1 Green’s first identity. Let f, g : Rn æ R be functions with continuous
partial derivatives. Let E and ˆE be as above. Then
⁄ ⁄ ⁄ ⁄ ⁄ ⁄
2
··· f Ò g dVn = ··· f Òg · n dVn≠1 ≠ ··· (Òf · Òg) dVn .
¸ ˚˙ E˝ ¸ ˚˙ ˆE˝ ¸ ˚˙ E˝
n times (n ≠ 1) times n times

Proof. We consider the divergence theorem in Rn with vector field F = f Òg. By the third
identity in Lemma 4.4.9, we know that

Ò · (f F) = (Òf ) · F + f Ò · F.

Thus

Ò · (f Òg) =Òf · Òg + f Ò · Òg
=Òf · Òg + f Ò2 g.

Therefore, the divergence theorem applied to F = f Òg becomes:

⁄ ⁄ 1 2 ⁄ ⁄
··· Òf · Òg + f Ò2 g dVn = ··· f Òg · n dVn≠1 ,
E ˆE
¸ ˚˙ ˝ ¸ ˚˙ ˝
n times (n ≠ 1) times

which is the statement of Green’s first identity. ⌅

CHAPTER 6. BEYOND ONE- AND TWO-FORMS 268

This may not be obvious at first, but Green’s first identity is essentially the equivalent of
integration by parts in higher dimension. Basically, integration by parts can be written as
⁄ b -b ⁄ b
-
f dg = f g - ≠ g df.
a a a

Green’s first identity generalizes this statement for the n-tuple integral of the function f Ò2 g
over a closed bounded region E µ Rn .
Lemma 6.4.2 Green’s second identity. Let f, g : Rn æ R be functions with continuous
partial derivatives. Let E and ˆE be as above. Then
⁄ ⁄ ⁄ ⁄
2 2
··· (f Ò g ≠ gÒ f ) dVn = ··· (f Òg ≠ gÒf ) · n dVn≠1 .
¸ ˚˙ E˝ ¸ ˚˙ ˆE˝
n times (n ≠ 1) times

Proof. Green’s second identity follows from the first identity. Using the first identity, we know
that
⁄ ⁄ ⁄ ⁄
2 2
··· (f Ò g ≠ gÒ f ) dVn = ··· (f Òg ≠ gÒf ) · n dVn≠1
¸ ˚˙ E˝ ¸ ˚˙ ˆE˝
n times (n ≠ 1) times
⁄ ⁄
≠ ··· (Òf · Òg ≠ Òg · Òf ) dVn .
¸ ˚˙ E˝
n times

But Òf · Òg = Òg · Òf , and hence the last term vanishes. The result is Green’s second
identity. ⌅
Green’s identities are quite useful in mathematics. There is in fact also a third Green’s
identity, but it is beyond the scope of this class. Have a look at the Wikipedia page on
“Green’s identities” if you are interested!

6.4.3 Exercises
1. Recall that a function g : U æ R with U ™ Rn is harmonic on U if it is a solution to
q ˆ2
the Laplace equation, that is, Ò2 g = 0 on U , where Ò2 = ni=1 ˆx 2 . Use Green’s first
i
identity to show that if g is harmonic on U , and E µ U (with E and ˆE as usual), then
⁄ ⁄
··· Òg · n dVn≠1 = 0.
ˆE
¸ ˚˙ ˝
(n ≠ 1) times

Solution. We consider Green’s identity with the constant function f = 1. It reads:

⁄ ⁄ ⁄ ⁄ ⁄ ⁄
··· Ò2 g dVn = ··· Òg · n dVn≠1 ≠ ··· (Ò(1) · Òg) dVn .
E ˆE E
¸ ˚˙ ˝ ¸ ˚˙ ˝ ¸ ˚˙ ˝
n times (n ≠ 1) times n times

But Ò(1) = 0, since the gradient of a constant function necessarily vanishes. Furthermore,
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 269

since we assume that g is harmonic, we know that Ò2 g = 0. Therefore we conclude that

⁄ ⁄
··· Òg · n dVn≠1 = 0.
ˆE
¸ ˚˙ ˝
(n ≠ 1) times

2. As in the previous exercise, let g be a harmonic function on U ™ Rn , with E µ U . Use

Green’s first identity to show that if g = 0 on the boundary space ˆE (with E and ˆE
as usual), then ⁄ ⁄
··· |Òg|2 dVn = 0.
¸ ˚˙ E˝
n times

Solution. We consider Green’s first identity again, but now with f = g. It reads:
⁄ ⁄ ⁄ ⁄ ⁄ ⁄
··· gÒ2 g dVn = ··· gÒg · n dVn≠1 ≠ ··· (Òg · Òg) dVn .
¸ ˚˙ E˝ ¸ ˚˙ ˆE˝ ¸ ˚˙ E˝
n times (n ≠ 1) times n times

We assume that g is harmonic, that is, Ò2 g = 0. Furthermore, we assume that the

function g vanishes on the boundary surface ˆE, therefore the integral
⁄ ⁄
··· gÒg · n dVn≠1
¸ ˚˙ ˆE˝
(n ≠ 1) times

vanishes, since the integrand is identically zero on the surface ˆE over which we are
integrating. As a result, Green’s first identity becomes
⁄ ⁄
··· (Òg · Òg) dVn = 0.
¸ ˚˙ E˝
n times

But Òg · Òg = |Ò(g)|2 , and we obtain

⁄ ⁄
··· |Òg|2 dVn = 0.
E
¸ ˚˙ ˝
n times

6.5 Integral theorems: when to use what

This is not really an independent section (or an independent lecture), but just a brief summary
of the typical usages of the various integral theorems that we have seen so far.

Objectives
You should be able to:
• Determine which integral theorem may be useful to evaluate certain type of line and
surface integrals.
CHAPTER 6. BEYOND ONE- AND TWO-FORMS 270

We have seen five integral theorems so far, all particular cases of the generalized Stokes’
theorem:

1. The Fundamental Theorem of calculus;

2. The Fundamental Theorem of line integrals;

3. Green’s theorem;

4. Stokes’ theorem;

5. The divergence theorem.

One of the main difficulties with the integral theorems of calculus is to determine which
theorem may be helpful in a given situation. In this section I list a few typical situations
for which integral theorems may be useful, highlighting the main applications of the integral
theorems. You can use this list as a rule of thumb.
Strategy 6.5.1 Integral theorems: when to use what.

• You want to evaluate a line integral along a curve in Rn for an exact one-form (or the
gradient of a function): Fundamental Theorem of line integrals (Section 3.4).

• You want to evaluate a line integral along a closed curve in R2 : Green’s theorem
(Section 5.7).

• You want to evaluate a line integral along a closed curve in R3 : Stokes’ theorem
(Section 5.8).

• You want to evaluate a surface integral along a surface in R3 for an exact two-form (or
the curl of a vector field): Stokes’ theorem (Section 5.8).

• You want to evaluate a surface integral along a closed surface in R3 : Divergence

theorem in R3 (Section 6.2).
Chapter 7

Unoriented line and surface integrals

7.1 Unoriented line integrals

We define unoriented line integrals of functions along parametric curves. As a particular case,
we study how to calculate the arc length of a parametric curve.

Objectives
You should be able to:

• Determine the arc length of a parametrized curve in Rn using an unoriented line integral.

• Evaluate the unoriented integral of a function along a parametrized curve in Rn .

7.1.1 Unoriented line integrals

In this course we developed a theory of integration along curves and surfaces using differential
forms. By construction, our theory was oriented, as integrals of differential forms naturally
depend on a choice of orientation on the space over which we are integrating.
However, not all integrals should be oriented. Sometimes we want to calculate a quantity
associated to a curve or a surface that should not depend on a choice of orientation. Typical
examples would be the length of a curve or the area of a surface: such quantities should not
depend on a choice of orientation. As integrals of differential forms are naturally oriented,
it follows that integrals calculating arc lengths or surface areas cannot be represented as
integrals of differential forms. We need to study unoriented line and surface integrals. In this
section we look at unoriented line integrals.
Before we define the concept of unoriented line integral of a function along a parametric
curve, let us review how we defined oriented line integrals. Let – : [a, b] æ Rn be a parametric
curve, with –(t) = (x1 (t), . . . , xn (t)). We defined the oriented line integral of a one-form Ê
along the parametric curve – via pullback (Definition 3.3.2). In terms of the vector field
F = (f1 , . . . , fn ) associated to the one-form Ê, by evaluating the pullback the oriented line
integral can be rewritten as (Lemma 3.3.7):
⁄ b
(F(–(t)) · T) dt,
a

271
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 272

where T is the tangent vector

T(t) = (xÕ1 (t), . . . , xÕn (t)).

