calculus iii - cognella academic publishing · preface this is the third volume of my calculus...

Calculus III

By Tunc Geveci

Included in this preview:

• Copyright Page• Table of Contents• Excerpt of Title

Section 12.1: Tangent vectors and Velocity

Section 12.7: Directional Derivatives and the Gradient

Section 13.2: Double Integrals over Non-Rectangular Regions

Section 14.2: Line Integrals

Section 14.5: Parametrized Surfaces and Tangent Planes

Section 14.6: Green’s Theorem

For additional information on adopting this book for your class, please contact us at 800.200.3908 x501 or via e-mail at

Tunc Geveci

Calculus III

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Copyright © 2011 by Tunc Geveci. All rights reserved. No part of this publication may be reprinted, reproduced, transmitted, or utilized in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information retrieval system without the written permission of University Readers, Inc.

First published in the United States of America in 2011 by Cognella, a division of University Readers, Inc.

Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

15 14 13 12 11 1 2 3 4 5

Printed in the United States of America

ISBN: 978-1-935551-45-4

Contents

11 Vectors 1

11.1 Cartesian Coordinates in 3D and Surfaces . . . . . . . . . . . . . . . . . . . . . . 1

11.2 Vectors in Two and Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 6

11.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

11.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

12 Functions of Several Variables 35

12.1 Tangent Vectors and Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

12.2 Acceleration and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

12.3 Real-Valued Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . 61

12.4 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

12.5 Linear Approximations and the Differential . . . . . . . . . . . . . . . . . . . . . 74

12.6 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

12.7 Directional Derivatives and the Gradient . . . . . . . . . . . . . . . . . . . . . . . 94

12.8 Local Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

12.9 Absolute Extrema and Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . 121

13 Multiple Integrals 131

13.1 Double Integrals over Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

13.2 Double Integrals over Non-Rectangular Regions . . . . . . . . . . . . . . . . . . . 138

13.3 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 144

13.4 Applications of Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

13.5 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

13.6 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . 167

13.7 Change of Variables in Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . 179

14 Vector Analysis 187

14.1 Vector Fields, Divergence and Curl . . . . . . . . . . . . . . . . . . . . . . . . . . 187

14.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

14.3 Line Integrals of Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . 210

14.4 Parametrized Surfaces and Tangent Planes . . . . . . . . . . . . . . . . . . . . . 220

14.5 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

14.6 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259

14.7 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

14.8 Gauss’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

K Answers to Some Problems 285

L Basic Differentiation and Integration formulas 309

iii

Preface

This is the third volume of my calculus series, Calculus I, Calculus II and Calculus III. This

series is designed for the usual three semester calculus sequence that the majority of science and

engineering majors in the United States are required to take. Some majors may be required to

take only the first two parts of the sequence.

Calculus I covers the usual topics of the first semester: Limits, continuity, the deriv-

ative, the integral and special functions such exponential functions, logarithms,

and inverse trigonometric functions. Calculus II covers the material of the second

semester: Further techniques and applications of the integral, improper integrals,

linear and separable first-order differential equations, infinite series, parametrized

curves and polar coordinates. Calculus III covers topics in multivariable calculus:

Vectors, vector-valued functions, directional derivatives, local linear approxima-

tions, multiple integrals, line integrals, surface integrals, and the theorems of Green,

Gauss and Stokes.

An important feature of my book is its focus on the fundamental concepts, essential

functions and formulas of calculus. Students should not lose sight of the basic concepts

and tools of calculus by being bombarded with functions and differentiation or antidifferentia-

tion formulas that are not significant. I have written the examples and designed the exercises

accordingly. I believe that "less is more". That approach enables one to demonstrate to the

students the beauty and utility of calculus, without cluttering it with ugly expressions. Another

important feature of my book is the use of visualization as an integral part of the expo-

sition. I believe that the most significant contribution of technology to the teaching of a basic

course such as calculus has been the effortless production of graphics of good quality. Numerical

experiments are also helpful in explaining the basic ideas of calculus, and I have included such

data.

Remarks on some icons: I have indicated the end of a proof by ¥, the end of an example by¤ and the end of a remark by ♦.

Supplements: An instructors’ solution manual that contains the solutions of all the prob-

lems is available as a PDF file that can be sent to an instructor who has adopted the book. The

student who purchases the book can access the students’ solutions manual that contains the

solutions of odd numbered problems via www.cognella.com.

Acknowledgments: ScientificWorkPlace enabled me to type the text and the mathematical

formulas easily in a seamless manner. Adobe Acrobat Pro has enabled me to convert the

LaTeX files to pdf files. Mathematica has enabled me to import high quality graphics to my

documents. I am grateful to the producers and marketers of such software without which I

would not have had the patience to write and rewrite the material in these volumes. I would

also like to acknowledge my gratitude to two wonderful mathematicians who have influenced

me most by demonstrating the beauty of Mathematics and teaching me to write clearly and

precisely: Errett Bishop and Stefan Warschawski.

v

vi PREFACE

Last, but not the least, I am grateful to Simla for her encouragement and patience while I spent

hours in front a computer screen.

Tunc Geveci ([email protected])

San Diego, January 2011

Chapter 12

Functions of Several Variables

In this chapter we will discuss the differential calculus of functions of several variables. We will

study functions of a single variable that take values in two or three dimensions. The images

of such functions are parametrized curves that may model the trajectory of an object. We

will calculate their velocity and acceleration. Geometrically, we will calculate tangents

to curves and their curvature. Then we will take up the differential calculus real-valued

functions of several variables. We will study their rates of change in different directions

and their maxima and minima. In the process we will generalize the ideas of local linear

approximation and the differential from functions of a single variable to functions of several

variables.

12.1 Parametrized Curves, Tangent Vectors and Velocity

Parametrized Curves

Definition 1 Assume that σ is a function from an interval J to the plane R2. If

σ (t) = (x (t) , y (t)), then x (t) and y (t) are the component functions of σ. The letter t isthe parameter. The image of σ, i.e., the set of points C = (x (t) , y (t)) : t ∈ J is said tobe the curve that is parametrized by the function σ. We say that the function σ is a

parametric representation of the curve C.

We will use the notation σ : J → R2 to indicate that σ is a function from J into R2. Wecan identify the point σ (t) = (x (t) , y (t)) with its position vector, and regard σ as a vector-valued function whose values are two-dimensional vectors. We may use the notation σ (t) =x (t) i + y (t) j. It is useful to imagine that σ (t) is the position of a particle in motion at thefictitious time t. Figure 1 illustrates a parametrized curve and the "motion of the particle" is

indicated by arrows. In many applications, the parameter does refer to time.

35

36 CHAPTER 12. FUNCTIONS OF SEVERAL VARIABLES

x

y

Σt

Figure 1: A parametrized curve in the plane

Example 1 Let σ (t) = (cos (t) , sin (t)), where t ∈ [0, 2π].The curve that is parametrized by the function σ is the unit circle, since

cos2 (t) + sin2 (t) = 1.

The arrows in Figure 2 indicate the motion of a particle that is at σ (t) at ”time" t. ¤

1-1

1

-1

x

y

Figure 2: σ (t) = (cos (t) , sin (t))

We have to distinguish between a vector-valued function and the curve that is

parametrized by that function, as in the following example.

Example 2 Let σ2 (t) = (cos (2t) , sin (2t)), where t ∈ [0, 2π].Since

cos2 (2t) + sin2 (2t) = 1,

the function σ2 parametrizes the unit circle, just as the function σ of Example 1. But the

function σ2 6= σ. For example,

σ2 (π/4) = (cos (π/2) , sin (π/2)) = (0, 1) ,

whereas,

σ (π/4) = (cos (π/4) , sin (π/4)) =

µ1√2,1√2

¶.

If we imagine that the position of a particle is determined by σ2 (t), the particle revolves aroundthe origin twice on the time interval [0, 2π], whereas, a particle whose position is determined byσ revolves around the origin only once on the ”time" interval [0, 2π]. ¤

12.1. TANGENT VECTORS AND VELOCITY 37

Lines are basic geometric objects. Let P0 = (x0, y0) be a given point and let v = v1i+ v2j bea given vector. With reference to Figure 2, a line in the direction of v that passes through the

point P0 can be parametrized by

σ (t) =−−−→OP 0 + tv = x0i+ y0j+ t (v1i+ v2j)

= (x0 + tv1) i+ (y0 + tv2) j.

Thus, the component functions of σ are

x (t) = x0 + tv1 and y (t) = y0 + tv2

Here, the parameter t varies on the entire number line.

x

y

P0

O

Σttv

Figure 3: A line through P0 in the direction of v

Example 3 Determine a parametric representation of the line in the direction of the vector

v = i− 2j that passes through the point P0 = (3, 4).Solution

σ (t) =−−−→OP 0 + tv = 3i+ 4j+ t (i−2j)

= (3 + t) i+ (4− 2t) j.Thus, the component functions of σ are

x (t) = 3 + t and y (t) = 4− 2t.Figure 4 shows the line.

3 6x

4

8

12

y

P0

O

tv

Σt

Figure 4


We can eliminate the parameter t and obtain the relationship between x and y:

t = x− 3⇒ y = 4− 2 (x− 3) = −2x+ 10.

Note that the same line can be represented parametrically by other functions. Indeed, the

direction can be specified by any scalar multiple of v, and we can pick an arbitrary point on the

line. For example, if we set t = 1 in the above representation, Q0 = (4, 2) is a point on the line,and w = 2v = 2i− 4j is a scalar multiple of v. The function

σ (u) =−−−→OQ0 + uw = 4i+ 2j+ u (2i− 4j)

= (4 + 2u) i+ (2− 4u) j

is another parametric representation of the same line. ¤

Example 4 Imagine that a circle of radius r rolls along a line, and that P (θ) is a point on thecircle, as in Figure 5 ( θ = 0 when P is at the origin).

x

y

rΘ, r

ΘP

rΘ, 0x

y

r

Figure 5

We have

x (θ) = rθ − r sin (θ) = r (θ − sin (θ))

and

y (θ) = r − r cos (θ) = r (1− cos (θ)) ,

as you can confirm. The parameter θ can be any real number. Let’s set

σ (θ) = (x (θ) , y (θ)) = (r (θ − sin (θ)) , r (1− cos (θ))) ,−∞ < θ < +∞.

The curve that is parametrized by the function σ : R→ R2 is referred to as a cycloid. Figure6 shows the part of the cycloid that is parametrized by

σ (θ) = (2θ − 2 sin (θ) , 2− 2 cos (θ)) , where θ ∈ [0, 8π] .

¤


x

y

Figure 6

Similarly, we can consider parametrized curves in three dimensions:

Definition 2 Assume that σ is a function from an interval J to the three-dimensional

space R3. If σ (t) = (x (t) , y (t) , z (t)), then x (t), y (t) and z (t) are the component func-tions of σ. The letter t is the parameter. The image of σ, i.e., the set of points C =(x (t) , y (t) , z(t)) : t ∈ J is said to be the curve that is parametrized by the functionσ.

We will use the notation σ : J → R3 to indicate that σ is a function from J into R3. Wecan identify the point σ (t) = (x (t) , y (t) , z (t)) with its position vector, and regard σ as a

vector-valued function whose values are three-dimensional vectors. We may use the notation

σ (t) = x (t) i+ y (t) j+ z (t)k.

You can imagine that σ (t) is the position at time t of a particle in motion.

Example 5 Let σ (t) = (cos (t) , sin (t) , t), where t ∈ R. The curve that is parametrized by σis a helix.

Since

cos2 (t) + sin2 (t) = 1,

the curve that is parametrized by σ lies on the cylinder x2+ y2 = 1. Figure 7 shows part of thehelix.¤

-10

1.0

-5

0

5

z

10

0.51.0

y0.0 0.5

x0.0-0.5

-0.5-1.0 -1.0

Figure 7


Tangent Vectors

Definition 3 The limit of σ : J → R2 at t0 is the vector w if

limt→t0

||σ (t)−w|| = 0

In this case we write

limt→t0

σ (t) = w.

Proposition 1 Assume that σ (t) = x (t) i+ y (t) j. We have

limt→t0

σ (t) = w = w1i+ w2j

if and only if limt→t0 x (t) = w1 and limt→t0 y(t) = w2. Thus,

limt→t0

(x (t) i+ y (t) j) =

µlimt→t0

x (t)

¶i+

µlimt→t0

y (t)

¶j.

Proof

Assume that

limt→t0

x (t) = w1 and limt→t0

y(t) = w2.

Since

||σ (t)−w||2 = (x (t)− w1)2 + (y (t)− w2)2 ,we have

limt→t0

||σ (t)−w||2 = limt→t0

(x (t)− w1)2 + limt→t0

(y (t)− w2)2 = 0.

Thus, limt→t0 σ (t) = w = w1i+ w2j.

Conversely, assume that limt→t0 σ (t) = w = w1i+ w2. Since

|x (t)− w1| ≤ ||σ (t)−w|| and |y (t)− w2| ≤ ||σ (t)−w|| ,

and limt→t0 ||σ (t)−w|| = 0, we have

limt→t0

|x (t)− w1| = 0 and limt→t0

|y (t)− w2| = 0.

Therefore,

limt→t0

x (t) = w1 and limt→t0

y(t) = w2.

¥

Definition 4 The derivative of σ : J → R2 at t is

lim∆t→0

σ (t+∆t)− σ (t)∆t

and will be denoted bydσ

dt,dσ

dt(t) ,

dσ (t)

dtor σ0 (t) .

If C is the curve that is parametrized by σ and σ0 (t) is not the zero vector, then σ0 (t) is avector that is tangent to C at σ (t).


x

y

Σt

Σt t Σt

Σt t

Figure 8

x

y

Σt

Figure 9

Proposition 2 If σ (t) = x (t) i+ y (t) j and x0 (t), y0 (t) exist, then

dσ

dt=dx

dti+

dy

dtj.

Proof

σ (t+∆t)− σ (t)∆t

=1

∆t(x (t+∆t) i+ y (t+∆t) j− x (t) i− y (t) j)

=x (t+∆t)− x (t)

∆ti+

y (t+∆t)− y (t)∆t

j.

Therefore,

dσ

dt= lim∆t→0

σ (t+∆t)− σ (t)∆t

= lim∆t→0

µx (t+∆t)− x (t)

∆t

¶i+ lim

∆t→0

µy (t+∆t)− y (t)

∆t

¶j

=dx

dti+

dy

dtj

¥

Example 6 Let σ (t) = cos (t) i+ sin(t)j. Determine σ0 (t).


Solution

σ0 (t) =µd

dtcos (t)

¶i+

µd

dtsin (t)

¶j = − sin (t) i+ cos (t) j.

Note that σ0 (t) is orthogonal to σ (t):

σ0 (t) · σ (t) = − sin (t) cos (t) + cos (t) sin (t) = 0.

¤

1 1x

1

1y

Σt

Σ't

Figure 10

Definition 5 Let C be the curve that is parametrized by the function σ : J → R2. If σ0 (t) 6= 0,the unit tangent to C at σ (t) is

T (t) =σ0 (t)||σ0 (t)||

The curve C is said to be smooth at σ (t) if σ0 (t) 6= 0.

With reference to the curve of Example 6, σ0 (t) = − sin (t) i+ cos (t) j is of unit length, since

||σ0 (t)|| =qsin2 (t) + cos2 (t) = 1.

Therefore, T (t) = σ0 (t) = − sin (t) i+ cos (t) j.

Definition 6 If C is the curve that is parametrized by σ : J → R2 and σ0 (t0) 6= 0, the line

that is in the direction of σ0 (t0) and passes through σ (t0) is the tangent line to C at σ (t0) .

Note that a parametric representation of the tangent line is the function L : R→ R2, where

L (u) = σ (t0) + uσ0 (t0) .

Another parametric representation is

l(u) = σ (t0) + uT (t0) .

Example 7 Let σ (θ) = (2θ − 2 sin (θ) , 2− 2 cos (θ)), as in Example 4, and let C be the curvethat is parametrized by σ.


a) Determine T (θ). Indicate the values of θ such that T (θ) exists.b) Determine parametric representations of the tangent line to C at σ (π/2).

Solution

a)

σ0 (θ) = (2− 2 cos (θ)) i+ 2 sin (θ) jTherefore,

||σ0 (θ)|| =q(2− 2 cos (θ))2 + 4 sin2 (θ)

= 2

q1− 2 cos (θ) + cos2 (θ) + sin2 (θ)

= 2√2p1− cos (θ).