From the point of view of vector fields, the orientation of the integral is encapsulated in the
choice of tangent vector T. However, the way it is formulated, the tangent vector includes
more information than just the orientation, as it also has a non-trivial norm specified by the
parametrization. To isolate the oriented nature of the integral, we normalize the tangent
vector, and define the unit tangent vector
T Ò
T̂ = , |T| = (xÕ1 (t))2 + . . . + (xÕn (t))2 .
|T|

We can then rewrite the oriented line integral as

⁄ b1 2 ⁄ b1 2
F(–(t)) · T̂ |T| dt = F(–(t)) · T̂ ds,
a a

where we defined the “line element”

Ò
ds = |T| dt = (xÕ1 (t))2 + . . . + (xÕn (t))2 dt.

With this formulation, we see that the choice of orientation is completely encapsulated in the
expression F(–(t)) · T̂(t), which is function of t which depends on the choice of direction on
the parametric curve.
We can now see how unoriented line integrals can be naturally defined: we simply replace
the function F(–(t)) · T̂(t), constructed out of a vector field and a choice of orientation on the
curve, by an arbitrary function f (–(t)) that does not depend on a choice of orientation. This
leads to the following definition.
Definition 7.1.1 Unoriented line integrals. Let – : [a, b] æ Rn be a parametric curve,
with image curve C = –([a, b]) µ Rn , and let f : C æ R be a continuous function. We define
the unoriented line integral of f along the curve C to be
⁄ ⁄ b ⁄ b Ò
f ds = f (–(t))|T(t)| dt = f (–(t)) (xÕ1 (t))2 + . . . + (xÕn (t))2 dt.
C a a

⌃
A similar calculation as in the proof of Lemma 3.3.5 shows that unoriented line integrals
are invariant under reparametrizations, regardless of whether the reparametrization preserves
the orientation or not (the integrals are unoriented). What this means is that the line integral
does not depend on the choice of parametrization, but only on the image curve C. This is
why we wrote ⁄
f ds
C
for the unoriented line integral of the function f along the curve C µ Rn , without specifying
the parametrization –, since the integral is independent of the choice of parametrization.
Remark 7.1.2 We note that even though the notation “ds” is similar to the notation we
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 273

used for one-forms, the line element is not a one-form. For instance,
⁄ ⁄
ds = ds,
C+ C≠

i.e. the integral remains the same if we change the orientation of the curve, which would not
be the case if ds was a one-form.
Example 7.1.3 An example of an unoriented line integral. Evaluate the unoriented
line integral ⁄
xy 6 ds,
C

where C is the right half of the circle x2 + y 2 = 4.

First, we parametrize the curve as – : [≠fi/2, fi/2] æ R2 with

–(◊) = (2 cos(◊), 2 sin(◊)).

As ◊ œ [≠fi/2, fi/2], we are parametrizing the right half of the circle, as required. We do not
need to check here whether the parametrization induces the right orientation on the curve, as
we do not care about the orientation whatsoever: the integral is unoriented.
To evaluate the unoriented line integral, we need the line element ds. We calculate:
Ò
ds = (xÕ (◊))2 + (y Õ (◊))2
Ò
= 4 sin2 (◊) + 4 cos2 (◊)
=2.

Using this parametrization, the unoriented line integral becomes:

⁄ ⁄ fi/2
6
xy ds = (2 cos(◊))(2 sin(◊))6 (2) d◊
C ≠fi/2
⁄ fi/2
=256 cos(◊) sin6 (◊) d◊
≠fi/2
⁄ 1
=256 u6 du
≠1
512
= .
7
Note that we used the substitution u = sin(◊) to evaluate the trigonometric integral. ⇤

7.1.2 Arc length of a curve

A particularly important example of an unoriented line integral calculates the arc length
of a curve C. This is the most trivial example, where we choose the function that we are
integrating to be the constant function f = 1. More precisely:
Definition 7.1.4 Arc length of a curve. Let – : [a, b] æ Rn be a parametric curve, with
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 274

image curve C = –([a, b]) µ Rn . The arc length of C is given by the unoriented line integral
⁄ ⁄ bÒ
ds = (xÕ1 (t))2 + . . . + (xÕn (t))2 dt.
C a

⌃
You may have seen this formula before for the arc length, at least in or R2
It is R3 .
straightforward to justify that it calculates the arc length of the curve, using the standard
slicing argument from integral calculus. Consider a small curve segment between two points
P (–(t)) and Q(–(t + dt)). The length ds of this curve segment can be approximated by the
length of the line joining the two points, which can be written as
- -
ds ƒ |–(t + dt) ≠ –(t)| ƒ -–Õ (t)- dt,
where on the right-hand-side we kept only terms of first-order in dt. As –Õ (t) = T(t), we
recover the formula above for the line element. Finally, we sum over line elements and take
the limit of an infinite number of line element of infinitesimal size, which turns the sum into
the definite integral
⁄ b ⁄ bÒ
|T(t)| dt = (xÕ1 (t))2 + . . . + (xÕn (t))2 dt.
a a

Note that this also justifies our definition of unoriented line integrals in general above; it is
constructed via the same slicing process, but where we also introduce a function f evaluated
the point –(t) where we calculate the line element. We then sum over slices and take the limit
of infinite number of infinitesimal slices as usual, and the integral of the function over the
parametric curve becomes Definition 7.1.1.
Example 7.1.5 Calculating the arc length of a parametric curve. Find the length of
the parametric curve – : [0, 2] æ R3 with
–(t) = (1, t2 , t3 ).
We first calculate the line element ds:
Ò
ds = (xÕ (t))2 + (y Õ (t))2 + (z Õ (t))2 dt

= 4t2 + 9t4 dt

=t 4 + 9t2 dt,
Ô
where in the last line we used t2 = t since we know that t œ [0, 2] and hence it is positive.
The arc length is thus given by
⁄ ⁄ 2 
ds = t 4 + 9t2 dt
C 0
⁄
1 40 Ô
= u du
18 4
1
= (403/2 ≠ 43/2 )
27
8
= (103/2 ≠ 1),
27
2
where we used the substitution u = 4 + 9t , du = 18t dt. ⇤
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 275

7.1.3 Exercises
1. Find the arc length of the circular helix – : [0, 3] æ R3 with

–(t) = (t, 2 cos(t), 2 sin(t)).

Solution. To find the arc length, we first calculate the line element ds. The tangent
vector is
T(t) = (1, ≠2 sin(t), 2 cos(t)).
Its norm is Ò Ô
|T(t)| = 1 + 4 sin2 (t) + 4 cos( t) = 5.
Thus the line element is Ô
ds = 5 dt.
We then calculate the arc length:
⁄ ⁄ 3Ô
ds = 5 dt
– 0
Ô
=3 5.
2. Show that the arc length of the curve C at the intersection of the surfaces x2 = 2y
and 3z = xy between the origin and the point (6, 18, 36) is the answer to the ultimate
question of life, the universe, and everything.
Solution. We first parametrize the curve as – : [0, 6] æ R3 with
A B
t2 t 3
–(t) = t, , .
2 6

The tangent vector is A B

t2
T(t) = 1, t, .
2
Its norm is Û Û
3 42
t4 t2 t2
|T(t)| = 1 + t2 + = 1+ =1+ ,
4 2 2
t2
since 1 + 2 > 0. The line element is then
A B
t2
ds = 1+ dt,
2

and the arc length is

⁄ ⁄ 6A B
t2
ds = 1+ dt
C 0 2
63
=6 +
6
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 276

=42,

which is of course the answer to the ultimate question of life, the univers, and everything!
:-)
3. Evaluate the unoriented line integral
⁄
(xz + e≠y ) ds,
C

where C is the line segment between the origin and the point (1, 2, 3).
Solution. We parametrize C as – : [0, 1] æ R3 with

–(t) = (t, 2t, 3t).

The tangent vector is

T(t) = (1, 2, 3),
with norm  Ô
|T(t)| = 1 + 22 + 32 = 14.
The line element is Ô
ds = 14 dt,
and the unoriented line integral can be evaluated:
⁄ ⁄ 1 Ô
(xz + e≠y ) ds = ((t)(3t) + e≠2t ) 14 dt
C 0
Ô ⁄ 1 2
= 14 (3t + e≠2t ) dt
A0 B
Ô e≠2 1
= 14 1 ≠ +
2 2
Ô
14
= (3 ≠ e≠2 ).
2
4. Evaluate the unoriented line integral
⁄
x3 y ds,
C

where C is the circular helix parametrized by – : [0, fi/2] æ R3 with

–(t) = (cos(2t), sin(2t), t).

Solution. The tangent vector to the parametric curve is

T(t) = (≠2 sin(2t), 2 cos(2t), 1),
with norm Ò Ô
|T(t)| = 4 sin2 (2t) + 4 cos2 (2t) + 1 = 5.
The line element is Ô
ds = 5 dt.
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 277

The unoriented line integral becomes:

⁄ Ô ⁄ fi/2
3
x y ds = 5 cos3 (2t) sin(2t) dt
C 0
Ô ⁄
5 ≠1 3
=≠ u du
2 1
=0,

where we did the substitution u = cos(2t).

5. In single-variable calculus, you saw that the length of the curve y = f (x) in R2 , with
x œ [a, b], is given by the definite integral
⁄ bÒ
1 + (f Õ (x))2 dx.
a

Show that this is consistent with our definition of arc length in this section.
Solution. From our point of view, we realize the curve as the parametric curve – :
[a, b] æ R2 with
–(t) = (t, f (t)).
Then the tangent vector is
T(t) = (1, f Õ (t)),
with norm Ò
|T(t)| = 1 + (f Õ (t))2 .
So our arc length formula is
⁄ ⁄ b ⁄ bÒ
ds = |T(t)| dt = 1 + (f Õ (t))2 dt,
C a a

which is indeed the formula that you obtained in single-variable calculus.

7.2 Unoriented surface integrals

We define unoriented surface integrals of functions along parametric surfaces in R3 . As a
special case, we study how to calculate the surface area of a surface in R3 .

Objectives
You should be able to:

• Determine the surface area of a parametrized surface in R3 using an unoriented surface

integral.

• Evaluate the unoriented integral of a function along a parametrized surface in R3 .

CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 278

7.2.1 Unoriented surface integrals

In the previous section we defined unoriented line integrals over parametric curves in Rn .
Those cannot be defined as integrals of differential forms, as the integrals are independent of
a choice of orientation on the curve. We can similarly define unoriented surface integrals, to
which we now turn to. Those will be useful to calculate quantities associated to parametric
surfaces in R3 that should not depend on a choice of orientation, such as the surface area of
the surface.
We proceed in a way similar to what we did for line integrals, by first recalling our
construction of oriented surface integrals using differential forms. Let – : D æ R3 be a
parametric surface, with –(u, v) = (x(u, v), y(u, v), z(u, v)). We defined the oriented surface
integral of a two-form Ê along – via pullback to D (Definition 5.6.1). In terms of the vector
field F = (f1 , f2 , f3 ) associated to the two-form Ê, the surface integral can be written as
⁄⁄
(F(–(u, v)) · n) dA,
D

where the normal vector n is

ˆ– ˆ–
n = Tu ◊ Tv , Tu = , Tv = .
ˆu ˆv
The orientation is now encapsulated in the choice of normal vector. But as was the case
for oriented line integrals, the normal vector here contains more information than just the
orientation, as it is not normalized. We define the unit normal vector as
n Tu ◊ Tv
n̂ = = .
|n| |Tu ◊ Tv |

We can then rewrite the surface integral as

⁄⁄ ⁄⁄
(F(–(u, v)) · n̂)|Tu ◊ Tv | dA = (F(–(u, v)) · n̂) dS,
D D

where we defined the “surface element”

dS = |Tu ◊ Tv | dA.

With this rewriting, we can think of the surface integral as integrating the expression
(F(–(u, v)) · n̂), which is a function of the parameters (u, v) that depends on the choice
of orientation via the unit normal vector n̂.
Unoriented surface integrals are then naturally defined by replacing the orientation-
dependent function (F(–(u, v)) · n̂) by an arbitrary function f (–(u, v)) that does not depend
on a choice of orientation on the surface. We obtain the following definition of unoriented
surface integrals.
Definition 7.2.1 Unoriented surface integrals. Let – : D æ R3 be a parametric surface,
with –(u, v) = (x(u, v), y(u, v), z(u, v)), and image surface S = –(D) µ R3 . The tangent
vectors are
ˆ– ˆ–
Tu = , Tv = .
ˆu ˆv
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 279

Let f : S æ R be a continuous function. We define the unoriented integral of f along

the surface S to be ⁄⁄ ⁄⁄
f dS = f (–(u, v))|Tu ◊ Tv | dA.
S D
⌃
A calculation similar to the proof of Lemma 5.6.3 shows that unoriented surface integrals
are invariant under arbitrary reparametrizations of the surface S µ R3 , regardless of whether
the orientation is preserved or not. This is why we wrote
⁄⁄
f dS
S
to denote the unoriented surface integral along the surface S, as the integral does not depend
on how we parametrize the surface.
Example 7.2.2 An example of an unoriented surface integral. Evaluate the surface
integral ⁄⁄
x2 z dS
S
over the cone z2= x2
+ y2
above the (x, y)-plane and below the plane z = 1.
We first parametrize the surface as – : D æ R3 with
D = {(r, ◊) œ R2 | r œ [0, 1], ◊ œ [0, 2fi]}
and
–(r, ◊) = (r cos(◊), r sin(◊), r).
To evaluate the surface integral, we need to calculate the surface element dS. The tangent
vectors are
Tr = (cos(◊), sin(◊), 1), T◊ = (≠r sin(◊), r cos(◊), 0).
The cross product is (we note here that the order does not matter, as the integral is unoriented;
we will calculate the norm of the cross product afterwards, which is the same regardless of
whether we take Tr ◊ T◊ or T◊ ◊ Tr ):
Tr ◊ T◊ = (≠r cos(◊), ≠r sin(◊), r).
Its norm is Ò Ô
|Tr ◊ T◊ | = r2 cos2 (◊) + r2 sin2 (◊) + r2 = 2r,
Ô
where we used the fact that r2 = r since r œ [0, 1] and hence is positive.
The unoriented surface integral then becomes
⁄⁄ ⁄⁄ Ô
x2 z dS = (r cos(◊))2 (r)( 2r) dA
S D
Ô ⁄ 1 ⁄ 2fi
= 2 r4 cos2 (◊)d◊dr
0 0
Ô ⁄ 1 4
= 2fi r dr
0
Ô
2fi
= .
5
⇤
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 280

7.2.2 Surface area of a parametric surface in R3

Just as evaluating the unoriented line integral of the constant function f = 1 gave the arc
length of the parametric curve, evaluating the unoriented surface integral of the constant
function f = 1 gives the surface area of the parametric surface.
Definition 7.2.3 Surface area of a parametric surface in R3 . Let – : D æ R3 be a
parametric surface, with –(u, v) = (x(u, v), y(u, v), z(u, v)), and image surface S = –(D) µ R3 .
The surface area of S is given by the unoriented surface integral
⁄⁄ ⁄⁄
dS = |Tu ◊ Tv | dA.
S D

⌃
The justification for this definition of the surface area comes from the standard slicing
process of integral calculus, as usual. Consider a small rectangle at position (u, v) within the
domain D, and with sides of length du and dv. The area of this small rectangle is dA = dudv.
The parametrization – maps the small rectangle to a small region dS within the surface S.
This region is now curved, and not necessarily rectangular. However, one can approximate its
area by replacing it by the parallelogram in the tangent plane at the point –(u, v) spanned by
the vectors duTu and dvTv . The area of this parallelogram is dS = |Tu ◊ Tv |dudv. Finally,
we sum over all small regions within the domain D, and take a limit of an infinite number
of infinitesimal regions, which turns the approximate calculation of the surface area into an
exact one. The result of this limit process is the double integral
⁄⁄
|Tu ◊ Tv | dA.
D
Admittedly, the justification is a little bit hand-wavy here, but it can be made precise.
This also justifies our general construction of unoriented surface integrals in Definition 7.2.1:
all that we add to the slicing process is a function on the surface S that we evaluate at each
point –(u, v) on S before summing over slices to turn the calculation into a double integral.
Example 7.2.4 Calculating the surface area of a parametric surface. Find the
surface area of the part of the plane x + 2y + z = 1 that lies within the cylinder x2 + y 2 = 4.
As we know that the region will be within the cylinder x2 +y 2 = 4, we use polar coordinates
to parametrize the surface. We write down the parametrization – : D æ R3 with

D = {(r, ◊) œ R2 | r œ [0, 2], ◊ œ [0, 2fi]}

and
–(r, ◊) = (r cos(◊), r sin(◊), 1 ≠ r cos(◊) ≠ 2r sin(◊)).
We need to find the surface element dS. The tangent vectors are

Tr = (cos(◊), sin(◊), ≠ cos(◊) ≠ 2 sin(◊)), T◊ = (≠r sin(◊), r cos(◊), r sin(◊) ≠ 2r cos(◊)).

The cross product is

Tr ◊ T◊ = (r, 2r, r),
whose norm is  Ô
|Tr ◊ T◊ | = r2 + 4r2 + r2 = 6r,
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 281
Ô
where we used the fact that r2 = r since r œ [0, 2] and hence is positive.
Finally, we evaluate the surface integral of dS to get the surface area of S:
⁄⁄ ⁄⁄
dS = |Tr ◊ T◊ | dA
S D
Ô ⁄ 2 ⁄ 2fi
= 6 r d◊dr
0 0
Ô ⁄ 2
=2 6fi r dr
0
Ô
=4 6fi.