Thus, σ0 (θ) = 0 if cos (θ) = 1, i.e., if θ = 2nπ, n = 0,±1,±2,±3, . . .. If θ is not an integermultiple of 2π, the unit tangent to C at σ (θ) is

T (θ) =σ0 (θ)||σ0 (θ)|| =

1

2√2p1− cos (θ) ((2− 2 cos (θ)) i+ 2 sin (θ) j) .

Note that

σ (2nπ) = (2θ − 2 sin (θ) , 2− 2 cos (θ))|θ=2nπ = (4nπ, 0) .A tangent line to C does not exist at such points. Figure 6 is consistent with this observation.

The curve C appears to have a cusp at (4nπ, 0). Let’s consider the case θ = 2π. If the curve isthe graph of the equation y = y (x), by the chain rule,

dy

dx=

dy

dθdx

dθ

=2 sin (θ)

2 (1− cos (θ)) .

Therefore,

limx→4π−

dy

dx= lim

θ→2π−2 sin (θ)

2 (1− cos (θ)) .

We have

limθ→2π

sin (θ) = 0 and limθ→2π

(1− cos (θ)) = 0.By L’Hôpital’s rule,

limθ→2π−

2 sin (θ)

2 (1− cos (θ)) = limθ→2π

2 cos (θ)

2 sin (θ).

Since sin (θ) < 0 if θ < 2π and θ is close to 2π, and

limθ→2π

2 cos (θ) = 2 > 0, limθ→2π−

2 sin (θ) = 0,

we have

limx→4π−

dy

dx= lim

θ→2π−2 sin (θ)

2 (1− cos (θ)) = limθ→2π

2 cos (θ)

2 sin (θ)= −∞.

Similarly,

limx→4π+

dy

dx= +∞.

Therefore, the curve has a cusp at (4π, 0).

b)


We have

σ (π/2) = (2θ − 2 sin (θ) , 2− 2 cos (θ))|θ=π/2 = (π − 2, 2) ,and

T (π/2) =1

2√2p1− cos (θ) ((2− 2 cos (θ)) i+ 2 sin (θ) j)

¯¯π/2

=1

2√2(2i+ 2j) =

1√2(i+ j) .

A parametric representation of the line that is tangent to C at σ (π/2) is

L (u) = σ (π/2) + uσ0 (π/2)= (π − 2) i+ 2j+ u (2 (i+ j))= (π − 2 + 2u) i+ (2 + 2u) j

¤

The above definitions extend to curves in the three-dimensional space in the obvious manner:

if σ : J → R3, the limit of σ is the vector w if

limt→t0

||σ (t)−w|| = 0,

and we write

limt→t0

σ (t) = w.

If σ (t) = x (t) i+ y (t) j+ z (t)k, then

limt→t0

σ (t) = w1i+ w2j+ w3k

if and only if

limt→t0

x (t) = w1, limt→t0

y (t) = w2 and limt→t0

z (t) = w3.

Thus,

limt→t0

(x (t) i+ y (t) j+ z (t)k) =

µlimt→t0

x (t)

¶i+

µlimt→t0

y (t)

¶j+

µlimt→t0

z (t)

¶k.

The derivative of σ is

dσ

dt= lim∆t→0

σ (t+∆t)− σ (t)∆t

=dx

dti+

dy

dtj+

dz

dtk,

and the unit tangent is

T (t) =1

||σ0 (t)||σ0 (t) ,

provided that ||σ0 (t)|| 6= 0.

Example 8 Let σ (t) = (cos (t) , sin (t) , t) as in Example 5, and let C be the curve that is

parametrized by σ.


a) Determine T (t). Indicate the values of t such that T (t)exists.b) Determine a parametric representation of the tangent line to C at σ (π/3).Solution

a)

σ0 (t) = − sin (t) i+ cos (t) j+ k.Therefore,

||σ0 (t)|| =qsin2 (t) + cos2 (t) + 1 =

√2.

Thus, T (t) exists for each t ∈ R and

T (t) =1

||σ0 (t)||σ0 (t) =

1√2(− sin (t) i+ cos (t) j+ k) .

b) We have

σ (π/3) = cos (π/3) i+ sin (π/3) j+π

3k =

1

2i+

√3

2j+

π

3k,

and

σ0 (π/3) = − sin (π/3) i+ cos (π/3) j+ k = −√3

2i+

1

2j+ k.

A parametric representation of the line that is tangent to C at σ (π/6) is

L (u) = σ (π/3) + uσ0 (π/3)

=1

2i+

√3

2j+

π

3k+ u

Ã−√3

2i+

1

2j+ k

!

=

Ã1

2−√3

2u

!i+

Ã√3

2+1

2u

!j+

³π3+ u

´k.

¤

Velocity

Assume that σ is a function from an interval on the number line into the plane or three-

dimensional space, and that a particle in motion is at the point σ (t) at time t. Let ∆t denotea time increment. The average velocity of the particle over the time interval determined by t

and t+∆t isdisplacement

elapsed time=σ (t+∆t)− σ (t)

∆t.

We define the instantaneous velocity at the instant t as the limit of the average velocity as ∆tapproaches 0:

Definition 7 If σ : J → R2 or σ : J → R3 and a particle is at the point σ (t) at time t, thevelocity v (t) of the particle at the instant t is

v (t) =dσ

dt= lim∆t→0

σ (t+∆t)− σ (t)∆t

.

Note that v (t)is a vector in the direction of the tangent to C at σ (t) if v (t) 6= 0.

Example 9 Let σ (t) = (2t− 2 sin (t)) i + (2− 2 cos (t)) j, as in Example 7, and let C be the

curve that is parametrized by σ. Determine v (3π/4), v (π) and v (2π).


Solution

As in Example 7,

v(t) = σ0 (t) = (2− 2 cos (t)) i+ 2 sin (t) j.Therefore,

v (3π/4) =

µ2− 2 cos

µ3π

4

¶¶i+ 2 sin

µ3π

4

¶j =

Ã2− 2

Ã−√2

2

!!i+ 2

Ã√2

2

!j

=³2 +√2´i+√2j,

v (π) = (2− 2 cos (π)) i+ 2 sin (π) j = 4i,and

v (2π) = (2− 2 cos (2π)) i+ 2 sin (2π) j = 0.Note that the direction of the motion is not defined at t = 2π since v (2π) = 0 (and the curveC is not smooth at σ (2π)). ¤

Problems

In problems 1-4 assume that a curve C is parametrized by the function σ.

a) Determine σ0 (t) , σ0 (t0) and the unit tangent T (t0) to C at σ (t0).b) Determine the vector-valued function L that parametrizes the tangent line to C at σ (t0)(specify the direction of the line by the vector σ0 (t0)).

1.

σ (t) =¡t, sin2 (4t)

¢, −∞ < t < +∞, t0 = π/3.

2.

σ (t) = (2 cos (3t) , sin (t)) , 0 ≤ t ≤ 2π, t0 = π/4.

3.

σ (t) =

µ3t

t2 + 1,3t2

t2 + 1

¶, −∞ < t <∞, t0 = 1.

4.

σ (t) = (4 cos (3t) , 4 sin (3t) , 2t) , −∞ < t <∞, t0 = π/2.

In problems 5 and 6, the position of a particle at time t is σ (t). Determine the velocity functionv (t), the velocity at the instant t0, and the speed at t0. Express the result in the ij-notation.5

σ (t) =¡e−t sin (t) , e−t cos (t)

¢, t0 = π

6.

σ (t) = (t, arctan (t)) , t0 = 1

12.2 Acceleration and Curvature

Acceleration

Acceleration is the rate of change of velocity:

Definition 1 Let v(t) be the velocity of an object at time t. The acceleration a(t) of the

object at time t is the derivative of v at that instant:

a(t) = v0(t) =dv

dt.


a) Determine zx (x, y) and zy (x, y) directly (i.e., without relying on a formula from the text).

b) Evaluate zx (1, 2) and zy (1, 2) and determine an equation for the plane that is tangent to thegraph of the given equation at (1, 2,−3).11. Assume that z (x, y) is defined implicitly by the equation

z2 + x2 − y2 − 9 = 0,and z (3, 6) = 6.

6

3x

z

6

y

a) Determine zx (x, y) and zy (x, y) directly (i.e., without relying on a formula from the text).

b) Evaluate zx (3, 6) and zy (3, 6). and determine an equation for the plane that is tangent tothe graph of the given equation at (3, 6, 6).

12.7 Directional Derivatives and the Gradient

Directional Derivatives

We will begin by considering functions of two variables. Assume that f has partial derivatives

in some open disk centered at (x, y). The partial derivative ∂xf (x, y) can be interpreted as therate of change of f in the direction of the standard basis vector i and the partial derivative

∂yf (x, y) can be interpreted as the rate of change of f in the direction of the standard basisvector j. We would like to arrive at a reasonable definition of the rate of f at (x, y) in thedirection of an arbitrary unit vector u = u1i+ u2j.

x

y

hu

P

Ph

Figure 1

Let h 6= 0. Let’s assume that −−→OPh = −−→OP + hu so thatPh = (x+ hu1, y + hu2) .

We define the average rate of change of f corresponding to the displacement−−→PPh as

f (Ph)− f (P )h

=f (x+ hu1, y + hu2)− f (x, y)

h

12.7. DIRECTIONAL DERIVATIVES AND THE GRADIENT 95

It is reasonable to define the rate of change of f at P in the direction of u as

limh→0

f (Ph)− f (P )h

.

Another name for this limit is ”directional derivative":

Definition 1 The directional derivative of f at (x, y) in the direction of the unit

vector u is

Duf (x, y) = limh→0

f (x+ hu1, y + hu2)− f (x, y)h

.

Note that Dif (x, y) is ∂xf (x, y) and Djf (x, y) is ∂yf (x, y). We can calculate the directionalderivative in an arbitrary direction in terms of ∂xf (x, y) and ∂yf (x, y):

Proposition 1 Assume that f is differentiable at (x, y) and that u = u1i+u2j is a unitvector. Then

Duf (x, y) =∂f

∂x(x, y)u1 +

∂f

∂y(x, y)u2.

Proof

Set

g (t) = f (x+ tu1, y + tu2) , t ∈ R.Then,

g0 (0) = limh→0

g (h)− g (0)h

= limh→0

f (x+ hu1, y + hu2)− f (x, y)h

= Duf (x, y)

Let’s set

v (x, t) = x+ tu1 and w (x, t) = y + tu2,

so that g (t) = f (v (x, t) , w (x, t)). By the chain rule,

d

dtg (t) =

d

dtf (x+ tu1, y + tu2)

=∂f

∂v(v,w)

dv

dt+

∂f

∂w(v, w)

dw

dt

=∂f

∂v(x+ tu1, y + tu2)

d

dt(x+ tu1) +

∂f

∂w(x+ tu1, y + tu2)

d

dt(y + tu2)

=∂f

∂v(x+ tu1, y + tu2)u1 +

∂f

∂w(x+ tu1, y + tu2)u2.

Therefore,

Duf (x, y) = g0 (0) =

∂f

∂v(x+ tu1, y + tu2)u1 +

∂f

∂w(x+ tu1, y + tu2)u2

¯t=0

=∂f

∂v(x, y)u1 +

∂f

∂w(x, y)u2

=∂f

∂x(x, y)u1 +

∂f

∂y(x, y)u2,

as claimed. ¥

Example 1 Let f (x, y) = 36 − 4x2 − y2 and let u be the unit vector in the direction of

v = −i+ 4j. Determine Duf (1, 3).


Solution

We have

∂f

∂x(x, y) =

∂

∂x

¡36− 4x2 − y2¢ = −8x and ∂f

∂y(x, y) =

∂

∂y

¡36− 4x2 − y2¢ = −2y.

Therefore,

∂f

∂x(1, 3) = −8 and ∂f

∂y(1, 3) = −6.

We have

||v|| =p1 + 42 =

√17.

Therefore,

u =v

||v|| =1√17(−i+ 4j) = − 1√

17i+

4√17j.

Thus,

Duf (1, 3) =∂f

∂x(1, 3)u1 +

∂f

∂y(1, 3)u2

= (−8)µ− 1√

17

¶+ (−6)

µ4√17

¶= − 16√

17.

¤

Remark 1 All directional derivatives may exist, but f may not be differentiable, as in the

following example:

Let

f (x, y) =

⎧⎨⎩ xy2

x2 + y4if (x, y) 6= (0, 0) ,

0 if (x, y) = (0, 0)

Figure 2 shows the graph of f .

2

x0

-0.52

y0

0.0z

-2-2

0.5

Figure 2

In Section 12.5 we saw that f is not continuous at (0, 0) so that it is not differentiable. Nev-ertheless, the directional derivatives of f exist in all directions at the origin. Indeed, let u be


an arbitrary unit vector, so that u = (cos (θ) , sin (θ)) for some θ. Let’s calculate Duf (0, 0),directly from the definition. We have

f (h cos (θ) , h sin (θ))

h=

h3 cos (θ) sin2 (θ)

h2 cos2 (θ) + h4 sin4 (θ)

h=

h3 cos (θ) sin2 (θ)

h3 cos2 (θ) + h5 sin4 (θ)

=cos (θ) sin2 (θ)

cos2 (θ) + h2 sin4 (θ).

Therefore,

Duf (0, 0) = limh→0

f (h cos (θ) , h sin (θ))

h=cos (θ) sin2 (θ)

cos2 (θ)=sin2 (θ)

cos (θ)

if cos(θ) 6= 0. The case cos(θ) = 0 corresponds to fy (0, 0) or −fy (0, 0). We have

fy (0, 0) = limh→0

f (0, h)− f (0, 0)h

= limh→0

0

h= 0.

Thus, the directional derivatives of f exist in all directions. ♦The gradient of a function is a useful notion within the context of directional derivatives, and

in many other contexts:

Definition 2 The gradient of f at (x, y) is the vector-valued function that assigns the vector

∇f (x, y) = ∂f

∂x(x, y) i+

∂f

∂y(x, y) j

to each (x, y) where the partial derivatives of f exist (read ∇f as "del f").We can express a directional derivative in terms of the gradient:

Proposition 2 The directional derivative of f at (x, y) in the direction of the unitvector u can be expressed as

Duf (x, y) =∇f (x, y) · uProof

By Proposition 1,

Duf (x, y) =∂f

∂x(x, y)u1 +

∂f

∂y(x, y)u2

By the definition of∇f (x, y), the right-hand side can be expressed as a dot product. Therefore,

Duf (x, y) =

µ∂f

∂x(x, y) i+

∂f

∂y(x, y) j

¶· (u1i+ u2j)

=∇f (x, y) · u¥Thus, the directional derivative of f at (x, y) in the direction of the unit vector u isthe component of ∇f (x, y) along u.

Example 2 Let f (x, y) = x2 + 4xy + y2.


a) Determine the gradient of f .

b) Calculate the directional derivative of f at (2, 1) in the directions of the vectors i + j andi− j.Solution

a) Figure 3 displays the graph of f .

0

4

50

z

100

24

y0 2

x0-2

-2-4 -4

Figure 3

We have∂f

∂x(x, y) =

∂

∂x

¡x2 + 4xy + y2

¢= 2x+ 4y

and∂f

∂y(x, y) =

∂

∂y

¡x2 + 4xy + y2

¢= 4x+ 2y.

Therefore,

∇f (x, y) = ∂f

∂x(x, y) i+

∂f

∂y(x, y) j =(2x+ 4y) i+ (4x+ 2y) j.

Figure 4 displays ∇f (x, y) at certain points (x, y) as arrows located at these points.

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

x

y

Figure 4

b) A unit vector in the direction of i+ j is

u1 =1√2i+

1√2j,

and a unit vector in the direction of i− j is

u2 =1√2i− 1√

2j.

We have

∇f (2, 1) = (2x+ 4y) i+ (4x+ 2y) j|x=2,y=1 = 8i+ 10j.


Therefore,

Du1f (2, 1) =∇f (2, 1) · u1= (8i+ 10j) ·

µ1√2i+

1√2j

¶=

8√2+10√2=18√2= 9√2,

and

Du2f (2, 1) =∇f (2, 1) · u2= (8i+ 10j) ·

µ1√2i− 1√

2j

¶=

8√2− 10√

2= − 2√

2= −√2.

Thus, at (2, 1) the function increases in the direction of u1 and decreases in the direction ofu2. Figure 5 shows some level curves of f , the vectors u1 and u2, and a unit vector along

∇f (2, 1) . ¤

13

13

5

5

0

0

0

0

55

13

13

4 2 2 4x

4

2

1

2

4

y

2,1fu1

u2

Figure 5

Proposition 3 The maximum value of the directional derivatives of a function f at

a point (x, y) is ||∇f (x, y)||, the length of the gradient of f at (x, y), and it is attainedin the direction of ∇f (x, y). The minimum value of the directional derivatives of f

at (x, y) is − ||∇f (x, y)|| and it is attained the direction of −∇f (x, y) .Proof

Since Duf (x, y) =∇f (x, y) · u, and ||u|| = 1, we haveDuf (x, y) = ||∇f (x, y)|| ||u|| cos (θ) = ||∇f (x, y)|| cos (θ) ,

where θ is the angle between ∇f (x, y) and the unit vector u. Since cos (θ) ≤ 1,Duf (x, y) ≤ ||∇f (x, y)|| .

The maximum value is attained when cos (θ) = 1, i.e., θ = 0. This means that u is the unitvector in the direction of ∇f (x, y) .


Similarly,

Duf (x, y) = ||∇f (x, y)|| cos (θ)attains its minimum value when cos (θ) = −1, i.e., θ = π. This means that u = −∇f (x, y).The corresponding directional derivative is − ||∇f (x, y)|| .¥

Example 3 Let f (x, y) = x2 + 4xy + y2, as in Example 2. Determine the unit vector alongwhich the rate of increase of f at (2, 1) is maximized and and the unit vector along which therate of decrease of f at (2, 1) is maximized. Evaluate the corresponding rates of change of f .

Solution

In Example 2 we showed that ∇f (2, 1) = 8i+ 10j. Therefore,

||∇f (2, 1)|| = ||(8i+ 10j)|| = √64 + 100 = √164 = 2√41.

Thus, the value of f increases at the maximum rate of 2√41 in the direction of the unit vector

1

||∇f (2, 1)||∇f (2, 1) =1

2√41(8i+ 10j) =

4√41i+

5

41j.

The value of the function decreases at the maximum rate of 2√41 (i.e., the rate of change is −

2√41) in the direction of the unit vector

− 1

||∇f (2, 1)||∇f (2, 1) = −4√41i− 5

41j.

¤

The Chain Rule and the Gradient

A special case of the chain rule says that

d

dtf (x (t) , y (t)) =

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt,

as In Proposition 1 of Section 12.6. The notion of the gradient enables us to provide a useful

geometric interpretation of the above expression. Let’s begin by expressing the chain rule as

follows:

Proposition 4 Assume that the curve C in the plane is parametrized by the function

σ where

σ (t) = (x(t), y(t)) .

Thend

dtf (σ (t)) =∇f (σ (t)) · dσ

dt

Proof

We have

d

dtf (σ (t)) =

d

dtf (x (t) , y (t)) ,

∇f (σ (t)) = ∂f

∂x(x (t) , y (t)) i+

∂f

∂y(x (t) , y (t)) ,


anddσ

dt=dx

dti+

dy

dtj.

Therefore,

∇f (σ (t)) · dσdt=

µ∂f

∂x(x (t) , y (t)) i+

∂f

∂y(x (t) , y (t))

¶·µdx

dti+

dy

dtj

¶=

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt.

By the chain rule,

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt=d

dtf (x (t) , y (t)) =

d

dtf (σ (t)) .

Thus,d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt,

as claimed. ¥

Remark 2 With the notation of Proposition 4, assuming that σ0 (t) 6= 0,

d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt=∇f (σ (t)) ·

dσ

dt(t)¯¯

dσ

dt(t)

¯¯ ¯¯dσdt(t)

¯¯

=

¯¯dσ

dt(t)

¯¯∇f (σ (t)) ·T (t) ,

where T (t) is the unit tangent to C at σ (t). The quantity∇f (σ (t))·T (t) is DT(t)f (σ (t)), thedirectional derivative of f at σ (t) in the direction of the unit tangent T (t), i.e., the componentof ∇f (σ (t)) along T (t). Thus,

d

dtf (σ (t)) =

¯¯dσ

dt(t)

¯¯DT(t)f (σ (t)) .

If σ (t) is the position of a particle at time t, ||σ0 (t)|| is its speed at t. In this case,d

dtf (σ (t)) = the speed of the object at t× component of ∇f (σ (t)) along T (t) .

♦

Proposition 5 Let C be the level curve of f that passes through the point (x0, y0).The gradient of f at (x0, y0) is perpendicular to C at (x0, y0), in the sense that ∇fis orthogonal to the tangent line to C at (x0, y0).

A Plausibility Argument for Proposition 5

We will assume that a segment of C near (x0, y0) can be parametrized by the function σ : J → R2such that σ (t0) = (x0, y0) and σ

0 (t0) 6= 0. By Proposition 4,d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt(t) .


Since C is a level curve of f , the value of f on C is a constant. Therefore,

d

dtf (σ (t)) = 0

for each t ∈ J . In particular,

0 =d

dtf (σ (t0)) =∇f (σ (t0)) · dσ

dt(t0) =∇f (x0, y0) · dσ

dt(t0) .

This implies that ∇f (x0, y0) is orthogonal to the vector σ0 (t0) that is tangent to C at σ (t0) =(x0, y0). ¥

Example 4

a) The point (1, 3) is on the curve C that is the graph of the equation x2 + 4xy + y2 = 22.Determine a vector that is orthogonal to C at (1, 3).b) Determine the line that is tangent to the C at (1, 3) .

Solution

a( We set f (x, y) = x2 + 4xy + y2, as in Example 2, so that C is a level curve of f . We have

∇f (1, 3) =∇f (1, 3) = (2x+ 4y) i+ (4x+ 2y) j|x=1,y=3 = 14i+ 10j.

Therefore, 14i+10j. is orthogonal to C at (1, 3). Figure 6 shows C and a vector along ∇f (1, 3).The picture is consistent with our claim.

1x

3

y

1,3

Figure 6

b) If P0 = (1, 3) and P = (x, y) is an arbitrary point on the line that is tangent to C at P0, wehave

∇f (1, 3) ·−−→P0P = 0.Therefore,

(14i+ 10j) · ((x− 1) i+ (y − 3)) = 0⇔

14 (x− 1) + 10 (y − 3) = 0.Figure 7 shows the tangent line.


1x

3

y

1,3

Figure 7

Functions of three or more variables

Our discussion of directional derivatives and the gradient extend to functions of more than two

variables in a straightforward fashion. Let’s consider functions of three variables to be specific.

Thus, assume that f is a scalar function of the variables x, y and z. If u = u1i+ u2j + u3kis a three-dimensional unit vector, the directional derivative of f at (x0, y0, z0) in the

direction of u is

Duf (x0, y0, z0) = limh→0

f (x0 + hu1, y0 + hu2, z0 + hu3)− f (x0, y0, z0)h

The gradient of f at (x, y, z) is

∇f (x, y, z) = ∂f

∂x(x, y, z) i+

∂f

∂y(x, y, z) j+

∂f

∂z(x, y, z)k.

If f is differentiable at (x0, y0, z0),

Duf (x0, y0, z0) =∂f

∂x(x0, y0, z0)u1 +

∂f

∂y(x0, y0, z0)u2 +

∂f

∂z(x0, y0, z0)u3

=∇f (x0, y0; z0) · uWe have

− ||∇f (x0, y0, z0)|| ≤ Duf (x0, y0, z0) ≤ ||∇f (x0, y0, z0)|| ,Thus, the rate of change of f at (x0, y0, z0) has the maximum value ||∇f (x0, y0, z0)|| in the direc-tion of∇f (x0, y0, z0). The function decreases at the maximum rate of ||∇f (x0, y0, z0)|| (i.e., therate of change is − ||∇f (x0, y0, z0)||) at the point (x0, y0, z0) in the direction of −∇f (x0, y0, z0).If σ : J → R3, and σ (t) = (x (t) , y (t) , z (t)), by the chain rule,

d

dtf (x (t) , y (t) , z (t))

=∂f

∂x(x (t) , y (t) , z (t))

dx

dt(t) +

∂f

∂y(x (t) , y (t) , z (t))

dy

dt(t) +

∂f

∂z(x (t) , y (t) , z (t))

dz

dt(t) .

so that.d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt.


If S is a level surface of f = f (x, y, z) so that f has a constant value on S, and (x0, y0, z0)lies on S, then ∇f (x0, y0, z0) is orthogonal to S at (x0,y0, z0), in the sense that ∇f (x0, y0, z0)is orthogonal at (x0, y0, z0) to any curve on S that passes through (x0, y0, z0). Indeed, if such acurve C is parametrized by σ : J → R3 andσ (t0) = (x0, y0, z0) we have

d

dtf (σ (t)) = 0 for t ∈ J,

since f (σ (t)) has a constant value. Therefore,

∇f (σ (t)) · dσdt(t) =

d

dtf (σ (t)) = 0.

In particular,

∇f (x0, y0, z0) · dσdt(t0) =∇f (σ (t0)) · dσ

dt(t0) = 0,

so that ∇f (x0, y0, z0) is orthogonal to the vector σ0 (t0) that is tangential to the curve C. Itis reasonable to declare that the plane that is spanned by all such tangent vectors and passes

through (x0, y0, z0) is the tangent plane to the level surface S of f . Since ∇f (x0, y0z0)is normal to that plane, the equation of the tangent plane is

∇f (x0, y0z0) · ((x− x0) i+ (y − y0) j+ (z − z0)k) = 0,

i.e.,∂f

∂x(x0, y0, z0) (x− x0) + ∂f

∂y(x0, y0, z0) (y − y0) + ∂f

∂z(x0, y0, z0) (z − z0) = 0.

Example 5 Let f be the function such that

f(x, y, z) =x2

9+y2

4+ z2

a) Determine∇f (x, y, z) and the directional derivative of f at (2, 1, 1) in the direction of (1, 1, 1).

b) Determine the tangent plane to the surface

x2

9+y2

4+ z2 =

61

36

at (2, 1, 1).

Solution

a) We have

∇f (x, y, z) = ∂f

∂x(x, y, z) i+

∂f

∂y(x, y, z) j+

∂f

∂z(x, y, z) =

2x

9i+

y

2j+ 2zk.

Therefore,

∇f (2, 1, 1) = 4

9i+

1

2j+ 2k.

The unit vector u along (1, 1, 1) is

1√3i+

1√3j+

1√3k.


Therefore,

Duf (2, 1, 1) =∇f (2, 1, 1) · u=

µ4

9i+

1

2j+ 2k

¶·µ1√3i+

1√3j+

1√3k

¶=

4

9√3+

1

2√3+

2√3=53

54

√3.

b) We can express the equation of the required plane as

∇f (2, 1, 1) · ((x− 2) i+ (y − 1) j+ (z − 1)k) = 0,

i.e., µ4

9i+

1

2j+ 2k

¶· ((x− 2) i+ (y − 1) j+ (z − 1)k) = 0.

Thus,4

9(x− 2) + 1

2(y − 1) + 2 (z − 1) = 0.

Figure 8 illustrates the surface and the above tangent plane. ¤

Figure 8

Problems

In problems 1-6,

a) Compute the gradient of f

b) Compute the directional derivative of f at P in the direction of the vector v.

1.

f (x, y) = 4x2 + 9y2, P = (3, 4) , v = (−2, 1)2.

f (x, y) = x2 − 4x− 3y2 + 6y + 1, P = (0, 0) , v = (1,−1)3.

f (x, y) = ex2−y2 , P = (2, 1) , v = (−1, 3)

4.

f (x, y) = sin (x) cos (y) , P = (π/3,π/4) , v = (2,−3)


5.

f (x, y, z) = x2 − y2 + 2z2, P = (1,−1, 2) , v = (1,−1, 1)6.

f (x, y, z) =1p

x2 + y2 + z2, P = (2,−2, 1) , v = (3, 4, 1)

In problems 7 and 8,

a) Determine a vector v such that the rate at which f increases at P has its maximum value in

the direction of v, and the corresponding rate of increase of f,

b) Determine a vector w such that the rate at which f decreases at P has its maximum value

in the direction of w, and the corresponding rate of decrease of f.

7.

f (x, y) =1p

x2 + y2, P = (2, 3)

8.

f (x, y) = ln³p

x2 + y2´, P = (3, 4)


a) Computed

dtf (σ (t))

¯t=t0

by making use of the chain rule,

b) Compute the directional derivative of f in the tangential direction to the curve that is

parametrized by σ at the point σ (t0) .

9.

f (x, y) = ex2−y2 , σ (t) = (2 cos (t) , 2 sin (t)) , t0 = π/6

10

f (x, y) = arctan

Ãyp

x2 + y2

!, σ (t) =

µ1− t21 + t2

,2t

1 + t2

¶, t0 = 2.

In problems 11-13,

a) Find a vector that is orthogonal at the point P to the curve that is the graph of the given

equation,

b) Determine an equation of the line that is tangent to that curve at P :

11.

2x2 + 3y2 = 35, (2, 3)

12.

x2 − y2 = 4,³3,√5´

13.

e25−x2−y2 = 1, (3, 4)

In problems 14-17,

a) Find a vector that is orthogonal at the point P to the surface that is the graph of the given

equation

b) Determine an equation of the plane that is tangent to the given surface at P.

14.

z − x2 + y2 = 0, P = (4, 3, 7)15.

x2 − y2 + z2 = 1, P = (2, 2, 1)

12.8. LOCAL MAXIMA AND MINIMA 107

16.

x2 − y2 − z2 = 1, P = (3, 2, 2)

17.

x− sin (y) cos (z) = 0, P = (1,π/2, 0)

12.8 Local Maxima and Minima

The Definitions and Preliminary Examples

Definition 1 A function f has a local maximum at the point (x0, y0) if there exists an opendisk D containing (x0, y0) such that f (x, y) ≤ f (x0, y0) for each (x, y) ∈ D. A function f hasa local minimum at the point (x0, y0) if there exists an open disk D containing (x0, y0) suchthat f (x, y) ≤ f (x0, y0) for each (x, y) ∈ D.

Definition 2 The absolute maximum of a function f on the set D ⊂ R2 is f (x0, y0) iff (x, y) ≤ f (x0, y0) for each (x, y) ∈ D. The absolute minimum of a function f on the set

D ⊂ R2 is f (x0, y0) if f (x, y) ≥ f (x0, y0) for each (x, y) ∈ D.

Example 1 Let Q (x, y) = x2 + y2. Since Q(x, y) ≥ 0 for each (x, y) ∈ R2 and Q (0, 0) = 0, Qattains its absolute minimum value 0 on R2 at (0, 0). Of course, the only point at which Q hasa local minimum is also (0, 0). The function does not attain an absolute maximum on /R

2since

it attains values of arbitrarily large magnitude. For example,

limx→+∞Q (x, x) = lim

x→+∞ 2x2 = +∞.

¤

0

5

10z

15

2

y20

x0

-2 -2

Figure 1

Example 2 Let Q (x, y) = −x2 − y2. Q attains its maximum value 0 at (0, 0) .The functiondoes not attain an absolute minimum on R2 since it attains negative values of arbitrarily large

magnitude. For example,

limx→+∞Q (x, x) = lim

x→+∞¡−2x2¢ = −∞.

¤

138 CHAPTER 13. MULTIPLE INTEGRALS

1. Z y=2

y=1

µZ x=3

x=0

¡x2 + 3xy2

¢dx

¶dy

2. Z x=1

x=0

ÃZ y=π/3

y=π/4

x sin (y) dy

!dx

3. Z y=4

y=2

µZ x=3

x==1

exyydx

¶dy

4. Z x=2

x=0

µZ y=3

y=1

xyp1 + y2dy

¶dx

In problems 5-8, calculate the double integral:

5. Z ZD

xy2

x2 + 1dA where D = (x, y) : 0 ≤ x ≤ 1, −2 ≤ y ≤ 2

6. Z ZD

xyex2ydA where D = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2

7. Z ZD

x2

1 + y2dA where D = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

8. Z ZD

y cos (x+ y) dA, D = (x, y) : 0 ≤ x ≤ π

3, 0 ≤ y ≤ π

2

13.2 Double Integrals over Non-Rectangular Regions

In this section we will discuss double integrals on regions that are not rectangular. Let D be

a bounded region in the Cartesian coordinate plane that has a boundary consisting of smooth

segments. As in the case of a rectangle, let’s begin with a function f that is continuous and

nonnegative on D. In order to approximate the volume of the region G in the xyz-space that is

between the graph of z = f (x, y) and D we form a mesh in the xy-plane that consists of vertical

and horizontal lines. In this case D is not necessarily the union of rectangles. Let’s consider

only those rectangles that have nonempty intersection with D and label them as

Rjk = [xj−1, xj ]× [yk−1, yk] , where j = 0, 1, 2, . . . ,m and k = 0, 1, 2, . . . , n.