7.2.3 Exercises
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 282

1. Find the surface area of the part of the surface z = 4 ≠ 2x2 + y over the triangle with
vertices (0, 0), (2, 0), (2, 2) in the (x, y)-plane.
Solution. First, we notice that the triangle in the (x, y)-plane can be realized as the
x-supported region x œ [0, 2], 0 Æ y Æ x. We can then parametrize the surface as
– : D æ R3 with
D = {(u, v) œ R2 | u œ [0, 2], 0 Æ v Æ u}
and
–(u, v) = (u, v, 4 ≠ 2u2 + v).
The tangent vectors are

Tu = (1, 0, ≠4u), Tv = (0, 1, 1).

The cross product is

Tu ◊ Tv = (4u, ≠1, 1).
Its norm is  Ô 
|Tu ◊ Tv | = 16u2 + 1 + 1 = 2 8u2 + 1.
The surface area becomes
⁄⁄ Ô ⁄ 2⁄ u
dS = 2 8u2 + 1 dvdu
S 0 0
Ô ⁄ 2 
= 2 u 8u2 + 1 du
0
Ô ⁄
2 33 Ô
= t dt
16
Ô 1
2 3/2
= (33 ≠ 1)
24Ô Ô
11 66 2
= ≠ .
8 24
where we did the substitution t = 8u2 + 1.
2. Show that the surface
Ô area of the lateral surface of a circular cone with radius R and
height H is A = fiR R2 + H 2 .
Solution. The equation of a circular cone with radius R and height H (with apex at
the origin) is
R2 2
z = x2 + y 2 , z Ø 0.
H2
We parametrize the cone as – : D æ R3 with

D = {(r, ◊) œ R2 | r œ [0, R], ◊ œ [0, 2fi]}

and 3 4
rH
–(r, ◊) = r cos(◊), r sin(◊), .
R
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 283

The tangent vectors are

3 4
H
Tr = cos(◊), sin(◊), , T◊ = (≠r sin(◊), r cos(◊), 0) .
R
The cross product is
3 4
Hr Hr
Tr ◊ T◊ = ≠ cos(◊), ≠ sin(◊), r .
R R
Its norm is
Û Û
H 2 r2 H 2 r2 H2
|Tr ◊ T◊ | = cos 2 (◊) + sin2 (◊) + r2 = r + 1,
R2 R2 R2

since r œ [0, R] and hence is positive. We then calculate the surface area:
Û
⁄⁄ ⁄ R ⁄ 2fi
H2
dS = +1 r d◊dr
S R2 0 0
Û
⁄ R
H2
=2fi +1 r dr
R2 0
Û
H2
=fiR2 +1
R2

=fiR H 2 + R2 .
3. Evaluate the unoriented surface integral
⁄⁄
xy dS,
S

where S is the part of the plane x + 2y + z = 4 that lies in the first octant.
Solution. We first need to parametrize the surface. We want the part of the plane that
lies in the first octant, so we must have x Ø 0, y Ø 0, and z Ø 0. The equation of the
plane is z = 4 ≠ x ≠ 2y. Since z Ø 0 and y Ø 0, we know that x cannot be more than 4.
So we take x œ [0, 4]. Then, since z Ø 0, we know that 4 ≠ x ≠ 2y Ø 0, which means that
2y Æ 4 ≠ x, that is, y Æ 2 ≠ x2 . So we know that 0 Æ y Æ 2 ≠ x2 .
Now that we identified the region in the (x, y)-plane over which the part of the plane
is, we can parametrize it as – : D æ R3 , with
u
D = {(u, v) œ R2 |u œ [0, 4], 0 Æ v Æ 2 ≠ },
2
and
–(u, v) = (u, v, 4 ≠ u ≠ 2v) .
The tangent vectors are

Tu = (1, 0, ≠1), Tv = (0, 1, ≠2),

with cross product

Tu ◊ Tv = (1, 2, 1),
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 284

whose norm is  Ô
|Tu ◊ Tv | = 1 + 22 + 1 = 6.
We calculate the unoriented surface integral:
⁄⁄ Ô ⁄ 4 ⁄ 2≠ u2
xy dS = 6 uv dvdu
S
Ô ⁄0 0 3 4
6 4 u 2
= u 2≠ du
2 0 2
Ô ⁄ A B
6 4 2 u3
= 4u ≠ 2u + du
2 0 4
Ô A 4
B
6 2 4
= 2(42 ) ≠ 43 +
2 3 16
Ô
8 6
= .
3
4. Consider the surface S = {y = f (x)} µ R3 , where f : R æ R is a smooth function, and
x œ [0, 2], z œ [0, 1]. Show that the surface area of S is equal to the arc length of the
curve y = f (x), x œ [0, 2], in the (x, y)-plane.
Solution. We can parametrize the surface S by – : D æ R3 with

D = {(u, v) œ R2 | u œ [0, 2], v œ [0, 1]},

and
–(u, v) = (u, f (u), v).
The tangent vectors are

Tu = (1, f Õ (u), 0), Tv = (0, 0, 1),

the cross product is

Tu ◊ Tv = (f Õ (u), ≠1, 0),
whose norm is Ò
|Tu ◊ Tv | = 1 + (f Õ (u))2 .
The surface area is then
⁄⁄ ⁄ 2⁄ 1Ò
dS = 1 + (f Õ (u))2 dvdu
S 0 0
⁄ 2Ò
= 1 + (f Õ (u))2 du.
0

But this is precisely the formula for the arc length of the parametric curve — : [0, 2] æ R2
with —(u) = (u, f (u)), which is the curve {y = f (x)} µ R2 with x œ [0, 2].
By the way, this result is not surprising. Indeed, the surface S is basically just the
curve y = f (x) extended uniformly in the z-direction. So the surface area should be
the arc length of the curve times the length of the surface in the z-direction. Since S
is defined with z œ [0, 1], the length in the z-direction is just 1, so we the surface area
should be equal to the arc length of the curve, as we obtained.
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 285

5. In single-variable calculus, you saw that the surface area of the solid of revolution obtained
by rotating the curve y = f (x), x œ [a, b] (and a Ø 0), about the y-axis is given by the
definite integral
⁄ b Ò
A = 2fi x 1 + (f Õ (x))2 dx.
a
Show that this is consistent with our formula for the surface area of parametric surfaces.
Solution. The surface S obtained by rotating the curve y = f (x), x œ [a, b], about the
y-axis, can be parametrized as follows. First, we can parametrize the curve y = f (x) in
the (x, y)-plane as (u, f (u), 0), u œ [a, b]. When we rotate the curve about the y-axis,
for a fixed value of u, the point (u, 0) in the (x, z)-plane gets rotated about the y-axis
around a circle of radius u. So we should replace (u, 0) by (u cos(◊), u sin(◊)). This gives
a parametrization for the surface S as – : D æ R3 with

D = {(u, ◊) œ R2 | u œ [a, b], ◊ œ [0, 2fi]}

with
–(u, ◊) = (u cos(◊), f (u), u sin(◊)).
The tangent vectors are

Tu = (cos(◊), f Õ (u), sin(◊)), T◊ = (≠u sin(◊), 0, u cos(◊)).

The cross product is

Tu ◊ T◊ = (uf Õ (u) cos(◊), ≠u, uf Õ (u) sin(◊)).

Its norm is
Ò Ò
|Tu ◊ T◊ | = u2 (f Õ (u))2 cos2 (◊) + u2 (f Õ (u))2 sin2 (◊) + u2 = u 1 + (f Õ (u))2 ,

since u œ [a, b], a Ø 0, and hence u is positive. The surface area is then:
⁄⁄ ⁄ b ⁄ 2fi Ò
dS = u 1 + (f Õ (u))2 d◊du
S a 0
⁄ b Ò
=2fi u 1 + (f Õ (u))2 du,
a

which is the formula that you obtained in single-variable calculus!

7.3 Applications of unoriented line and surface integrals

While most applications of integration over curves and surfaces involve the oriented line and
surface integrals that we studied throughout this course, unoriented line and surface integrals
can also be useful in applications, when we want to calculate a quantity associated to a
curve or a surface that should not depend on the orientation. In this section we study a few
applications of unoriented line and surface integrals, such as calculating the centre of mass of
a wire and a thin sheet of material.
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 286

Objectives
You should be able to:

• Determine and evaluate unoriented line and surface integrals in the context of applications
in science.