We set

∆xj = xj − xj−1 and ∆yk = yk − yk−1.We can approximate the volume of G by a sum

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆Ajk =

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆xj ∆yk,

where¡x∗j , y

∗k

¢ ∈ Rjk..The double integral of f on D is the numberZ ZD

f (x, y) dA =

Z ZD

f (x, y) dxdy

13.2. DOUBLE INTEGRALS OVER NON-RECTANGULAR REGIONS 139

that can be approximated by such sums as accurately as desired provided that the maximum

of ∆Ajk is small enough. Such a number exists for any continuous function f on D even if the

sign of f may vary.

There are counterparts of the Fubini Theorem that allowed us to calculated a double integral

on a rectangle as an iterated integral. Assume that the region D in the xy-plane is bounded by

the lines x = a, x = b and the graphs of functions f (x) and g (x), as in Figure 1. We will referto such a region as an x-simple region, or as a region of type 1.

Figure 1

Assume that F is continuous on D. We can evaluate the double integral as an iterated integral:Z ZD

F (x, y) dA =

Z x=b

x=a

ÃZ y=g(x)

y=f(x)

F (x, y) dy

!dx.

We will leave the proof to a course in advanced calculus. The statement should be plausible, as

in the case of rectangular regions: As x varies from a to b,.the region is swept by rectangles of

"infinitesimal" thickness dx that extend from y = g (x) to y = f (x) .The roles of x and y may be reversed: A region D is a y-simple region, or a region of type

2, if it is bounded by the lines y = a, y = b and the graphs of x = f (y) and x = g (y), as inFigure 2. We have Z Z

D

F (x, y) dA =

Z y=b

y=a

ÃZ x=g(y)

x=f(y)

F (x, y) dx

!dxy.

x

y

y b

y a

x fy x gy

Figure 2

Example 1 Evaluate Z ZD

p1 + x2dxdy,

where D is the triangle that is bounded by y = x, x = 1 and the x-axis.


Solution

Figure 3 shows the triangle D. The integral is the volume of the region between the graph of

z =√1 + x2 and the triangle D in the xy-plane, as shown in Figure 4.

0 x1

0

1

02

2y1

z

2

Figure 4

The region D is both x-simple and y-simple. We will treat it as an x-simple region. Thus,

Z ZD

p1 + x2dxdy =

Z 1

x=0

µZ x

y=0

p1 + x2dy

¶dx =

Z 1

x=0

p1 + x2

µZ x

y=0

1dy

¶dx

=

Z 1

x=0

p1 + x2 (y|x0) dx

=

Z x=1

x=0

p1 + x2xdx.

We set u = 1 + x2, so that du = 2xdx. Therefore,Z x=1

x=0

p1 + x2xdx =

1

2

Z 2

u=1

√udu =

1

2

Ãu3/2

3/2

¯2u=1

!=23/2 − 13

.

¤


x sin (y) dA,

where D is the region bounded by y = x, y = x2, x = 0 and x = 1.

Solution

Figure 5 shows the region D.

1x

1

y

Figure 5


This is a region of type 1 and of type 2. We will treat it as a region of type I, and integrate

with respect to y first:

Z ZD

x sin (y) dA =

Z x=1

x=0

µZ y=x

y=x2x sin (y) dy

¶dx =

Z x=1

x=0

x

µZ y=x

y=x2sin (y) dy

¶dx

=

Z x=1

x=0

x³− cos (y)|y=xy=x2

´dx

=

Z x=1

x=0

x¡− cos (x) + cos ¡x2¢¢ dx

= −Z 1

0

x cos (x) dx+

Z 1

0

x cos¡x2¢dx.

The first integral requires integration by parts: If we set u = x and dv = cos (x), then

du = dx and v =

Zcos (x) dx = sin (x) .

Thus, Zx cos (x) dx =

Zudv = uv −

Zvdu

= x sin (x)−Zsin (x) dx

= x sin (x) + cos (x) .

Therefore, Z 1

0

x cos (x) dx = sin(1) + cos (1)− cos (0) = sin(1) + cos (1)− 1.

As for the second integral, we set u = x2 so that du = 2xdx. Thus,Zx cos

¡x2¢dx =

1

2

Zcos (u) du =

1

2sin¡x2¢.

Therefore, Z 1

0

x cos¡x2¢dx =

1

2sin (1) .

Thus, Z ZD

x sin (y) dA = −Z 1

0

x cos (x) dx+

Z 1

0

x cos¡x2¢dx

= − sin (1)− cos (1) + 1 + 12sin (1)

= 1− cos (1)− 12sin (1) .

¤


yexdA

where D is the region bounded by x = y2/4, x = 1, y = 1 and y = 2.


Solution

Figure 6 shows the region D.

14 1 2

x

1

2

y

Figure 6

This is a region of type 1 and of type 2. We will treat it as a region of type 2, and integrate

with respect to x first:

Z ZD

yex2

dA =

Z y=2

y=1

ÃZ x=1

x=y2/4

yexdx

!dy =

Z y=2

y=1

y

ÃZ x=1

x=y2/4

exdx

!dy

=

Z y=2

y=1

y³ex|1y2/4

´dy

=

Z y=2

y=1

y³e− ey2/4

´dy

=e

2y2 − 2ey2/4

¯21

= 2e− 2e− e2+ 2e1/4

= −e2+ 2e1/4.

¤

In some cases, changing the order of integration enables us to evaluate an integral that is

intractable in its original form, as in the following example.

Example 4 Evaluate the iterated integralZ 1

x=0

µZ y=2

y=2x

e−y2

dy

¶dx.

Solution

We cannot evaluate the given iterated integral in terms of familiar antiderivatives. The first

thing to do is to sketch the region of integration:


1 2x

1

2

y

y = 2x

Figure 7

We will reverse the order of integration:Z 1

x=0

µZ 2

y=2x

e−y2

dy

¶dx =

Z y=2

y=0

ÃZ x=y/2

x=0

e−y2

dx

!dy =

Z y=2

y=0

e−y2

ÃZ x=y/2

x=0

dx

!dy

=

Z y=2

y=0

e−y2³x|y/20

´dy

=

Z y=2

y=0

e−y2 y

2dy

= −14e−y

2

¯20

= −14e−4 +

1

4.

¤

Problems

In problems 1 and 2 evaluate the given iterated integral:

1. Z y=4

y=0

Z x=√y

x=0

xy2dxdy

2. Z θ=π/2

θ=0

Z r=cos(θ)

r=0

esin(θ)drdθ

In problems 3-6, sketch the region D and evaluate the given double integral:

3. Z ZD

4x3ydA,

where D is the region bounded by the graphs of y = x2 and y = 2x.

4. Z ZD

y√xdA,

where D is the region bounded by the graphs of y = x2, y = 0, x = 0 and x = 4.

5. Z ZD

y2exydA,

where D is the region bounded by the graphs of y = x, y = 1 and x = 0.

6. Z ZD

y2 sin¡x2¢dA,


where D is the region bounded by the graphs of y = −x1/3, y = x1/3, x = 0 and x = √π.In problems 7-9 determine the volume of the region D in R3:

7. D is bounded by the paraboloid z = x2+y2+1, the xy-plane, the surfaces y = x2 and y = 1.

8. D is bounded by the graphs of z = ex, z = −ey, x = 0, x = 1, y = 0 and y = 1.9. D is the tetrahedron bounded by the plane 3x+ 2y + z = 6 and the coordinate planes.

In problems 10-12,

a) Sketch the region of integration,

b) Evaluate the integral by reversing the order of integration:

10. Z y=1

y=0

Z 3

x=3y

ex2

dxdy

11. Z y=√π/2

y=0

Z x=√π/2

x=y

cos¡x2¢dxdy.

12. Z y=1

y=0

Z π/2

x=arcsin(y)

cos (x)p1 + cos2 (x)dxdy.

13.3 Double Integrals in Polar Coordinates

In some cases it is convenient to evaluate a double integral by converting it to an integral in

polar coordinates.

Assume that D is the set of points in the plane with polar coordinates (r, θ) such that α ≤ θ ≤ β

and a ≤ r ≤ b,and we would like to approximateZ ZD

f (x, y) dA

by a sum that will be constructed as follows: Let

α = θ0 < θ1 < θ2 < · · · < θj−1 < θj < · · · < θm = β,

be a partition of the interval [α,β], and let

a = r0 < r1 < r2 < · · · < rk−1 < rk < · · · < rn = b

be a partition of [a, b]. We set

∆θj = θj − θj−1 and ∆rk = rk − rk−1.

The rays θ = θj and the arcs r = rk form a grid that covers D, as illustrated in Figure 1.

194 CHAPTER 14. VECTOR ANALYSIS

14.2 Line Integrals

The Integral of a Scalar Function with respect to arc length

Let C be a curve in the plane that is parametrized by the path σ : [a, b] → R2. Assumethat f is a real-valued function of two variables. You may imagine that C is a wire, and

f (σ (t)) = f (x (t) , y (t)) is the lineal density of the wire. How can we calculate the mass of thewire? It is reasonable to approximate mass by sums of the form

nXk=1

f (σ (tk))∆sk =nXk=1

f (x (tk) , y (tk))∆sk,

where t0, t1, . . . , tn−1, tn is a partition of [a, b] and ∆sk is the length of the part of C corre-

sponding to the interval [tk−1, tk]. Thus

∆sk =

Z tk

tk−1||σ0 (t)|| dt.

By the Mean Value Theorem for integrals,Z tk

tk−1||σ0 (t)|| dt = σ0 (t∗k) (tk − tk−1) = σ0 (t∗k)∆tk,

where t∗k is some point between tk−1 and tk. Therefore,

nXk=1

f (σ (tk))∆sk =nXk=1

f (σ (tk)) ||σ0 (t∗k)||∆tk

This is almost a Riemann sum for the integralZ b

a

f (σ (t)) ||σ0 (t)|| dt

and approximates the integral with desired accuracy provided that f and σ0 (t) are continuous.This leads to the following definition:

Definition 1 The integral of the scalar function f with respect to arc length on the

curve C that is parametrized by the function σ : [a, b]→ R2 is defined asZ b

a

f (σ (t)) ||σ0 (t)|| dt

Symbolically,

ds =ds

dtdt =

¯¯dσ

dt

¯¯dt,

so that we can express the integral as ZC

fds.

In terms of the coordinate functions x (t) and y (t) of σ (t),

ZC

fds =

Z b

a

f (x (t) , y (t))

sµdx

dt

¶2+

µdy

dt

¶2dt.

14.2. LINE INTEGRALS 195

Example 1 Evaluate ZC

x2y2ds,

where C is the semicircle x2 + y2 = 4 that is parametrized so that C is traversed from (2, 0) to(−2, 0).Solution

We can set

σ (t) = (2 cos (t) , 2 sin (t)) , 0 ≤ t ≤ π.

Then σ parametrizes C as required.

2 1 1 2x

1

2y

Figure 1

Therefore,

σ0 (t) = −2 sin (t) i+ 2 cos (t) j⇒ ||σ0 (t)|| =q4 sin2 (t) + 4 cos2 (t) = 2.

Thus, ZC

x2y2ds =

Z π

0

¡4 cos2 (t)

¢ ¡4 sin2 (t)

¢ ||σ0 (t)|| dt = 16Z π

0

cos2 (t) sin2 (t) (2) dt

= 32

Z π

0

cos2 (t) sin2 (t) dt.

As we saw in Section 7.3,Zcos2 (t) sin2 (t) dt =

1

8x− 1

8sin (x) cos3 (x) +

1

8sin3 (x) cos (x) .

Therefore,

32

Z π

0

cos2 (t) sin2 (t) dt = 4x− 4 sin (x) cos3 (x) + 4 sin3 (x) cos (x) .¯π0= 4π

¤A curve C can be parametrized by different functions and may be traversed from point A to

point B or vice versa. Thus, there is a notion of "the orientation" of a curve. Assume that

C is a curve in the plane that is parametrized initially by σ : [a, b] → R2 and σ (t) traces Cfrom "the initial point" P1 to "the terminal point" P2 as t increases from a to b. We

will say that σ : [α,β] → R2 is an orientation preserving parametrization of C if σ (τ)traces C from the initial point P1 to the terminal point P2 as the new parameter τ increases

from α to β. We will say that σ : [α,β]→ R2 is an orientation reversing parametrizationof C if σ (τ) traces C from the terminal point P2 to the iniital point P1 as the new parameter

τ increases from α to β. The positive orientation of C is the orientation provided by the


initial parametrization σ and the negative orientation of C is the orientation provided by

any orientation reversing parametrization.You can find the precise defintions at the end of this

section. The line integral of a scalar function is does not change if the curve is parametrized by

any orientation preserving or orientation reversing parametrization:

Proposition 1 If C is parametrized by σ : [a, b] → R2 and σ : [α,β] → R2 is an orientationpreserving or orientation reversing parametrization of C, thenZ

C

fds =

Z t=b

t=a

f (σ (t))

¯¯dσ

dt

¯¯dt =

Z τ=β

τ=α

f (σ (τ))

¯¯dσ

dτ

¯¯dτ

You can find the proof of Proposition 1 at the end of this section.

Example 2 Let P1 = (1, 2) and P2 = (2, 4), and let C be the directed line segment P1P2. We

can parametrize C by

σ (t) = OP1 + tP1P2 = (1, 2) + t (1, 2) = (1 + t, 2 + 2t) , 0 ≤ t ≤ 1.

If we set

σ (τ) = OP1 + 2τP1P2 = (1, 2) + 2τ (1, 2) = (1 + 2τ , 2 + 4τ) , 0 ≤ τ ≤ 12,

then σ is an orientation preserving parametrization of C.

If we set

σ∗ (μ) = OP2 + μτP2P1 = (2, 4) + μ (−1,−2) = (2− μ, 4− 2μ) , 0 ≤ μ ≤ 1,

then σ∗ is an orientation reversing parametrization of C. Confirm that the statements of Propo-sition 1 are valid in the special case Z

C

sin³π2(x− y)

´ds

Solution

We havedσ

dt= (1, 2) so that

¯¯dσ

dt

¯¯=√1 + 4 =

√5.

Therefore ZC

sin³π2(x− y)

´ds =

Z 1

0

sin³π2((1 + t)− (2 + 2t))

´ ¯¯dσdt

¯¯dt

=

Z 1

0

sin³π2(−t− 1)

´√5dt

=√5

Z 1

0

− sin³π2(t+ 1)

´dt.

We set

u =π

2(t+ 1)⇒ du =

π

2dt.

Thus,

√5

Z 1

0

− sin³π2(t+ 1)

´dt =

2√5

π

Z π

π/2

− sin (u) du = 2√5

π

³cos (u)|u=πu=π/2

´= −2

√5

π.


Now let’s use the orientation preserving parametrization σ. We have

dσ

dτ= (2, 4) so that

¯¯dσ

dt

¯¯=√4 + 16 =

√20 = 2

√5.

Therefore ZC

sin³π2(x− y)

´ds =

Z 1/2

0

sin³π2((1 + 2τ)− (2 + 4τ))

´ ¯¯dσdτ

¯¯dτ

=

Z 1/2

0

sin³π2(−2τ − 1)

´2√5dτ

= −2√5

Z 1/2

0

sin³π2(2τ + 1)

´dτ .

We set

u =π

2(2τ + 1)⇒ du = πdτ .

Thus

2√5

Z 1/2

0

− sin³π2(2τ + 1)

´dτ =

2√5

π

Z π

π/2

− sin (u) du = −2√5

π,

as in the case of the parametrization by σ.

σ∗ (μ) = OP2 + μτP2P1 = (2, 4) + μ (−1,−2) = (2− μ, 4− 2μ) , 0 ≤ μ ≤ 1,

Now let’s use the orientation reversing parametrization σ∗. We have

dσ∗

dμ= (−1,−2) so that

¯¯dσ∗

dμ

¯¯=√5.

ThereforeZ 1

0

sin³π2((2− μ)− (4− 2μ))

´ ¯¯dσ∗dμ

¯¯dμ =

Z 1

0

sin³π2(μ− 2)

´√5dμ

=√5

Z 1

0

− sin³π2(2− μ)

´dμ.

We set

u =π

2(2− μ)⇒ du = −π

2dμ.

Thus

√5

Z 1

0

− sin³π2(2− μ)

´dμ =

2√5

π

Z π/2

π

sin (u) du =2√5

π.

Therefore Z 1

0

sin (σ∗ (μ))¯¯dσ∗

dμ

¯¯dμ = −

ZC

fds,

as predicted by Proposition 1


The Line Integral of a Vector Field in the Plane

Assume that F (x, y) = M (x, y) i + N (x, y) j is a two-dimensional force field, σ : [a, b] → R2,and a particle is at σ (t) = (x (t) , y (t)) at time t. Let C be the curve that is parametrized by σ.How should we calculate the work done by the force F as the particle moves from (x (a) , y (a))to (x (b) , y (b)) along C? Our starting point is the definition of the work done by a constantforce F on a particle that moves along a line. If the displacement of the particle is represented

by the vector w, the work done is

F ·w = ||F|| ||w|| cos (θ) ,

where θ is the angle between F and w. Let us subdivide the interval [a, b] into subintervals bya partition

P = a = t0 < t1 < · · · < tk−1 < tk < · · · < tn−1 < tn = b .As usual we set ∆tk = tk − tk−1.Let ∆σk = σ (tk)−σ (tk−1). If the norm of the partition P is

small, we can approximate the work done by F as the particle moves from σ (tk−1) to σ (tk) by

F (σ (tk)) ·∆σk = F (σ (tk)) · (σ (tk)− σ (tk−1)) ∼= F (σ (tk)) · σ (tk)− σ (tk−1)∆tk

∆tk

∼= F (σ (tk)) · dσdt(tk)∆tk

Therefore, the total work done is approximately

nXk=1

F (σ (tk)) · dσdt(tk)∆tk

This is a Riemann sum for the integralZ b

a

F (σ (t)) · dσdtdt.

We will define the work done by the force F on the particle as it moves along the

curve C that is parametrized by σ by this integral. In more general terms, this is a line integral:

Definition 2 Assume that the curve C in the plane is parametrized by σ : [a, b] → R2 and Fis a continuous vector field in the plane. The line integral of F on C isZ

C

F·dσ =Z b

a


The notationRCF·dσ is appropriate since the line integral is approximated by sums of the form

nXk=1

F (σ (tk)) ·∆σk.

Symbolically,

dσ =dσ

dtdt.


ΣtΣ't

FΣt

Figure 2

Example 3 LetF (x, y) = 2xi+yj and assume that C is parametrized by σ (t) = (cosh (t) , sinh (t)) ,where 0 ≤ t ≤ 3. Calculate Z

C

F·dσ

Solution

Figure 3 shows the curve C.

1 3 6 9x

2

4

6

8

y

Figure 3

We havedσ

dt= sinh (t) i+ cosh (t) j.

Therefore, ZC

F·dσ =Z 3

0

(2 cosh (t) i+ sinh (t) j) · (sinh (t) i+ cosh (t) j) dt

=

Z 3

0

(2 cosh (t) sinh (t) + sinh (t) cosh (t)) dt

=

Z 3

0

3 cosh (t) sinh (t) dt.

If we set u = cosh (t) then du = sinh (t) dt, so thatZ 3

0

3 cosh (t) sinh (t) dt. = 3

Z cosh(3)

cosh(0)

udu = 3

Ãu2

2

¯cosh(3)1

!=3

2

¡cosh2 (3)− 1¢

¤


We stated that the integral of a scalar function with respect to arc length does not depend on

an orientation preserving or orientation reversing parametrization of C. In the case of the line

integral of a vector field, an orientation reversing reparametrization introduces a minus sign:

Proposition 2 Assume that the curve C is parametrized by σ : [a, b]→ R2. If σ : [α,β]→ R2is an orientation preserving parametrization of C, thenZ β

α

F (σ (τ)) · dσ (τ)dτ

dτ =

Z b

a


If σ : [α,β]→ R2 is an orientation reversing parametrization of C, then

Z β

α


dτ = −Z b

a

F (σ (t)) · dσdtdt

You can find the proof of Proposition 2 at the end of this section.

Remark Assume that the curve C is parametrized by σ : [a, b]→ R2. and that σ : [α,β]→ R2is an orientation reversing parametrization of C. We will setZ

−CF · d σ =

Z β

α


dτ .

By Proposition 2, Z−CF · dσ = −

Z b

a

F (σ (t)) · dσdtdt = −

ZC

F · dσ.

If C is a piecewise smooth curve that can be expressed as a sequence of smooth curves C1, C2, . . . , Cm,

we will write

C = C1 + C2 + · · ·+ Cm.With this understanding, we setZ

C

F·dσ =ZC1+C2+···+Cm

F·dσ =ZC1

F·dσ +ZC2

F·dσ + · · ·+ZCm

F·dσ

♦

Example 4 Let

F (x, y) = −yi+ xj.Evaluate Z

C1+C2

F · dσ,

where C1 is the line segment from (−1, 0) to (1, 0) on the x-axis, and C2 is the semicircle ofradius 1 centered at the origin traversed in the counterclockwise direction

Solution

1 1x

1y

C1

C2

Figure 4


We can parametrize the line segment by σ1 (t) = (t, 0), where t ∈ [−1, 1]. The semicircle can beparametrized by σ2 (t) = (cos (t) , sin (t)), where t ∈ [0,π]. Therefore,Z

C1

F · d σ =Z 1

−1F (σ (t)) · dσ

dtdt =

Z 1

−1tj · idt = 0,

andZC2

F · dσ =Z π

0

F (σ (t)) · dσdtdt =

Z π

0

(− sin (t) i+ cos (t) j) · (− sin (t) i+ cos (t) j) dt

=

Z π

0

¡sin2 (t) + cos2 (t)

¢dt =

Z π

0

1dt = π.

Thus, ZC1+C2

F · d σ =ZC1

F ·dσ +ZC2

F·dσ = π.

¤

The Line Integral as an Integral with respect to Arc Length

We can express the line integral of F on the curve C as the integral of the tangential

component of F with respect to arc length (Recall that T (t) denotes the unit tangentto C at σ (t)):

ZC

F · dσ =Z b

a


Z b

a

F (σ (t)) ·dσ

dt¯¯dσ

dt

¯¯ ¯¯dσdt

¯¯dt =

Z b

a

F (σ (t)) ·T (t)¯¯dσ

dt

¯¯dt

=

Z b

a

F (σ (t)) ·T (t) dsdtdt

=

ZC

F ·Tds.

Thus, we have the following fact:

Proposition 3 ZC

F · d σ =ZC

F ·Tds.

Example 5 Let

F (x, y) = −yi+ xjand let C be the circle of radius 2 that is centered at the origin and is traversed counterclockwise.

Evaluate the line integral of F on C as an integral with respect to arc length.

Solution

We can parametrize C via the function σ (t) = (2 cos (t) , 2 sin (t)), where 0 ≤ t ≤ 2π. We haveσ0 (t) = −2 sin (t) i+ 2 cos (t) j,

so that

F (σ (t)) = F (2 cos (t) , 2 sin (t)) = −2 sin (t) i+ 2 cos (t) j,


ds = ||σ0 (t)|| dt =q4 sin2 (t) + 4 cos2 (t)dt = 2dt,

and

T (t) =σ0 (t)||σ0 (t)|| = − sin (t) i+ cos (t) j.

Therefore,ZC

F · d σ =ZC

F ·Tds =Z 2π

0

F (σ (t)) ·T (t) ds

=

Z 2π

0

(−2 sin (t) i+ 2 cos (t) j) · (− sin (t) i+ cos (t) j) 2dt

=

Z 2π

0

¡4 sin2 (t) + 4 cos2 (t)

¢dt

=

Z 2π

0

4dt = 8π.

¤

The Differential Form Notation

There is still another useful way to express a line integral. If F (x, y) = M (x, y) i + N (x, y) jand the curve C is parametrized by σ (t) = (x (t) , y (t)) , where a ≤ t ≤ b,Z

C

F · dσ =Z b

a

(M (x (t) , y (t)) i+N (x (t) , y (t)) j) ·µdx

dti+

dy

dtj

¶dt

=

Z b

a

µM (x (t) , y (t))

dx

dt+N (x (t) , y (t))

dy

dt

¶dt

=

Z b

a

M (x (t) , y (t))dx

dtdt+

Z b

a

N (x (t) , y (t))dy

dtdt.

Using the formalism,

dx =dx

dtdt and dy =

dy

dtdt,

we can express the integrals asZ b

a

M (x (t) , y (t)) dx+

Z b

a

N (x (t) , y (t)) dy.

This motivates the notation ZC

Mdx+Ndy

for the line integral of F =M i+N j on the curve C.

Definition 3 If M and N are functions of x and y, and C is a smooth curve in the plane that

is parametrized by σ : [a, b]→ R2 we setZC

Mdx+Ndy =

Z b

a

M (x (t) , y (t))dx

dtdt+

Z b

a

N (x (t) , y (t))dy

dtdt.


Thus, ZC

F · d σ =ZC

Mdx+Ndy

if F =M i+N j. The expression Mdx+Ndy is referred to as a differential form, and we willrefer to the above expression for a line integral as the integral of a differential form on the

curve C.

Example 6 Evaluate the line integral of F (x, y) = y3i+x2j on the ellipse that is parametrizedby

σ (t) = (4 cos (t) , sin (t)) , 0 ≤ t ≤ 2π,by expressing the line integral as the integral of a differential form.

4-4

-1

1

x

y

Figure 5

Solution

ZC

F·dσ =ZC

y3dx+ x2dy =

Z 2π

0

µsin3 (t)

dx

dt+ 16 cos2 (t)

dy

dt

¶dt

=

Z 2π

0

¡sin3 (t) (−4 sin (x)) + 16 cos2 (t) (cos (t))¢ dt

=

Z 2π

0

¡−4 sin4 (t) + 16 cos3 (t)¢ dt = −3π(Check: You will have to refresh your memory with respect to the integrals of powers of sines

and cosines as we discussed in Section 7.3). ¤

Example 7 Let C be the boundary of the square [−1, 1] × [−1, 1] that is traversed in thecounterclockwise direction. Calculate Z

C

−ydx+ xdy.

1-1

1

-1

x

y

Figure 6


Solution

We can express C as C1 + C2 + C3 + C4,as indicated in Figure 4.We can parametrize C1 by setting x (t) = t and y (t) = −1, where −1 ≤ t ≤ 1. Then,Z

C1

−ydx+ xdy =Z 1

−1

µdx

dt+ tdy

dt

¶dt =

Z 1

−1dt = 2

We can express C3 as −C3, where −C3 is parametrized by setting x (t) = t and y (t) = 1, where−1 ≤ t ≤ 1. Then,Z

C3

−ydx+ xdy = −Z−C3−ydx+ xdy = −

Z 1

−1

µ−dxdt+ tdy

dt

¶dt =

Z 1

−1dt = 2

We can parametrize C2 by setting x (t) = 1 and y (t) = t, where −1 ≤ t ≤ 1. Then,ZC2

−ydx+ xdy =Z 1

−1

µ−tdxdt+dy

dt

¶dt =

Z 1

−1dt = 2

Similarly, C4 = −C2, where −C2 is parametrized by x (t) = −1 and y (t) = t, −1 ≤ t ≤ 1.Then, Z

C4

−ydx+ xdy = −Z−C2−ydx+ xdy = −

Z 1

−1

µ−dydt

¶dt =

Z 1

−1dt = 2

Therefore, ZC

−ydx+ xdy. =4Xk=1

ZCk

−ydx+ xdy = 2 + 2 + 2 + 2 = 8.

¤

Curves in R3

Our discussion of integrals with respect to arc length and line integrals extends to curves in

three dimensions. Thus, assume that σ : [a, b]→ R3 is a smooth function and parametrizes thecurve C. If f is a continuous scalar function of three variables, we define the integral of f with

respect to arc length on C as ZC

fds =

Z b

a

f (σ (t)) ||σ0 (t)|| dt.

If σ (t) = (x (t) , y (t) , z (t)),ZC

fds =

Z b

a

f (x (t) , y (t) , z (t))

sµdx

dt

¶2+

µdy

dt

¶2+

µdz

dt

¶2dt.

As in the case of curves in the plane, the above integral is the same for orientation preserving

and orientation reversing parametrizations of C.

If F is a vector field inR3 so that

F(x, y, z) =M (x, y, z) i+N (x, y, z) j+ P (x, y, z)k,

the line integral of F on the curve C isZC

F·dσ =Z b

a



In the differential form notation,ZC

F·dσ =ZC

Mdx+Ndy + Pdz

=

Z b

a

µM (x (t)) , y (t) , z (t)

dx

dt+N (x (t)) , y (t) , z (t)

dy

dt+ P (x (t)) , y (t) , z (t)

dz

dt

¶dt.

The line integral can be expressed as an integral with respect to arc length:ZC

F·d σ =ZC

F ·Tds,

where T denotes the unit tangent to the curve C. As in the case of two-dimensional vector

fields, ZC

F·dσ

is unchanged by an orientation preserving parametrization of C and multiplied by −1 if theparametrization reverses the orientation.

Example 8 Let C be the helix that is parametrized by

σ (t) = (cos (t) , sin (t) , t) , 0 ≤ t ≤ 4π,

and let F (x, y, z) = −yi+ xj+zk. Evaluatea) Z

C

zds,

b) ZC

F · dσ

Solution

The picture shows the helix.

z

yx

Figure 7

a) We have

σ0 (t) = − sin (t) i+ cos (t) j+ kTherefore,

ds = ||σ0 (t)|| =qsin2 (t) + cos2 (t) + 1 =

√1 + 1 =

√2.


Thus, ZC

zds =

Z 4π

0

t ||σ0 (t)|| dt =Z 4π

0

t√2dt =

√2

Ãt2

2

¯4π0

!=√2

µ16π2

2

¶= 8√2π2

b) ZC

F · dσ =Z 4π

0

(− sin (t) i+ cos (t) j+ tk) · (− sin (t) i+ cos (t) j+ k) dt

=

Z 4π

0

¡sin2 (t) + cos2 (t) + t

¢dt

=

Z 4π

0

(1 + t) dt = t+t2

2

¯4π0

= 4π +16π2

2= 4π + 8π2.

¤

Precise Definitions and Proofs

Let’s begin by stating the precise definitions of orientation preserving and orientation reversing

parametrizations of a curve:

Definition 4 Assume that C is a curve in the plane that is parametrized by σ : [a, b] → R2.We say that σ : [α,β] → R2 is an orientation preserving parametrization of C if there

exists a differentiable increasing function h : [α,β] → [a, b] such that σ (τ) = σ (h (τ)) foreach τ ∈ [α,β]. The function σ is an orientation reversing parametrization of C if thereexists a differentiable decreasing function h : [α,β]→ [a, b] such that σ (τ) = σ (h (τ)) for eachτ ∈ [α,β].Now let’s prove Proposition 1:

Assume that C is a curve in the plane that is parametrized by σ : [a, b] → R2 and thatσ : [α,β]→ R2 is an orientation preserving or reversing parametrization of C. ThenZ b

a

f (σ (t))

¯¯dσ

dt

¯¯dt =

Z β

α

f (σ (τ))

¯¯dσ

dt

¯¯dτ .

Proof

Assume that σ (τ) = σ (h (τ)), where h is an increasing or decreasing and differentiable functionfrom [α;β] onto [a, b]. By the chain rule,

dσ

dτ=dh

dτ

dσ

dt(h (τ)) .

Therefore, ¯¯dσ

dτ

¯¯=

¯dh

dτ

¯ ¯¯dσ

dt(h (τ))

¯¯Thus, Z β

α

f (σ (h (τ)))

¯¯dσ

dτ

¯¯dτ =

Z β

α

f (σ (h (τ)))

¯dh

dτ

¯ ¯¯dσ

dt(h (t))

¯¯dτ

If dh/dτ > 0, Z β

α

f (σ (h (τ)))

¯dh

dτ

¯ ¯¯dσ

dt

¯¯dτ =

Z β

α

f (σ (h (τ)))dh

dτ

¯¯dσ

dt

¯¯dτ .


By the substitution rule,

Z β

α

f (σ (h (τ)))dh

dτ

¯¯dσ

dt

¯¯dτ =

Z β

α

f (σ (h (τ)))

¯¯dσ

dt

¯¯dh

dτdτ =

Z h(β)

h(α)

f (t)

¯¯dσ

dt

¯¯dt

=

Z b

a

f (t)

¯¯dσ

dt

¯¯dt.

Thus, Z β

α

f (σ (h (τ)))

¯¯dσ

dτ

¯¯dτ =

Z b

a

f (t)

¯¯dσ

dt

¯¯dt,

as claimed.