7.3.1 Centre of mass of a wire

In the previous sections we studied unoriented line and surface integrals of functions. Those
are useful to calculate quantities associated to curves and surfaces that should not depend on
a choice of orientation. The prototypical examples were the arc length of a curve and surface
area of a surface, but there are many other applications.
Our first application concerns the calculation of the centre of mass of a wire in Rn . Let us
first recall the physical concept of centre of mass.
Suppose that there are k point particles of masses m1 , . . . , mk at positions X1 , . . . , Xk œ R
on a line. The centre of mass of the system of particles is located at
qk
i=1 mi Xi
x̄ = ,
m
q
where m = ki=1 mi is the total mass of the system. The numerator is sometimes called the
“first moment of the system about the origin” and written as M .
If we are given instead a rod in R between x = a and x = b with mass density fl(x), then
to get its centre of mass we use the slicing principle. We slice the rod into small line segments.
Let dx be of a typical line segment located at the point x. Its mass is dm = fl(x) dx, and its
first moment about the origin is dM = xfl(x) dx. We then sum over slices and take the limit
of an infinite number of infinitesimal slices. This turns the calculation of the total mass and
first moment of the rod as a definite integral, and its centre of mass is:
sb ⁄ b
xfl(x) dx1
x̄ = s b
a
= xfl(x) dx.
a fl(x) dx
m a

Now what if we want to calculate the centre of mass of a wire that is bent and twisted in
Rn ? We apply the same idea, but thinking of the wire as a parametric curve – : [a, b] æ Rn ,
with –(t) = (x1 (t), . . . , xn (t)) and image curve C = –([a, b]). Let fl : C æ R be the mass
density of the wire, which we assume to be continuous. We slice the wire (the image curve C)
into small curve segments. Let ds (the line element from Subsection 7.1.1) be the length of
a typical curve segment located at –(t). Its mass is dm = fl(–(t)) ds. By summing over all
curve segments and taking the limit of an infinite number of segments of infinitesimal length,
we calculate the total mass of the wire as an unoriented line integral:
⁄ ⁄ b
m= fl ds = fl(–(t))|T(t)| dt.
C a

To get the centre of mass, we also need to calculate the first moments of the wire. Here, we
are working in Rn . There are n first moments, one in each coordinate x1 , . . . , xn . The first
moments of the curve segment are given by

xk (t)fl(–(t)) ds, k = 1, . . . , n.
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 287

Summing over curve segments, and taking the limit as usual, we obtain that the position of
the centre of mass of the wire is given by the point in Rn with coordinates x̄1 , . . . , x̄n , with:
⁄ ⁄
1 1 b
x̄k = xk fl ds = xk (t)fl(–(t))|T(t)| dt, k = 1, . . . , n.
m C m a
In other words, to find the centre of mass of the wire, we need to evaluate the unoriented line
integrals corresponding to the total mass of the wire and its n first moments.
Note that the centre of mass of the wire will not generally be located on the wire itself,
since the wire is twisted and bent in the ambient space Rn ; this will be clear in the next
example.
Example 7.3.1 Finding the centre of mass of a wire in R2 . A wire is bent into the
semi-circle x2 + y 2 = 4, y Ø 0. Its mass density is given by the function

fl(x, y) = 4 ≠ y

on the wire. Find the centre of mass of the wire.

Let us see what we expect first. The wire is heavier near its base (y = 0) than at the top
(y = 2). The mass density is however symmetric about the y-axis (it does not vary in x). We
thus expect the centre of mass to be located at a point (0, ȳ), with ȳ < 2, since it should be
below the top of the wire.
To calculate the centre of mass, we will need to calulate unoriented line integrals along
the curve, so we first parametrize the curve as – : [0, fi] æ R2 with

–(◊) = (2 cos(◊), 2 sin(◊)).

The tangent vector is

T◊ = (≠2 sin(◊), 2 cos(◊)),
whose norm is Ò
|T◊ | = 4 sin2 (◊) + 4 cos2 (◊) = 2.
The line element is then
ds = |T◊ |d◊ = 2d◊.
We first calculate the total mass of the wire. It is given by the unoriented line integral:
⁄
m= fl ds
⁄Cfi
= (4 ≠ 2 sin(◊))2d◊
0
=4(2fi ≠ 2)
=8(fi ≠ 1).

We then calculate the coordinates x̄ and ȳ of the centre of mass.

⁄
1
x̄ = xfl ds
m C
⁄ fi
1
= (2 cos(◊))(4 ≠ 2 sin(◊))2 d◊
8(fi ≠ 1) 0
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 288

=0,

since the trigonometric integral vanishes. As for ȳ, we get:

⁄
1
ȳ = yfl ds
m C
⁄
1 fi
= (2 sin(◊))(4 ≠ 2 sin(◊))2 d◊
8(fi ≠ 1) 0
3 4
8 fi
= 4≠
8(fi ≠ 1) 2
8≠fi
=
2(fi ≠ 1)

Therefore, the centre of mass of the wire is located at the point

3 4
8≠fi
(x̄, ȳ) = 0, .
2(fi ≠ 1)

To make sure that this is consistent with our expectation, we can find a numerical value for
the location of the centre of mass. We get:

(x̄, ȳ) ƒ (0, 1.134).

This is consistent with our expectation. Phew! ⇤

7.3.2 Centre of mass of a thin sheet

We can do a very similar calculation to obtain the centre of mass of a thin sheet of material
(such as aluminium foil, or paper) in R3 . Suppose that the sheet of material takes the shape
of a parametric surface – : D æ R3 , with –(u, v) = (x(u, v), y(u, v), z(u, v)) and image surface
S = –(D). Suppose that fl : S æ R is the mass density of the sheet (mass per unit area).
Using the same slicing approach, but with small pieces of surface of area dS, we obtain that
the total mass of the sheet can be written as the unoriented surface integral
⁄⁄ ⁄⁄
m= fl dS = fl(–(u, v))|Tu ◊ Tv | dA.
S D
We calculate the centre of mass as before, by calculating the first moments in the three
coordinates x, y, z. Using the slicing process, those become unoriented surface integrals. The
result is that position (x̄, ȳ, z̄) of the centre of mass is given by
⁄⁄ ⁄⁄
1 1
x̄ = xfl dS = x(u, v)fl(–(u, v))|Tu ◊ Tv | dA,
m S m D
⁄⁄ ⁄⁄
1 1
ȳ = yfl dS = y(u, v)fl(–(u, v))|Tu ◊ Tv | dA,
m S m D
⁄⁄ ⁄⁄
1 1
z̄ = zfl dS = z(u, v)fl(–(u, v))|Tu ◊ Tv | dA.
m S m D
Note that, as was the case for the wire, the centre of mass of the sheet is not generally expected
to lie on the sheet itself, since the sheet can be bent and twisted in the ambient space R3 .
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 289

Example 7.3.2 Finding the centre of mass of a sheet in R3 . Suppose the a sheet of
paper is bent in the shape of the cylinder x2 + y 2 = 9, with z œ [0, 1]. Suppose that its mass
density is given by the function
fl(x, y, z) = z + 1.
Find the centre of mass of the cylinder.
First, let us see what we expect. The cylinder is heavier at the top than at the bottom.
However, its mass density has circular symmetry above the z-axis, as it only depends on
the height z on the cylinder. Thus we expect the centre of mass to be in the middle of the
cylinder, i.e. with coordinates (0, 0, z̄). Furthermore, we expect its z-coordinate to be a little
bit higher than half-way up the cylinder, since the cylinder is heavier at the top than at the
bottom. So we expect 0.5 < z̄ < 1.
To calculate the required unoriented surface integrals, we first parametrize the surface as
– : D æ R3 with
D = {(u, ◊) œ R2 | u œ [0, 1], ◊ œ [0, 2fi]}
and
–(u, ◊) = (3 cos(◊), 3 sin(◊), u).
The tangent vectors are

Tu = (0, 0, 1), T◊ = (≠3 sin(◊), 3 cos(◊), 0),

and the cross product is

Tu ◊ T◊ = (≠3 cos(◊), ≠3 sin(◊), 0).

Its norm is Ò
|Tu ◊ T◊ | = 9 cos2 (◊) + 9 sin2 (◊) = 3,
and thus the surface element is
dS = 3dud◊.
We calculate the total mass of the sheet of paper. We get:
⁄⁄
m= fl dS
S
⁄ 1 ⁄ 2fi
= (u + 1)3 d◊du
0 0
⁄ 1
=6fi (u + 1) du
0
=9fi.

As for the coordinates of the centre of mass, we get:

⁄⁄
1
x̄ = xfl dS
m S
⁄ ⁄
1 1 2fi
= (3 cos(◊))(u + 1)3 d◊du
9fi 0 0
=0,
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 290

since the integral over ◊ is zero. Similarly,

⁄⁄
1
ȳ = yfl dS
m S
⁄ ⁄
1 1 2fi
= (3 sin(◊))(u + 1)3 d◊du
9fi 0 0
=0.