Similarly, if dh/dτ < 0,

Z β

α

f (σ (h (τ)))

¯¯dσ

dτ

¯¯dτ =

Z β

α

f (σ (h (τ)))

¯dh

dτ

¯ ¯¯dσ

dt

¯¯dτ

= −Z β

α

f (σ (h (τ)))dh

dτ

¯¯dσ

dt

¯¯dτ

=

Z α

β

f (σ (h (τ)))dh

dτ

¯¯dσ

dt

¯¯dτ

=

Z h(α)

h(β)

f (σ (t))

¯¯dσ

dt

¯¯dt =

Z b

a

f (σ (t))

¯¯dσ

dt

¯¯dt.

¥

Now we will prove Proposition 2 with regard to the effect of reparametrization on the line

integral of a vector field. Recall the statement:

Assume that the curve C is parametrized by σ : [a, b]→ R2. If σ : [α,β]→ R2 is an orientationpreserving parametrization of C, then

Z β

α


dτ =

Z b

a


If σ : [α,β]→ R2 is an orientation reversing parametrization of C, then

Z β

α


dτ = −Z b

a


Proof

a) By the chain rule,

dσ (h (τ))

dτ=dh

dτ

dσ

dt(h (τ)) .


Therefore, Z β

α


dτ =

Z β

α

F (σ (h (τ))) · dσ (h (τ))dτ

dτ

=

Z β

α

F (σ (h (τ))) ·µdh

dτ

dσ

dt(h (τ))

¶dτ

=

Z β

α

F (σ (h (τ))) · dσdt(h (τ))

dh

dτdτ

=

Z h(β)

h(α)


=

Z b

a

F (σ (t)) · dσdtdt,

by the substitution rule.

b) As in part a)Z β

α


dτ =

Z β

α

F (σ (h (τ))) · dσdt(h (τ))

dh

dτdτ

=

Z h(β)

h(α)


Since h (τ) is a decreasing function, we have h (α) = b and h (β) = a. Therefore,Z h(β)

h(α)


Z a

b

F (σ (t)) · dσdtdt = −

Z b

a

F (σ (t)) · dσdtdt,

as claimed. ¥

Problems

In problems 1-5 evaluate the given integral of a scalar function with respect to arc length:

1. ZC

y3ds,

where C is parametrized by

σ (t) =¡t3, t

¢, 0 ≤ t ≤ 2.

2. ZC

xy2ds,

where C is parametrized by

σ (t) = (2 cos (t) , 2 sin (t)) , −π2≤ t ≤ π

2.

3. ZC

xeyds,

where C is the line segment traversed from (2, 1) to (4, 5) .

4. ZC

y

x2 + y2ds,


where C is the semicircle of radius 2 that is centered at (4, 3) and traversed from (6, 3) to (2, 3)in the counterclockwise direction.

5. ZC

(x− 2) e(y−3)(z−4)ds,

where C is the line segment from (2, 3, 4) to (3, 5, 7).

In problems 6-10 evaluate the line integral ZC

F·dσ

6.

F (x, y) = x2i+ y2j

and C is the line segment from (1, 2) to (3, 4).

7.

F (x, y) = yi− xjand C is the part of the unit circle traversed from (0, 1) to (−1, 0) in the counterclockwisedirection.

8.

F (x, y) = ln (y) i− exjand C is parametrized by

σ (t) =¡ln (t) , t3

¢, 1 ≤ t ≤ e.

9.

F (x, y, z) = cos (y) i+ sin (z) j+ xk

and C is parametrized by

σ (t) = (cos (t) , t, t) ,π

4≤ t ≤ π

2

10.

F (x, y) = xyi+ (x− y) jand C = C1 + C2, where C1 is the line segment from (0, 0) to (2, 0) and C2 is the line segmentfrom (2, 0) to (3, 2).

In problems 11 and 12 evaluate ZC

F ·Tds

11.

F (x, y) = −2xyi+ (y + 1) jand C is the part of the circle of radius 2 centered at the origin traversed from (−2, 0) to (0, 2)in the clockwise direction.

12.

F (x, y, z) = ex+yi+ xzj+ yk

and C is the line segment from (1, 2, 3) to (−1,−2,−3) .

In problems 13 and 14


a) Express the line integral ZC

F·dσ

in the differential form notation,

b) Evaluate the expression that you obtained in part a).

13.

F (x, y) = −xyi+ 1

x2 + 1j

and C is parametrized by

σ (t) =¡t, t2

¢, −4 ≤ t ≤ −1

14.

F (x, y) = yi− xjand C is the part of the unit circle that is traversed from (0,−1) to (0, 1) in the counterclockwisedirection.

In problems 15-18 evaluate the given line integral:

15. ZC

3x2dx− 2y3dy

where C is the part of the unit circle traversed from (1, 0) to (0, 1).

16. ZC

exdx+ eydy,

where C is the part of the ellipse x2 + 4y2 = 4 traversed from (0, 1) to (2, 0) in the clockwisedirection.

Hint: Parametrize C by a function of the form (a cos (θ) , b sin (θ)).

17. ZC

− sin (x) dx+ cos (x) dy,

where C is the part of the parabola y = x2 traversed from (0, 0) to¡π,π2

¢.

18. ZC

x3dx+ y2dy + zdz,

where C is the line segment from the origin to (2, 3, 4).

14.3 Line Integrals of Conservative Vector Fields

In this section we will see that the line integral of the gradient of a scalar function on a curve

is simply the difference between the values of the function at the endpoints of the curve. Thus,

such a line integral does not depend on the particular path that connects two given points. We

will discuss conditions under which a vector field is the gradient of a scalar function.

14.5. SURFACE INTEGRALS 239

-5

0

5

z

5

y0

2

x-5 1

0

8.

Ω (x, θ) = (x cos (θ) , x sin (θ) , ex) ,

where 0 ≤ x ≤ 2 and 0 ≤ θ ≤ 2π,P = Ω (1,π/2)

Note that S is part of the surface of revolution that is obtained by revolving the graph of z = ex

about the z-axis.

2

y0

2

-2

4z

x

6

0-22

14.5 Surface Integrals

In this section we will compute the area of a parametrized surface and define the integrals of

scalar functions and vector fields on such a surface. We will consider only orientable surfaces,

as we discussed in Section 14.4.

Surface Area

We will motivate the definition of the area of a surface. Assume that the surface M is parame-

trized by the function Φ : D ⊂ R2 → R3, such that

Φ(u, v) = (x (u, v) , y (u, v) , z (u, v)) , (u, v) ∈ D.

We assume that Φ is smooth, i.e., the tangent vectors

∂Φ

∂u(u, v) =

∂x

∂u(u, v) i+

∂y

∂u(u, v) j+

∂z

∂u(u, v)k

and∂Φ

∂v(u, v) =

∂x

∂v(u, v) i+

∂y

∂v(u, v) j+

∂z

∂v(u, v)k

are continuous on D, and the normal vector

N (u, v) =∂Φ

∂u(u, v)× ∂Φ

∂v(u, v) 6= 0


for each (u, v) ∈ D. Let (u, v) be a point in the interior of D and assume that ∆u and ∆v arepositive numbers that are small enough so that the rectangle D (u, v,∆u,∆v) determined bythe points (u, v) and (u+∆u, v +∆v) is contained in the interior of D. The vectors

∆u∂Φ

∂u(u, v) = ∆u

∂x

∂u(u, v) i+∆u

∂y

∂u(u, v) j+∆u

∂z

∂u(u, v)k,

and

∆v∂Φ

∂v(u, v) = ∆v

∂x

∂v(u, v) i+∆v

∂y

∂v(u, v) j+∆v

∂z

∂v(u, v)k

are tangent to S at Φ (u, v). Since ∆u and ∆v are small,

Φ(u+∆u, v)−Φ(u, v) ∼= ∆u∂Φ∂u

(u, v)

and

Φ(u, v +∆v)−Φ(u, v) ∼= ∆v∂Φ∂v

(u, v) .

Figure 1

It is reasonable to approximate the area of the part of M corresponding to the rectangle

D (u, v,∆u,∆v) by the area of the part of the tangent plane at Φ(u, v) that is spanned by

∆u∂Φ

∂u(u, v) and ∆v

∂Φ

∂v(u, v) .

Recall that this area can be expressed as¯¯∆u

∂Φ

∂u(u, v)× ∆v

∂Φ

∂v(u, v)

¯¯=

¯¯∂Φ

∂u(u, v)× ∂Φ

∂v(u, v)

¯¯∆u∆v

= ||N (u, v)||∆u∆v.Therefore, it makes sense to define the area of the surface as the integral of the magnitude of

the normal vector:

Definition 1 Assume that the surface M is parametrized by the smooth function Φ : D ⊂R2 → R3. The area of M isZ Z

D

||N (u, v)|| dudv =Z Z

D

¯¯∂Φ

∂u(u, v)× ∂Φ

∂v(u, v)

¯¯dudv.


Example 1 Confirm that the area of a sphere of radius ρ0 is 4πρ20.

-22

0z

2

2

y0

x0

-2 -2

Figure 2

Solution

We can center the sphere M at the origin and parametrize it by the function

Φ (φ, θ) = (ρ0 sin (φ) cos (θ) , ρ0 sin (φ) sin (θ) , ρ0 cos (φ)) ,

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2. As in Example 9 of Section 14.4,

N (φ, θ) = ρ20 sin (φ) (sin (φ) cos (θ) i+ sin (φ) sin (θ) j+ cos (φ)k)

Therefore,

||N (φ, θ)|| = ρ20 sin (φ)

qsin2 (φ) cos2 (θ) + sin2 (φ) sin2 (θ) + cos2 (φ)

= ρ20 sin (φ)qsin2 (φ)

¡cos2 (θ) + sin2 (θ)

¢+ cos2 (φ)

= ρ20 sin (φ)

qsin2 (φ) + cos2 (φ)

= ρ20 sin (φ) .

Let D = [0,π]× [0, 2π] denote the domain of Φ in the φθ-plane. The area of the sphere isZ ZD

||N (φ, θ)|| dφdθ =Z Z

D

ρ20 sin (φ) dφdθ = ρ20

Z φ=π

φ=0

Z θ=2π

θ=0

sin (φ) dθdφ

= ρ20

Z φ=π

φ=0

2π sin (φ) dφ

= 2πρ20 (− cos (φ)|π0 )= 2πρ20 (2) = 4πρ

20,

as claimed. ¤

Remark 1 Assume that M is the graph of f : D → R, where D is a region in the plane. M

can be parametrized by Φ : D→ R3, where

Φ (x, y) = (x, y, f (x, y))

for each (x, y) ∈ D. In Section 14.4 we noted that

N (x, y) = −∂f∂x(x, y) i− ∂f

∂y(x, y) j+ k.


Therefore,

||N (x, y)|| =sµ

∂f

∂x

¶2+

µ∂f

∂y

¶2+ 1.

Therefore, the area of S can be expressed asZ ZD

sµ∂f

∂x

¶2+

µ∂f

∂y

¶2+ 1 dxdy.

♦

Example 2 Let f (x, y) = x2 − y2, and let M be the part of the graph of f corresponding to

the unit disk in the xy-plane (i.e., the disk of radius 1 centered at the origin). Determine the

area of M.

5

-10

0z

10

5

y0

x0

-5 -5

Figure 3

Solution

We have∂f

∂x(x, y) = 2x and

∂f

∂y(x, y) = −2y.

Therefore,sµ∂f

∂x

¶2(x, y) +

µ∂f

∂y(x, y)

¶2(x, y) + 1 =

q(2x)2 + (−2y)2 + 1 =

p4x2 + 4y2 + 1

Thus, the area of S is Z ZD

p4x2 + 4y2 + 1dxdy,

whereD is the unit disk. In polar coordinates r and θ,Z ZD

p4x2 + 4y2 + 1dxdy =

Z 2π

θ=0

Z 1

r=0

p4r2 + 1rdrdθ

= 2π

Z 1

r=0

p4r2 + 1rdr

= 2π

Ã1

12

¡4r2 + 1

¢3/2 ¯10

!= 2π

µ5

12

√5− 1

12

¶Example 3 Let M be the part of the cylinder that is parametrized by

Φ (θ, z) = (a cos (θ) , a sin (θ) , z) ,

where 0 ≤ θ ≤ 2π and 0 ≤ z ≤ h. Determine the area of S.


-52

0z

5

2

y0

x0

-2 -2

Figure 4

Solution

As in Example 11 of Section 14.4,

N (θ, z) = a cos (θ) i+ a sin (θ) j.

Therefore,

||N (θ, z)|| =qa2 cos2 (θ) + a2 sin2 (θ) = a.

Thus, the area of S is Z 2π

θ=0

Z h

z=0

adzdθ = 2πah,

as it should be (the perimeter of the base times height). ¤

Example 4 Assume that the torus M is parametrized by

Φ (θ,φ) = ((a+ b cos (φ)) cos (θ) , (a+ b cos (φ)) sin (θ) , b sin (φ)) ,

where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ 2π. Determine the area of S.

x

z

y

Figure 5

Solution

We have

∂Φ

∂θ(θ,φ) =

∂

∂θ((a+ b cos (φ)) cos (θ)) i+

∂

∂θ((a+ b cos (φ)) sin (θ)) j+ b

∂

∂θsin (φ)k

= − (a+ b cos (φ)) sin (θ) i+ (a+ b cos (φ)) cos (θ) j,and

∂Φ

∂φ(θ,φ) =

∂

∂φ((a+ b cos (φ)) cos (θ)) i+

∂

∂φ((a+ b cos (φ)) sin (θ)) +ג b

∂

∂φsin (φ)k

= −b sin (φ) cos (θ) i− b sin (φ) sin (θ) j+ b cos (φ)k.


Therefore,

N (θ,φ) =∂Φ

∂θ(θ,φ)× ∂Φ

∂φ(θ,φ)

=

¯¯ i j k

− (a+ b cos (φ)) sin (θ) (a+ b cos (φ)) cos (θ) 0−b sin (φ) cos (θ) −b sin (φ) sin (θ) b cos (φ)

¯¯

= b (a+ b cos (φ)) (cos (θ) cos (φ) i+ sin (θ) cos (φ) j+ sin (φ)k)

Thus,

||N (θ,φ)||2 = b2 (a+ b cos (φ))2 ¡cos2 (θ) cos2 (φ) + sin2 (θ) cos2 (φ) + sin2 (φ)¢= b2 (a+ b cos (φ))

2 ¡cos2 (φ) + sin2 (φ)

¢= b2 (a+ b cos (φ))2 .

Therefore,

||N (θ,φ)|| = b (a+ b cos (φ))Thus, the area of the torus isZ Z

D

||N (θ,φ)|| dθdφ =Z Z

D

b (a+ b cos (φ)) dθdφ,

where D is the square [0, 2π]× [0, 2π] in the φθ-plane. Therefore,Z ZD

b (a+ b cos (φ)) dθdφ =

Z φ=2π

φ=0

Z θ=2π

θ=0

b (a+ b cos (φ)) dθdφ

= 2πb

Z φ=2π

φ=0

(a+ b cos (φ)) dφ

= 2πb³aφ+ b sin (φ)|2π0

´= 2πb2πa = 4πab.

¤

Surface Integrals of Scalar Functions

Definition 2 Assume that the (orientable) surface M is parametrized by the function Φ : D ⊂R2 → R3, and that the scalar function f is continuous on M . We define the integral of f on Mas

Z ZD

f (Φ (u, v)) ||N (u, v)|| dudv.

We will denote the integral of f on M as Z ZM

fdS.

Symbolically, dS is “the element of area" ||N (u, v)|| dudv. You can imagine that dS representsthe area of an “infinitesimal" portion of the surface M . If the surface M is a thin shell, and f

is its mass density per unit area, the integralZ ZM

fdS


yields the total mass of the shell. Soon we will discuss other contexts for the appearance of

surface integrals.

Example 5 Let M be the hemisphere sphere of radius 2 that is parametrized by

Φ (φ, θ) = (2 sin (φ) cos (θ) , 2 sin (φ) sin (θ) , 2 cos (φ)) ,

where 0 ≤ φ ≤ π/2 and 0 ≤ θ ≤ 2π Assume that M is a thin shell with mass density

f (x, y, z) = 4− z.Determine the total mass of M .

Solution

Let D = [0,π]× [0, 2π] denote the domain of Φ in the φθ-plane. As in Example 1,||N (φ, θ)|| = 22 sin (φ) = 4 sin (φ) .

Therefore, the mass of M isZ ZM

fdS =

Z(4− z) ds =

Z(4− 2 cos (φ)) ||N (φ, θ)|| dφdθ

=

Z ZD

(4− 2 cos (φ)) (4 sin (φ)) dφdθ

= 8

Z φ=π/2

φ=0

Z θ=2π

θ=0

(2 sin (φ)− cos (φ) sin (φ)) dφdθ

= 16π

Z φ=π/2

φ=0

(2 sin (φ)− cos (φ) sin (φ)) dφ

= 16π

Ã−2 cos (φ) + 1

2cos2 (φ)

¯π/20

!