As for z̄, we get:

⁄⁄
1
z̄ = zfl dS
m S
⁄ ⁄
1 1 2fi
= (u)(u + 1)3 d◊du
9fi 0 0
⁄
2 1 2
= (u + u) du
3 0
5
= .
9
As a result, the centre of mass of the sheet of paper is located at the point
3 4
5
(x̄, ȳ, z̄) = 0, 0, .
9

Since 59 ƒ 0.556,, this is consistent with our expectation that the centre of mass should be on
the z-axis, a little bit higher than half-way up the cylinder, which is at z = 0.5. ⇤

7.3.3 Exercises
1. Find the centre of mass of a wire that is bent in the shape of a circle of radius R centered
at the origin in R2 , with mass density fl(x, y) = 2R ≠ y. Is your result consistent with
your expectations?
Solution. Let us first see what we expect. As the wire is bent in a circle centered at
the origin, if the mass density was constant (or symmetric about the x- and y-axes), the
centre of mass would be at the origin. However, the mass density is not constant here: it
decreases linearly with y. It is constant in the x-direction, so we expect that centre of
mass to lie on the y-axis. As the mass decreases as y increases, we expect the centre of
mass to be at a position (0, ȳ), with ≠R < ȳ < 0.
We parametrize the circle of radius R as – : [0, 2fi] æ R2 with –(◊) = (R cos(◊), R sin(◊)).
The tangent vector is
T(◊) = (≠R sin(◊), R cos(◊)).
Its norm is Ò
|T(◊)| = R2 sin2 (◊) + R2 cos2 (◊) = R.
The total mass of the wire is
⁄ ⁄ 2fi
m= fl ds = (2R ≠ R sin(◊))Rd◊
C 0
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 291

=4fiR2 .

The x̄-coordinate of the centre of mass is

⁄
1
x̄ = xfl ds
m C
⁄ 2fi
1
= R cos(◊)(2R ≠ R sin(◊))R d◊
4fiR2 0
=0,

as expected. As for the ȳ-coordinate, we get:

⁄
1
ȳ = yfl ds
m C
⁄
1 2fi
= R sin(◊)(2R ≠ R sin(◊))R d◊
4fiR 02
1
= (≠R3 fi)
4fiR2
R
=≠ .
4
Therefore, the position of the centre of mass is
3 4
R
(x̄, ȳ) = 0, ≠ .
4
As expected, it is on the y-axis, and the y-coordinate is between ≠R and 0.
2. Given a wire with density fl(x, y) that lies on a curve C µ R2 , its moments of inertia
about the x- and the y-axes are defined by
⁄ ⁄
2
Ix = y fl ds, Iy = x2 fl ds.
C C

Find the moments of inertia of a wire that lies along the line 2x + y = 5 between x = 0
and x = 1, with density fl(x, y) = x.
Solution. We parametrize the line as – : [0, 1] æ R2 with

–(t) = (t, 5 ≠ 2t).

The tangent vector and its norm are:

 Ô
T(t) = (1, ≠2), |T(t)| = 1 + 22 = 5.

We can then evaluate the moments of inertia. First,

⁄
Ix = y 2 fl ds
C
Ô ⁄ 1
= 5 (5 ≠ 2t)2 t dt
0
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 292

Ô ⁄ 1
= 5 (25t ≠ 20t2 + 4t3 ) dt
0
Ô 3 25 20 4
= 5 ≠ +1
2 3
Ô
41 5
= .
6
Second,
⁄
Iy = x2 fl ds
C
Ô ⁄ 1 3
= 5 t dt
0
Ô
5
= .
4
3. Consider a sheet of aluminium foil that is bent in the shape of a paraboloid z = x2 + y 2
with z œ [0, 1]. Suppose that it mass density is fl = k where k is a constant. Find the
total mass of the sheet and its centre of mass. Does it agree with your expectation?
Solution. As the mass density is constant, and the paraboloid has circular symmetry
about the z-axis, we expect the centre of mass to lie on the z-axis, somewhere between
z = 0 and z = 1.
We parametrize the paraboloid as – : D æ R3 with

D = {(r, ◊) œ R2 | r œ [0, 1], ◊ œ [0, 2fi]}

and
–(r, ◊) = (r cos(◊), r sin(◊), r2 ).
The tangent vectors are

Tr = (cos(◊), sin(◊), 2r), T◊ = (≠r sin(◊), r cos(◊), 0).

The cross product is

1 2
Tr ◊ T◊ = ≠2r2 cos(◊), ≠2r2 sin(◊), r .

Its norm is
Ò 
|Tr ◊ T◊ | = 4r4 cos2 (◊) + 4r4 sin2 (◊) + r2 = r 4r2 + 1.

We first calculate the total mass. We get:

⁄⁄
m= fl dS
S
⁄ 1 ⁄ 2fi 
=k r 4r2 + 1 d◊r
0 0
⁄ 1 
=2fik r 4r2 + 1 dr
0
CHAPTER 7. UNORIENTED LINE AND SURFACE INTEGRALS 293
⁄
fik 5 Ô
= u du
4 1
fik 1 3/2 2
= 5 ≠1 .
6
In the process of evaluating the integral we did the substitution u = 4r2 + 1.
Next we calculate the coordinates of the centre of mass. First,
⁄⁄
1
x̄ = xfl dS
m S
⁄ 1 ⁄ 2fi 
k
= r2 cos(◊) 4r2 + 1 d◊r
m 0 0
=0,

as expected. Next,
⁄⁄
1
ȳ = yfl dS
m S
⁄ 1 ⁄ 2fi 
k
= r2 sin(◊) 4r2 + 1 d◊r
m 0 0
=0,

as expected. Finally,
⁄⁄
1
z̄ = zfl dS
m S
⁄ 1 ⁄ 2fi
6k 
= ! " r3 4r2 + 1 d◊r
fik 53/2 ≠ 1 0 0
⁄ 1 
12
= ! 3/2 " r3 4r2 + 1 dr
5 ≠1 0
⁄ 5
3 Ô
= ! 3/2 " (u ≠ 1) u du
8 5 ≠1 1
3 4
3 2 5/2 2 2 3/2 2
= ! " 5 ≠ ≠ 5 +
8 53/2 ≠ 1 5 5 3 3
1 55/2 + 1
= .
10 53/2 ≠ 1

As before, to evaluate the integral we did the substitution u = 4r2 + 1.

Therefore, the centre of mass is located at
A B
1 55/2 + 1
(x̄, ȳ, z̄) = 0, 0, ƒ (0, 0, 0.559).
10 53/2 ≠ 1

As expected, it is located on the z-axis, somewhere between z = 0 and z = 1.

Appendix A

List of results

Chapter 1 A preview of vector calculus

Chapter 2 One-forms and vector fields

Lemma 2.2.10 Exact one-forms in R2 are closed
Lemma 2.2.11 Screening test for conservative vector fields in R2
Lemma 2.2.15 Exact one-forms in R3 are closed
Lemma 2.2.16 Screening test for conservative vector fields in R3
Lemma 2.4.4 The pullback of dx
Lemma 2.4.5 The pullback of a one-form

Chapter 3 Integrating one-forms: line integrals

Lemma 3.1.5 Integrals of one-forms over intervals are invariant under orientation-preserving reparametri
Lemma 3.1.6 Integrals of one-forms over intervals pick a sign under orientation-reversing reparametrizat
Lemma 3.2.6 Parametric curves are oriented
Lemma 3.2.9 Reparametrizations of a curve
Lemma 3.2.11 Orientation-preserving reparametrizations
Lemma 3.3.5 Line integrals are invariant under orientation-preserving reparametrizations
Lemma 3.3.7 Line integrals in terms of vector fields
Theorem 3.4.1 The Fundamental Theorem of line integrals
Corollary 3.4.2 The line integrals of an exact form along two curves that start and end at the same points
Corollary 3.4.3 The line integral of an exact one-form along a closed curve vanishes
Theorem 3.4.5 The Fundamental Theorem of line integrals for vector fields
Theorem 3.6.1 Poincare’s lemma, version I
Theorem 3.6.3 Equivalent formulations of exactness on Rn
(Cont