= 16π

µ3

2

¶= 24π.

(we made use of the substitution u = cos (φ)). ¤The integral of a scalar function on a surface M is independent of the parametrization of M .

In particular, the calculation of the area of a surface does not change under an orientation

preserving or reversing parametrization of M . You can find the proof at the end of this section.

Remark 2 As we noted before, if M is the graph of the function g : D ⊂ R2 → R, then S canbe parametrized by

Φ (x, y) = (x, y, g (x, y)) , (x, y) ∈ D,N (x, y) = −∂g

∂x(x, y) i− ∂g

∂y(x, y) j+ k,

and

||N (x, y)|| =sµ

∂g

∂x

¶2+

µ∂g

∂y

¶2+ 1.

Therefore, the integralR R

Mfds takes the formZ Z

D

f (x, y, g (x, y)) ||N (x, y)|| dxdy =Z Z

D

f (x, y, g (x, y))

sµ∂g

∂x

¶2+

µ∂g

∂y

¶2+ 1dxdy.

The roles of the coordinates can be interchanged, of course. ♦


Example Let M be the part of the graph of z = g (x, y) = 4 − x2 − y2 over the xy-plane.Evaluate Z Z

M

1

1 + 4 (x2 + y2)dS.

20

2z

-2

4

y0

x0

-22

Figure 7

Solution

We have

4− x2 − y = 0⇔ x2 + y2 = 4.

Thus, the intersection of M with the xy-plane is the circle x2 + y2 = 4, and M is the graph of

g over the disk D of radius 2 centered at the origin.

We have∂g

∂x(x, y) = −2x and ∂g

∂y(x, y) = −2y.

Therefore,Z ZM

1

1 + 4 (x2 + y2)dS =

Z ZD

1

1 + 4 (x2 + y2)

q(−2x)2 + (−2y2) + 1dxdy

=

Z ZD

1

1 + 4 (x2 + y2)

p4x2 + 4y2 + 1dxdy

=

Z ZD

1p1 + 4 (x2 + y2)

dxdy.

It is convenient to transform the integral to polar coordinates:Z ZD

1p1 + 4 (x2 + y2)

dxdy =

Z θ=2π

θ=0

Z r=2

r=0

1√1 + 4r2

rdrdθ

= 2π

Z r=2

r=0

1√1 + 4r2

rdr.

If we set u = 1 + 4r2 then du = 8rdr, so thatZ r=2

r=0

1√1 + 4r2

rdr =1

8

Z u=17

u=1

1√udu =

1

8

Z u=17

u=1

u−1/2du

=1

8

µ2u1/2

¯171

¶=1

4

³√17− 1

´.


Therefore, Z ZS

1

1 + 4 (x2 + y2)dS = 2π

Z r=2

r=0

1√1 + 4r2

rdr

= 2π

µ1

4

³√17− 1

´¶=

π

2

³√17− 1

´.

¤

Flux Integrals

In order to motivate the definition of a flux integral, let us imagine that v (x, y, z) is the velocityof a fluid particle at (x, y, z) ∈ R3, and that M is a surface in R3. We would like to calculatethe flow across S per unit time. Assume that a small patch of the surface is almost planar, has

area dS, and n is the unit normal to that piece which points in the direction of the flow. The

component of the velocity vector in the direction of n is v · n. If the mass density of the fluidhas the constant value δ, the mass of fluid that flows across that small patch is approximately

δv · ndS. Therefore it is reasonable to calculate the total flow across S per unit time asZ ZM

δv · ndS.

This leads to the definition of the flux integral of an arbitrary vector field over a surface:

Definition 3 Assume that F is a continuous vector field in R3 and M is a smooth surface that

is parametrized by the function Φ : D ⊂ R2 → R3. The flux integral of F over M is the

integral of the normal component of F on MS, i.e.Z ZM

F · ndS

where n is the unit normal to M.

The “vectorial element of area" is the symbol

dS = ndS,

so that the flux integral of F over M can be denoted as.Z ZM

F · dS.

If

N (u, v) =∂Φ

∂u(u, v)× ∂Φ

∂v(u, v) ,

then

n =N

||N|| and dS = ||N|| dudv.

Thus, Z ZM

F · ndS =Z Z

D

F (Φ (u, v)) · N (u, v)

||N (u, v)|| ||N (u, v)|| dudv

=

Z ZD

F (Φ (u, v)) ·N (u, v) dudv

=

Z ZD

F (Φ (u, v)) ·µ∂Φ

∂u(u, v)× ∂Φ

∂v(u, v)

¶dudv.


Example 6 Consider the cylinder of radius 2 whose axis is along the z-axis. Let M be the

portion of the cylinder between z = 0 and z = 4. As in Example 3, M can be parametrized

by Φ (θ, z) = (2 cos (θ) , 2 sin (θ) , z), where 0 ≤ θ ≤ 2π and 0 < z < 4. Let F (x, y, z) = xi.

Calculate the flux of F over M .

02

2z

4

2

y0

x0

-2 -2

Figure 8

Solution

We have

F (Φ (θ, z)) = F (2 cos (θ) , 2 sin (θ) , z) = 2 cos (θ) i.

As in Example 3,

N (θ, z) = 2 cos (θ) i+ 2 sin (θ) j.

Therefore,

F (Φ (θ, z)) ·N (θ, z) = (2 cos (θ) i) · (2 cos (θ) i+ 2 sin (θ) j) = 4 cos2 (θ) .

Thus, ZM

F · ndS =Z Z

D

F (Φ (θ, z)) ·N (θ, z) dθdz =Z 4

z=0

Z 2π

θ=0

4 cos2 (θ) .dθdz

= 4

Z 4

z=0

ÃZ θ=2π

θ=0

cos2 (θ) dθ

!dz

= 16

Z θ=2π

θ=0

cos2 (θ) dθ

= 16

Ã1

2cos (θ) sin (θ) +

1

2θ

¯2π0

!= 16π.

¤

Example 7 Let M be the sphere of radius ρ0 that is centered at the origin. M can be para-

metrized by

Φ (φ, θ) = (ρ0 sin (φ) cos (θ) , ρ0 sin (φ) sin (θ) , ρ0 cos (φ)) ,

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π.CalculateZ ZM

(xi+ yj+ zk) · dS


Solution

Z ZM

(xi+ yj+ zk) · dS =Z 2π

θ=0

Z π

φ=0

Φ (φ, θ) ·N (φ, θ) dφdθ,

where

N (φ, θ) = ρ20 sin (φ) (sin (φ) cos (θ) i+ sin (φ) sin (θ) j+ cos (φ)k)

= ρ0 sin (φ)Φ (φ, θ) ,

as in Example 1. Therefore,Z 2π

θ=0

Z π

φ=0

Φ (φ, θ) ·N (φ, θ) dφdθθ =Z 2π

θ=0

Z π

φ=0

Φ (φ, θ) · ρ0 sin (φ)Φ (φ, θ) dφdθ

= ρ0

Z 2π

θ=0

Z π

φ=0

sin (φ)Φ (φ, θ) ·Φ (φ, θ) dφdθ

= ρ0

Z 2π

θ=0

Z π

φ=0

sin (φ) ||Φ (φ, θ)||2 dφdθ

= ρ0

Z 2π

θ=0

Z π

φ=0

sin (φ) ρ20dφdθ

= ρ0¡4πρ20

¢= 4πρ30.

¤

Proposition 1 Assume that M is an orientable surface. The flux integralZM

F · ndS

is independent of an orientation-preserving parametrization ofM . The flux integral is multiplied

by −1 under an orientation-reversing parametrization of M.

The proof of Proposition 2can be found at the end of this section.

Remark 3 If M is the graph of the function g (x, y) over the domain D in the xy-plane, we

can parametrize M by

Φ (x, y) = (x, y, g (x, y)) ,

where (x, y) ∈ D. As we noted above,

N (x, y) = −gx (x, y) i− gy (x, y) j+ k,

and

||N (x, y)|| =sµ

∂g

∂x

¶2+

µ∂g

∂y

¶2+ 1.


Therefore,Z ZM

F · dS =Z Z

M

F · ndS

=

Z ZM

F · N

||N||dS

=

Z ZD

F (x, y, g (x, y)) · N (x, y)||N (x, y)|| ||N (x, y)|| dxdy

=

Z ZD

F (x, y, g (x, y)) ·N (x, y) dxdy

=

Z ZD

F (x, y, g (x.y)) ·µ−∂g∂x(x, y) i− ∂g

∂y(x, y) j+ k

¶dxdy..

♦

Example 8 Let F (x, y, z) = zk and letM be the hemisphere of radius a centered at the origin.

Calculate the flux integral of F across M in the direction of the outward normal.

Solution

The surface M is the graph of the function

g (x, y) =pa2 − x2 − y2,

where (x, y) is in the disk Da of radius a centered at the origin. A normal that points outwardis

−gx (x, y) i− gy (x, y) j+ k. = xpa2 − x2 − y2 i+

ypa2 − x2 − y2 j+ k

On the hemisphere M ,

F (x, y, z) = F (x, y, g (x, y)) = g (x, y)k =pa2 − x2 − y2k.

Therefore,Z ZM

F · dS =Z Z

Da

g (x, y)k · (−gx (x, y) i− gy (x, y) j+ k) dxdyZ ZDa

pa2 − x2 − y2k·

Ãxp

a2 − x2 − y2 i+yp

a2 − x2 − y2 j+ k!dxdy

=

Z ZDa

pa2 − x2 − y2dxdy.

In polar coordinates,Z ZDa

pa2 − x2 − y2dxdy =

Z 2π

θ=0

Z a

r=0

pa2 − r2rdrdθ

= 2π

µZ a

r=0

pa2 − r2rdr

¶= 2π

µa3

3

¶=2πa3

3.

¤


Example 9 Assume that the electrostatic field due to the charge q at the origin is

F (x, y, z) = qxi+ yj+ zk

(x2 + y2 + z2)3/2= q

r

||r||3

Calculate the flux of F out of the sphere M of radius ρ0 centered at the origin.

Solution

Z ZM

F · ndS =Z Z

M

qr

||r||3 ·r

||r||dS

= q

Z ZM

||r||2||r||4 dS = q

Z ZM

1

||r||2 dS =q

ρ20

Z ZM

dS =q

ρ20

¡4πρ20

¢= 4πq

¤

Example 10 Let

F (x, y, z) =−yi+ xjx2 + y2

.

Determine the flux of F in the direction of j across the rectangle M in the xz-plane that has

the vertices (1, 0, 0) and (3, 0, 2)

Solution

The unit normal to M is j. We haveZ ZM

F · ndS =Z 2

z=0

Z 3

x=1

1

xdxdz = 2 ln (3)

¤

Alternative Notations and some Proofs (Optional)

At the end of Section 14.4. we saw that

N (u, v) =∂ (y, z)

∂ (u, v)i+

∂ (z, x)

∂ (u, v)j+

∂ (x, y)

∂ (u, v)k,

where∂ (y, z)

∂ (u, v)=

¯yu zuyv zv

¯,∂ (z, x)

∂ (u, v)=

¯zu xuzv xv

¯,∂ (x, y)

∂ (u, v)=

¯xu yuxv yv

¯Therefore, we can express the area of the surface that is parametrized by (x (u, v) , y (u, v) , z (u, v)),where (u, v) ∈ D, as

Z ZD

||N (u, v)|| dudv =Z Z

D

sµ∂ (y, z)

∂ (u, v)

¶2+

µ∂ (z, x)

∂ (u, v)

¶2+

µ∂ (x, y)

∂ (u, v)

¶2dudv.

Example 11 Consider the cone that is parametrized by

Φ (ρ, θ) = (ρ sin (φ0) cos (θ) , ρ sin (φ0) sin (θ)) , ρ cos (φ0) .

Here φ0 is a given angle, ρ ≥ 0 and 0 ≤ θ ≤ 2π. Let M be the part of the cone of slant height

l, i.e., 0 ≤ ρ ≤ l. Determine the area of M .


0

1

2

2z

3

1

y0

x0

-2 -1

Figure 9

Solution

As in Example 14 of Section 14.4,

N (ρ, θ) =∂ (y, z)

∂ (ρ, θ)i+

∂ (z, x)

∂ (ρ, θ)j+

∂ (x, y)

∂ (ρ.θ)k

= −ρ sin (φ0) cos (φ0) cos (θ) i− ρ sin(φ0) cos (φ0) sin (θ) j+ ρ sin2 (φ0)k.

Therefore,

||N (ρ, θ)|| =qρ2 sin2 (φ0) cos

2 (φ0) cos2 (θ) + ρ2 sin2

¡φ20¢cos2 (φ0) sin

2 (θ) + ρ2 sin4 (φ0)

=

qρ2 sin2 (φ0) cos

2 (φ0) + ρ2 sin4 (φ0)

= ρ sin (φ0)

qcos2 (φ0) + sin

2 (φ) = ρ sin (φ0)

Thus, the area of S isZ 2π

θ=0

Z l

ρ=0

ρ sin (φ0) dρdθ = 2π sin (φ0)

Z l

ρ=0

ρdρ = 2π sin (φ0)

µl2

2

¶= πl2 sin (φ0) .

Note that the radius of the base is l sin (φ0), so that the above expression is π×slant height×radiusof the base. ¤

Proposition 2 The surface integral Z ZM

fdS

is independent of an orientation-preserving or orientation reversing parametrization of the sur-

face M .

Proof

Assume that M is parametrized by Φ,

Φ (u, v) = (x (u, v) , y (u, v) .z (u, v)) , for each (u, v) ∈ D,and

u = u (u∗, v∗) , v = v (u∗, v∗)

for each (u∗, v∗) ∈ D∗. Setx (u∗, v∗) = x (u (u∗, v∗) , v (u∗, v∗)) , y (u∗, v∗) = y (u (u∗, v∗) , v (u∗, v∗))z (u∗, v∗) = z (u (u∗, v∗) , v (u∗, v∗)) ,


and

Φ∗ (u∗, v∗) = (x (u∗, v∗) , y (u∗, v∗) .z (u∗, v∗)) , where (u∗, v∗) ∈ D∗.If we make use of the parametrization Φ∗, the surface integral can be expressed asZ Z

D∗f (Φ∗ (u∗, v∗)) ||N (u∗, v∗)|| du∗dv∗

=

Z ZD∗f (Φ∗ (u∗, v∗))

sµ∂ (y, z)

∂ (u∗, v∗)

¶2+

µ∂ (z, x)

∂ (u∗, v∗)

¶2+

µ∂ (x, y)

∂ (u∗, v∗)

¶2du∗dv∗

As a consequence of the chain rule,

∂ (y, z)

∂ (u∗, v∗)=

∂ (y, z)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗),

∂ (z, x)

∂ (u∗, v∗)=

∂ (z, x)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗),

∂ (x, y)

∂ (u∗, v∗)=

∂ (x, y)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗).

Therefore,Z ZD∗f (Φ∗ (u∗, v∗)) ||N (u∗, v∗)|| du∗dv∗

=

Z ZD∗

ZD∗f (Φ∗ (u∗, v∗))

sµ∂ (y, z)

∂ (u∗, v∗)

¶2+

µ∂ (z, x)

∂ (u∗, v∗)

¶2+

µ∂ (x, y)

∂ (u∗, v∗)

¶2du∗dv∗

=

Z ZD∗

ZD∗f (Φ∗ (u∗, v∗))

sµ∂ (y, z)

∂ (u, v)

¶2+

µ∂ (z, x)

∂ (u, v)

¶2+

µ∂ (x, y)

∂ (u, v)

¶2 ¯∂ (u, v)

∂ (u∗, v∗)

¯du∗dv∗

=

Z ZD

f (Φ (u, v))

sµ∂ (y, z)

∂ (u, v)

¶2+

µ∂ (z, x)

∂ (u, v)

¶2+

µ∂ (x, y)

∂ (u, v)

¶2dudv,

by the rule for the change of variables in double integrals. Note thatsµ∂ (y, z)

∂ (u, v)

¶2+

µ∂ (z, x)

∂ (u, v)

¶2+

µ∂ (x, y)

∂ (u, v)

¶2= ||N (u, v)|| .

Therefore, Z ZD∗||N (u∗, v∗)|| du∗dv∗ =

Z ZD

||N (u, v)|| dudv.