294
APPENDIX A. LIST OF RESULTS 295

Theorem 3.6.4 Poincare’s lemma, version II

Chapter 4 k-forms
Lemma 4.1.6 Antisymmetry of basic k-forms
Lemma 4.2.6 Comparing Ê · ÷ to ÷ · Ê
Lemma 4.2.7 The wedge product of two one-forms is the cross-product of the associated vector fields
Lemma 4.2.8 The wedge product of a one-form and a two-form is the dot product of the associated vect
Lemma 4.3.2 The exterior derivative in R3
Lemma 4.3.6 The graded product rule for the exterior derivative
Lemma 4.3.9 d2 = 0
Lemma 4.4.9 Vector calculus identities, part 1
Lemma 4.4.10 Vector calculus identities, part 2
Lemma 4.4.11 Vector calculus identities, part 3
Lemma 4.4.12 Vector calculus identities, part 4
Lemma 4.6.3 Exact k-forms are closed
Theorem 4.6.4 Poincare’s lemma for k-forms, version 1
Theorem 4.6.5 Poincare’s lemma for k-forms, version II
Lemma 4.7.1 The pullback of a k-form
Lemma 4.7.4 The pullback commutes with the exterior derivative
Lemma 4.7.7 The pullback of a top form in Rn in terms of the Jacobian determinant
Theorem 4.7.8 Inverse Function Theorem
Theorem 4.7.9 Implicit Function Theorem
Lemma 4.7.11 An explicit formula for the pullback of a basic one-form
Lemma 4.7.13 The pullback commutes with the wedge product
Corollary 4.7.14 An explicit formula for the pullback of a basic k-form
Lemma 4.7.15 The pullback of a basic n-form in Rn
Lemma 4.8.8 The Laplace-Beltrami operator and the Laplacian of a function
Lemma 4.8.9 The Laplace-Beltrami operator and the Laplacian of a vector field
Lemma 4.8.10 Vector calculus identities, part 5

Chapter 5 Integrating two-forms: surface integrals

Theorem 5.1.7 The Fundamental Theorem of Calculus
Theorem 5.1.8 The Fundamental Theorem of line integrals
Lemma 5.3.7 Integrals of two-forms over regions in R2 are invariant under orientation-preserving reparam
Lemma 5.5.3 Parametric surfaces are oriented
Lemma 5.5.7 Orientation-preserving reparametrizations
Lemma 5.6.3 Surface integrals are invariant under orientation-preserving reparametrizations
Lemma 5.6.4 The pullback of a two-form along a parametric surface in terms of vector fields
Corollary 5.6.5
Theorem 5.7.1 Green’s theorem
(Cont
APPENDIX A. LIST OF RESULTS 296

Theorem 5.8.1 Stokes’ theorem

Corollary 5.8.2 The surface integrals of an exact two-form along two surfaces that share the same oriented
Corollary 5.8.3 The surface integral of an exact two-form along a closed surface vanishes
Theorem 5.8.11 Stokes’ theorem for vector fields

Chapter 6 Beyond one- and two-forms

Theorem 6.1.1 The generalized Stokes’ theorem
Lemma 6.2.5 Integrals of three-forms over regions in R3 are oriented and reparametrization-invariant
Theorem 6.2.6 The divergence theorem in R3
Lemma 6.3.1 Rewriting the left-hand-side
Lemma 6.3.2 Rewriting the right-hand-side
Theorem 6.3.3 Divergence theorem in Rn
Lemma 6.4.1 Green’s first identity
Lemma 6.4.2 Green’s second identity

Chapter 7 Unoriented line and surface integrals

Appendix B

List of definitions

Chapter 1 A preview of vector calculus

Chapter 2 One-forms and vector fields

Definition 2.1.1 One-forms
Definition 2.1.2 Vector fields
Definition 2.2.1 Differential of a function
Definition 2.2.5 Exact one-forms
Definition 2.2.6 Conservative vector fields
Definition 2.2.9 Closed one-forms in R2
Definition 2.2.14 Closed one-forms in R3
Definition 2.3.3 The pullback of a function on R
Definition 2.3.4 The pullback of a one-form on R
Definition 2.4.1 The pullback of a function

Chapter 3 Integrating one-forms: line integrals

Definition 3.1.1 The integral of a one-form over [a, b]
Definition 3.1.3 The orientation of an interval
Definition 3.1.4 The oriented integral of a one-form
Definition 3.2.1 Parametric curves
Definition 3.2.2 Closed parametric curves
Definition 3.2.4 The tangent vector to a parametric curve
Definition 3.2.5 Orientation of a curve
Definition 3.3.2 (Oriented) line integrals

(Continued on next page)

297
APPENDIX B. LIST OF DEFINITIONS 298

Chapter 4 k-forms
Definition 4.1.1 The basic one-forms
Definition 4.1.3 Basic two-forms
Definition 4.1.4 Basic three-forms
Definition 4.1.5 Basic k-forms
Definition 4.1.8 k-forms in R3
Definition 4.2.1 The wedge product
Definition 4.3.1 The exterior derivative of a k-form
Definition 4.4.1 The gradient of a function
Definition 4.4.2 The curl of a vector field
Definition 4.4.3 The divergence of a vector field
Definition 4.6.1 Exact and closed k-forms
Definition 4.7.5 The Jacobian
Definition 4.7.6 Top form
Definition 4.7.10 The pullback of a basic one-form with respect to a linear map
Definition 4.7.12 The pullback of a basic k-form with respect to a linear map
Definition 4.8.1 The Hodge star dual of a k-form in Rn
Definition 4.8.7 The codifferential and the Laplace-Beltrami operator

Chapter 5 Integrating two-forms: surface integrals

Definition 5.1.1 Oriented points
Definition 5.1.2 Integral of a zero-form over an oriented point
Definition 5.1.4 The orientation of an interval
Definition 5.1.5 The integral of a one-form over an oriented interval [a, b]±
Definition 5.1.6 (Oriented) line integrals
Definition 5.2.1 Orientation of Rn
Definition 5.2.2 Canonical orientation of Rn
Definition 5.2.6 Regions in R2
Definition 5.2.8 Orientation of a closed bounded region in R2
Definition 5.2.9 Induced orientation on the boundary of a region in R2
Definition 5.3.1 Integral of a two-form over an oriented closed bounded region in R2
Definition 5.3.6 Orientation-preserving reparametrizations of regions in R2
Definition 5.4.1 Parametric surfaces in Rn
Definition 5.4.2 Closed parametric surfaces
Definition 5.4.8 Tangent planes to a parametric surface
Definition 5.4.9 Normal vectors to a parametric surface in R3
Definition 5.5.1 Orientable surfaces and orientation
Definition 5.5.5 Induced orientation on the boundary of a parametric surface
Definition 5.6.1 Surface integrals
Definition 5.9.1 The flux of a vector field across a surface
(Continued on next page)
APPENDIX B. LIST OF DEFINITIONS 299

Chapter 6 Beyond one- and two-forms

Definition 6.2.1 Orientation of a region in R3 and induced orientation on the boundary
Definition 6.2.3 Integral of a three-form over a closed bounded region in R3

Chapter 7 Unoriented line and surface integrals

Definition 7.1.1 Unoriented line integrals
Definition 7.1.4 Arc length of a curve
Definition 7.2.1 Unoriented surface integrals
Definition 7.2.3 Surface area of a parametric surface in R3
Appendix C

List of examples

Chapter 1 A preview of vector calculus

Chapter 2 One-forms and vector fields

Example 2.1.4 A one-form and its associated vector field
Example 2.2.3 The differential and gradient of a function
Example 2.2.7 An exact one-form and its associated conservative vector field
Example 2.2.8 The gravitational force field is conservative
Example 2.2.12 Exact one-forms are closed
Example 2.2.13 Closed one-forms are not necessarily exact
Example 2.3.2 An example of a change of variables
Example 2.3.5 Change of variables as pullback
Example 2.4.2 The pullback of a function from R3 to R
Example 2.4.3 The pullback of a function from R3 to R2
Example 2.4.6 The pullback of a one-form from R3 to R
Example 2.4.7 The pullback of a one-form from R3 to R2
Example 2.4.8 Consistency check: the pullback of a one-form from R to R

Chapter 3 Integrating one-forms: line integrals

Example 3.1.2 An example of an integral of a one-form over an interval
Example 3.2.3 Parametrizing the unit circle
Example 3.2.8 Parametrizing the unit circle counterclockwise
Example 3.2.12 Two parametrizations of the unit circle
Example 3.2.14 Parametrizing a triangle
Example 3.3.1 Pulling back along a circle
(Continued on next pag

300
APPENDIX C. LIST OF EXAMPLES 301

Example 3.3.3 An example of a line integral

Example 3.3.6 How line integrals change under reparametrizations
Example 3.4.4 An example of a line integral of an exact one-form
Example 3.5.1 Work done by a (non-conservative) force field
Example 3.6.2 Closed forms are exact
Example 3.6.5 An example of a closed one-form that is not exact

Chapter 4 k-forms
Example 4.2.2 The wedge product of two one-forms
Example 4.2.3 The wedge product of a one-form and a two-form
Example 4.2.4 The wedge product of a zero-form and a k-form
Example 4.3.3 The exterior derivative of a zero-form on R3
Example 4.3.4 The exterior derivative of a one-form on R3
Example 4.3.5 The exterior derivative of a two-form on R3
Example 4.3.7 The exterior derivative of the wedge product of two one-forms
Example 4.4.7 Maxwell’s equations
Example 4.5.1 The direction of steepest slope
Example 4.5.4 The curl of the velocity field of a moving fluid
Example 4.5.5 An irrotational velocity field
Example 4.5.6 Another irrotational velocity field
Example 4.5.7 The divergence of the velocity field of an expanding fluid
Example 4.5.8 An imcompressible velocity field
Example 4.5.9 Another incompressible velocity field
Example 4.6.2 Exact and closed one-forms in R3
Example 4.7.2 The pullback of a two-form
Example 4.7.3 The pullback of a three-form
Example 4.8.2 The action of the Hodge star in R
Example 4.8.3 The action of the Hodge star in R2
Example 4.8.4 The action of the Hodge star in R3
Example 4.8.5 An example of the Hodge star action in R3
Example 4.8.6 Maxwell’s equations using differential forms (optional)

Chapter 5 Integrating two-forms: surface integrals

Example 5.1.3 Integral of a zero-form at points
Example 5.2.3 Orientation of R and choice of positive or negative direction
Example 5.2.4 Orientation of R2 and choice of counterclockwise or clockwise rotation
Example 5.2.5 Orientation of R3 and choice of right-handed or left-handed twirl
Example 5.2.10 Closed disk in R2
Example 5.2.11 Closed square in R2
Example 5.2.12 Annulus in R2
(Continued on next pag
APPENDIX C. LIST OF EXAMPLES 302

Example 5.3.3 Integral of a two-form over a rectangular region with canonical orientation
Example 5.3.4 Integral of a two-form over an x-supported (or type I) region with canonical orientation
Example 5.3.8 Area of a disk
Example 5.4.4 The graph of a function in R3
Example 5.4.5 The sphere
Example 5.4.6 The cylinder
Example 5.4.7 Grid curves on the sphere
Example 5.5.6 Upper half-sphere
Example 5.6.2 An example of a surface integral
Example 5.6.7 An example of a surface integral of a vector field
Example 5.7.4 Using Green’s theorem to calculate line integrals
Example 5.7.5 Area of an ellipse
Example 5.8.5 Using Stokes’ theorem to evaluate a surface integral by transforming it into a line integral
Example 5.8.7 Using Stokes’ theorem to evaluate a surface integral by using a simpler surface
Example 5.8.8 Using Stokes’ theorem to evaluate a line integral by transforming it into a surface integral
Example 5.9.2 The electric flux and net charge of a point source

Chapter 6 Beyond one- and two-forms

Example 6.2.2 Solid region bounded by a sphere in R3
Example 6.2.4 Integral of a three-form over a recursively supported region
Example 6.2.8 Using the divergence theorem to evaluate the flux of a vector field over a closed surface in R

Chapter 7 Unoriented line and surface integrals

Example 7.1.3 An example of an unoriented line integral
Example 7.1.5 Calculating the arc length of a parametric curve
Example 7.2.2 An example of an unoriented surface integral
Example 7.2.4 Calculating the surface area of a parametric surface
Example 7.3.1 Finding the centre of mass of a wire in R2
Example 7.3.2 Finding the centre of mass of a sheet in R3
Appendix D

List of exercises

Chapter 1 A preview of vector calculus

Chapter 2 One-forms and vector fields

Exercise 2.1.3.1
Exercise 2.1.3.2
Exercise 2.1.3.3
Exercise 2.1.3.4
Exercise 2.2.5.1
Exercise 2.2.5.2
Exercise 2.2.5.3
Exercise 2.2.5.4
Exercise 2.2.5.5
Exercise 2.2.5.6
Exercise 2.2.5.7
Exercise 2.2.5.8
Exercise 2.2.5.9
Exercise 2.3.3.1
Exercise 2.3.3.2
Exercise 2.3.3.3
Exercise 2.4.3.1
Exercise 2.4.3.2
Exercise 2.4.3.3
Exercise 2.4.3.4
Exercise 2.4.3.5
Exercise 2.4.3.6
(Continued on next page)

303
APPENDIX D. LIST OF EXERCISES 304

Chapter 3 Integrating one-forms: line integrals

Exercise 3.1.5.1
Exercise 3.1.5.2
Exercise 3.1.5.3
Exercise 3.2.6.1
Exercise 3.2.6.2
Exercise 3.2.6.3
Exercise 3.2.6.4
Exercise 3.2.6.5
Exercise 3.2.6.6
Exercise 3.3.5.1
Exercise 3.3.5.2
Exercise 3.3.5.3
Exercise 3.3.5.4
Exercise 3.3.5.5
Exercise 3.4.3.1
Exercise 3.4.3.2
Exercise 3.4.3.3
Exercise 3.4.3.4
Exercise 3.4.3.5
Exercise 3.5.3.1
Exercise 3.5.3.2
Exercise 3.5.3.3
Exercise 3.6.3.1
Exercise 3.6.3.2
Exercise 3.6.3.3
Exercise 3.6.3.4

Chapter 4 k-forms
Exercise 4.1.5.1
Exercise 4.1.5.2
Exercise 4.1.5.3
Exercise 4.1.5.4
Exercise 4.1.5.5
Exercise 4.2.3.1
Exercise 4.2.3.2
Exercise 4.2.3.3
Exercise 4.2.3.4
Exercise 4.2.3.5
(Continued on next page)
APPENDIX D. LIST OF EXERCISES 305

Exercise 4.2.3.6
Exercise 4.3.4.1
Exercise 4.3.4.2
Exercise 4.3.4.3
Exercise 4.3.4.4
Exercise 4.3.4.5
Exercise 4.3.4.6
Exercise 4.3.4.7
Exercise 4.3.4.8
Exercise 4.3.4.9
Exercise 4.4.5.1
Exercise 4.4.5.2
Exercise 4.4.5.3
Exercise 4.4.5.4
Exercise 4.4.5.5
Exercise 4.4.5.6
Exercise 4.4.5.7
Exercise 4.5.4.1
Exercise 4.5.4.2
Exercise 4.5.4.3
Exercise 4.5.4.4
Exercise 4.5.4.5
Exercise 4.5.4.6
Exercise 4.6.3.1
Exercise 4.6.3.2
Exercise 4.6.3.3
Exercise 4.6.3.4
Exercise 4.6.3.5
Exercise 4.6.3.6
Exercise 4.7.5.1
Exercise 4.7.5.2
Exercise 4.7.5.3
Exercise 4.7.5.4
Exercise 4.7.5.5
Exercise 4.7.5.6
Exercise 4.7.5.7
Exercise 4.8.4.1
Exercise 4.8.4.2
Exercise 4.8.4.3
Exercise 4.8.4.4
Exercise 4.8.4.5
(Continued on next page)
APPENDIX D. LIST OF EXERCISES 306

Exercise 4.8.4.6

Chapter 5 Integrating two-forms: surface integrals

Exercise 5.1.4.1
Exercise 5.1.4.2
Exercise 5.1.4.3
Exercise 5.1.4.4
Exercise 5.1.4.5
Exercise 5.1.4.6
Exercise 5.2.3.1
Exercise 5.2.3.2
Exercise 5.2.3.3
Exercise 5.2.3.4
Exercise 5.2.3.5
Exercise 5.3.3.1
Exercise 5.3.3.2
Exercise 5.3.3.3
Exercise 5.3.3.4
Exercise 5.3.3.5
Exercise 5.4.4.1
Exercise 5.4.4.2
Exercise 5.4.4.3
Exercise 5.4.4.4
Exercise 5.5.4.1
Exercise 5.5.4.2
Exercise 5.5.4.3
Exercise 5.5.4.4
Exercise 5.6.4.1
Exercise 5.6.4.2
Exercise 5.6.4.3
Exercise 5.6.4.4
Exercise 5.6.4.5
Exercise 5.7.3.1
Exercise 5.7.3.2
Exercise 5.7.3.3
Exercise 5.7.3.4
Exercise 5.7.3.5
Exercise 5.8.3.1
Exercise 5.8.3.2
Exercise 5.8.3.3
(Continued on next page)
APPENDIX D. LIST OF EXERCISES 307

Exercise 5.8.3.4
Exercise 5.8.3.5
Exercise 5.8.3.6
Exercise 5.9.3.1
Exercise 5.9.3.2
Exercise 5.9.3.3
Exercise 5.9.3.4

Chapter 6 Beyond one- and two-forms

Exercise 6.2.3.1
Exercise 6.2.3.2
Exercise 6.2.3.3
Exercise 6.2.3.4
Exercise 6.2.3.5
Exercise 6.4.3.1
Exercise 6.4.3.2

Chapter 7 Unoriented line and surface integrals

Exercise 7.1.3.1
Exercise 7.1.3.2
Exercise 7.1.3.3
Exercise 7.1.3.4
Exercise 7.1.3.5
Exercise 7.2.3.1
Exercise 7.2.3.2
Exercise 7.2.3.3
Exercise 7.2.3.4
Exercise 7.2.3.5
Exercise 7.3.3.1
Exercise 7.3.3.2
Exercise 7.3.3.3