Thus, the surface integral Z ZM

fdS

can be calculated by making use of either parametrization of S, as claimed .¥The Differential Form Notation for a Flux Integral

As we saw before, if Φ (u, v) = (x (u, v) , y (u, v) , z (u, v)) then

N (u, v) =∂ (y, z)

∂ (u, v)i+

∂ (z, x)

∂ (u, v)j+

∂ (x, y)

∂ (u, v)k.

Therefore, if F (x, y, z) =M (x, y, z) i+N (x, y, z) j+ P (x, y, z)k then


F (Φ (u, v)) ·N (u, v) =M (Φ (u, v))∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

Therefore,Z ZM

F · ndS =Z Z

D

F (Φ (u, v)) ·N (u, v) dudv

=

Z ZD

µM (Φ (u, v))

∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

¶dudv.

Symbolically, we can set

dydz =∂ (y, z)

∂ (u, v)dudv, dzdx =

∂ (z, x)

∂ (u, v)dudv, dxdy =

∂ (x, y)

∂ (u, v)dudv,

and writeZ ZD

µM (Φ (u, v))

∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

¶dudv

=

Z ZM

Mdydz +Ndzdx+ Pdxdy.

This is the differential form notation for the flux integral of

F (x, y, z) =M (x, y, z) i+N (x, y, z) j+ P (x, y, z)k

over the surface M .

Example 12 Let M be the torus that is parametrized by

Φ (θ,φ) = ((a+ b cos (φ)) cos (θ) , (a+ b cos (φ)) sin (θ) , b sin (φ)) ,

where a > b > 0, 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ 2π. Determine the flux integralZ ZM

xdydz + ydzdx+ zdxdy

Solution

We have

xdydz = (a+ b cos (φ)) cos (θ)∂ (y, z)

∂ (θ,φ)dθdφ

= (a+ b cos (φ)) cos (θ) b (a+ b cos (φ)) cos (θ) cos (φ) dθdφ

= b (a+ b cos (φ))2 cos2 (θ) cos (φ) dθdφ,

ydzdx = (a+ b cos (φ)) sin (θ)∂ (z, x)

∂ (θ,φ)dθdφ

= b (a+ b cos (φ)) sin (θ) (a+ b cos (φ)) sin (θ) cos (φ) dθdφ

= b (a+ b cos (φ))2sin2 (θ) cos (φ) dθdφ,


and

zdxdy = z∂ (x, y)

∂ (u, v)dθdφ

= b sin (φ) (a+ b cos (φ)) sin (φ) dθdφ

= b (a+ b cos (φ)) sin2 (φ) dθdφ.

Therefore,


= b (a+ b cos (φ))¡(a+ b cos (φ)) cos2 (θ) cos (φ) + (a+ b cos (φ)) sin2 (θ) cos (φ) + sin2 (φ)

¢dθdφ

= b (a+ b cos (φ))¡(a+ b cos (φ)) cos (φ) + sin2 (φ)

¢dθdφ

Thus Z ZS


=

Z θ=2π

θ=0

Z φ=π

φ=0

b (a+ b cos (φ))¡(a+ b cos (φ)) cos (φ) + sin2 (φ)

¢dθdφ

= 2π

µb2aπ +

1

2baπ

¶= 2ab2π2 + π2ab.

¤

The Proof of Proposition 2:

Assume that M is an orientable surface. The flux integralZM

F · ndS

is independent of an orientation-preserving parametrization ofM . The flux integral is multiplied

by −1 under an orientation-reversing parametrization of M.As in the proof of Proposition 1, assume that M is parametrized by Φ,

Φ (u, v) = (x (u, v) , y (u, v) .z (u, v)) , for each (u, v) ∈ D,

and

u = u (u∗, v∗) , v = v (u∗, v∗)

for each (u∗, v∗) ∈ D∗. Set

x (u∗, v∗) = x (u (u∗, v∗) , v (u∗, v∗)) , y (u∗, v∗) = y (u (u∗, v∗) , v (u∗, v∗))z (u∗, v∗) = z (u (u∗, v∗) , v (u∗, v∗)) ,

and

Φ∗ (u∗, v∗) = (x (u∗, v∗) , y (u∗, v∗) .z (u∗, v∗)) , where (u∗, v∗) ∈ D∗.Let’s assume that the transformation (u∗, v∗)→ (u, v) is orientation preserving so that

∂ (u, v)

∂ (u∗, v∗)> 0


on D∗. We haveZ ZD∗(M (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u∗, v∗)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u∗, v∗)

+P (Φ∗ (u∗, v∗))∂ (x, y)

∂ (u∗, v∗))du∗dv∗

=

Z ZD∗(M (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗)

+P (Φ∗ (u∗, v∗))∂ (x, y)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗))du∗dv∗

=

Z ZD∗

µM (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u, v)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u, v)+ (Φ∗ (u∗, v∗))

∂ (x, y)

∂ (u, v)

¶∂ (u, v)

∂ (u∗, v∗)du∗dv∗

=

Z ZD

µM (Φ (u, v))

∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

¶dudv

=

Z ZM

F · ndS

by the rule for the change of variables in double integrals.

If the transformation is orientation reversing so that

∂ (u, v)

∂ (u∗, v∗)< 0

on D∗, we haveZ ZD∗

µM (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u, v)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u, v)+ (Φ∗ (u∗, v∗))

∂ (x, y)

∂ (u, v)

¶∂ (u, v)

∂ (u∗, v∗)du∗dv∗

= −Z Z

D∗

µM (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u, v)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u, v)+ (Φ∗ (u∗, v∗))

∂ (x, y)

∂ (u, v)

¶ ¯∂ (u, v)

∂ (u∗, v∗)

¯du∗dv∗

= −Z Z

D

µM (Φ (u, v))

∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

¶dudv

= −ZM

F · ndS.

¥

Problems

In problems 1-5, determine the area of the surface that is parametrized by the given function

(leave the relevant integral in a simplified form, if that is indicated)

1 (a cylinder)

Φ (u, v) = (v, 3 cos (u) , 3 sin (u)) , −4 ≤ v ≤ 4, 0 ≤ u ≤ 2π.

2 (a paraboloid)

Φ (u, v) =¡u2, u cos (v) , u sin (v)

¢, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2π.


3 (a cone)

Φ (ρ, θ) =

Ã1

2ρ cos (θ) ,

1

2ρ sin (θ) ,

√3

2ρ

!where 0 ≤ ρ ≤ 4 and 0 ≤ θ ≤ 2π.4 (an ellipsoid) Let

Φ (φ, θ) =

µ2 sin (φ) cos (θ) , sin (φ) sin (θ) ,

1

3cos (φ)

¶,

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π. Express the area of the surface as an integral. Simplify theexpression as much as possible. Do not attempt to evaluate the integral.

5 (a surface of revolution)

Ω (x, θ) = (x, ex sin (θ) , ex cos (θ)) ,

where 0 ≤ x ≤ 2, 0 ≤ θ ≤ 2π.

6. Let S be the hemisphere that is parametrized by


where 0 ≤ φ ≤ π and 0 ≤ θ ≤ π EvaluateZ ZS

¡x2 + y2

¢dS

7. Let S be the surface that is parametrized by

Φ (u, v) = (v, 2 cos (u) , 2 sin (u)) , −4 ≤ v ≤ 4, 0 ≤ u ≤ π.

Evaluate Z ZS

y2dS

(this is part of the cylinder y2 + z2 = 1).8. Let S be the helicoid that is parametrized by

Φ (r, θ) = (r cos (θ) , r sin (θ) , θ) , 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π.

Evaluate Z ZS

px2 + y2dS

9. Let S be the part of the plane

x+ y + z = 1

in the first octant (i.e., x ≥ 0, y ≥ 0 and z ≥ 0). EvaluateZ ZS

xdS.


10. Let S be the part of the sphere

x2 + y2 + z2 = 1

that is in the first octant (i.e., x ≥ 0, y ≥ 0 and z ≥ 0). EvaluateZ ZS

z2dS.

11. Let S be the surface that is parametrized by

Φ (u, v) = (cos (u) , v, sin (u)) , −2 ≤ v ≤ 2, 0 ≤ u ≤ π

(this is part of the cylinder x2 + z2 = 1). EvaluateZ ZS

(xi+ zk) · dS.

12. Let S be the planar surface that is parametrized by

Φ (y, z) = (2, y, z) , 0 ≤ y ≤ 3, 0 ≤ z ≤ 4.Evaluate Z Z

S

(yi+ xj+ zk) · dS.

13. Let S be the hemisphere of radius 3 that is parametrized by


where 0 ≤ φ ≤ π and 0 ≤ θ ≤ π. EvaluateZ ZS

xi+ yj+ zk

(x2 + y2 + z2)· dS.

14. Let S be the portion of the plane

z = 1− 12x− 1

2y

in the first octant. Evaluate Z ZS

(xi− zk) ·dS,

if S is oriented so that the normal points upward.

15. Let S be the hemisphere

z =p4− x2 − y2.

Evaluate Z ZS

(xi+yj+ zk) · dS,


16. Let S be the portion of the paraboloid

z = 1− x2 − y2

14.6. GREEN’S THEOREM 259

above the xy-plane. Evaluate Z ZS

(yj+ k) · dS


17. Assume that S consists of the hemisphere S1

z =p4− x2 − y2

and the disk

S2 =©(x, y, 0) : x2 + y2 ≤ 4ª

Evaluate Z ZS

(yzi+ xzj+ xyk) · dS,

if S is oriented so that the normal points towards the exterior of the region enclosed by S.

14.6 Green’s Theorem

In this section we will discuss Green’s Theorem that establishes a link between a line integral on

a closed curve and a double integral and some of the applications of the theorem. For example,

Green’s theorem will be used to show that the necessary condition for the existence of a potential

for a vector field is also sufficient if the vector field is smooth in a domain without holes.

Green’s Theorem in Simply Connected Regions

Assume that the boundary of a region D in the plane is a simple closed curve C, i.e., a closed

curve without self intersections, and that C is oriented so that D is to the left as C is traversed.

In this case we will say that C is the positively oriented boundary of D and refer to D as

the interior of C. The region D is said to be simply connected. There are no holes in

D. You can imagine that any simple closed curve that lies in the interior of D can be shrunk to

a point in D without leaving D.

D

C

Figure 1: A simply connected region D and its positively oriented boundary C

Green’s Theorem relates a line integral on C to a double integral on D:

Theorem 1 (Green’s Theorem) Let the simple closed curve C be the positively oriented

boundary of the region D. We have

ZC

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy,


provided that the partial derivatives of M and N are continuous in D ∪ C.We will not give the proof of Green’s Theorem for an arbitrary region, but we will confirm the

statement of the theorem in the case of a region that is both of type I and type II. We will refer

to such a region as a simple region. The theorem will be confirmed by showing thatZC

Mdx = −Z Z

D

∂M

∂ydxdy,

and ZC

Ndy =

Z ZD

∂N

∂xdxdy.

Let’s establish the first equality. Since D is of type I, D can be expressed as the set of point

(x, y) such that a ≤ x ≤ b and f (x) ≤ y ≤ g (x), as illustrated in Figure 2.

x

y

y gx

C

y fx

C

a b

Figure 2

By Fubini’s Theorem, Z ZD

∂M

∂ydxdy =

Z b

a

ÃZ y=g(x)

y=f(x)

∂M

∂y(x, y) dy

!dx.

By the Fundamental Theorem of Calculus,Z y=g(x)

y=f(x)

∂M

∂y(x, y) dy =M (x, g (x))−M (x, f (x)) .

Therefore, Z ZD

∂M

∂ydxdy =

Z b

a

(M (x, g (x))−M (x, f (x))) dx

=

Z b

a

M (x, g (x)) dx−Z b

a

M (x, f (x)) dx.

Note that Z b

a

M (x, g (x)) dx = −ZC+

Mdx,

and Z b

a

M (x, f (x)) dx =

ZC−Mdx.


Therefore, ZC

Mdx =

ZC+

Mdx+

ZC−Mdx

= −Z b

a

M (x, g (x)) dx+

Z b

a

M (x, f (x)) dx

= −ÃZ b

a

M (x, g (x)) dx−Z b

a

M (x, f (x)) dx

!

= −Z Z

D

∂M

∂ydxdy,

as claimed.

Similarly, since D is also of type II, we can express D as the set of points (x, y) such thatc ≤ y < d and F (y) ≤ x < G (y), as illustrated in Figure 3.

x

y

x GyCx Fy C

a b

Figure 3

By Fubini’s Theorem, Z ZD

∂N

∂xdxdy =

Z d

c

ÃZ x=G(y)

x=F (x)

∂N

∂x(x, y) dx

!dy.

By the Fundamental Theorem of Calculus,Z x=G(y)

x=F (x)

∂N

∂x(x, y) dx = N (G(y), y)−N (F (y), y) .

Therefore, Z ZD

∂N

∂xdxdy =

Z d

c

(N (G(y), y)−N (F (y), y)) dy

=

Z d

c

N (G(y), y) dy −Z d

c

N (F (y), y) dy

Note that Z d

c

N (G(y), y) dy =

ZC+

Ndy,

and

−Z d

c

N (F (y), y) dy =

ZC−Ndy.


Therefore, ZC

Ndy =

ZC+

Ndy +

ZC−Ndy

=

Z d

c

N (G(y), y) dy −Z d

c

N (F (y), y) dy =

Z ZD

∂N

∂xdxdy,

as claimed. ¥

Remark Note that ZC

Mdx+Ndy =

ZC

(M i+N j) · dσ.Thus, Green’s Theorem can be stated asZ

C

F · dσ =Z Z

D

µ∂N

∂x− ∂M

∂y

¶dxdy,

where F =M i+Nj.

Example 1 Let D be the unit disk (x, y) : x2 + y2 ≤ 1 and let C be its positively oriented

boundary. Confirm Green’s Theorem for the line integralZC

ydx+ ln¡x2 + y2 + 1

¢dy.

Solution

Let’s evaluate the line integral directly. We can parametrize the unit circle C by σ (t) =(cos (t) , sin (t)), t ∈ [0, 2π]. Thus,Z

C

ydx+ ln¡x2 + y2 + 1

¢dy =

Z 2π

0

(sin (t) (− sin t) + ln (2) cos (t)) dt

=

Z 2π

0

¡− sin2 (t) + ln (2) cos (t)¢ dt = −π.By Green’s Theorem,Z

C

ydx+ ln¡x2 + y2 + 1

¢dy =

Z ZD

µ∂

∂xln¡x2 + y2 + 1

¢− ∂

∂y(y)

¶dxdy

=

Z ZD

µ1

x2 + y2 + 1(2x)− 1

¶dxdy.

In polar coordinates,Z ZD

µ1

x2 + y2 + 1(2x)− 1

¶dxdy =

Z 2π

θ=0

Z 1

r=0

µ2r cos (θ)

r2 + 1− 1¶rdrdθ

=

Z 1

r=0

2r2

r2 + 1

Z 2π

θ=0

cos (θ) dθdr −Z θ=2π

θ=0

Z 1

r=0

rdrdθ

= 0− 2πµ1

2

¶= −π.

Thus, we have confirmed thatZC

ydx+ ln¡x2 + y2 + 1

¢dy =

Z ZD

µ1

x2 + y2 + 1(2x)− 1

¶dxdy.

¤


Green’s Theorem in Multiply Connected Regions

Assume that D is a region whose boundary consists of the closed curves C1 and C2, as in Fig-

ure 4, with the orientations of the curves as indicated (if you imagine that you walk along the

boundary you should see the region D to your left). Such a region is doubly connected: You

can consider that there is a hole in the interior of C1 that is bounded by the inner curve C2.

x

y

C1C2

Figure 4

In this case Green’s Theorem takes the following form:ZC1

Mdx+Ndy +

ZC2

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy.

We can provide a plausibility argument as follows: Imagine that a cut is made in D so that the

resulting region D0 is bounded by the simple closed curve C1 + C3 + C2 + C4, as in Figure 5(you can imagine that the cut has width that can be made arbitrarily small, so that C3 and C4correspond to the same curve traversed in opposite directions).

x

y

C1C2

C3C4

Figure 5

Since Green’s Theorem is valid in the simply connected domain D0, we haveZC1

Mdx+Ndy+

ZC3

Mdx+Ndy+

ZC2

Mdx+Ndy+

ZC4

Mdx+Ndy =

Z ZD0

µ∂N

∂x− ∂M

∂y

¶dxdy

Since ZC3

Mdx+Ndy +

ZC4

Mdx+Ndy = 0

due to the opposite orientations of C3 and C4.ZC1

Mdx+Ndy +

ZC2

Mdx+Ndy =

Z ZD0

µ∂N

∂x− ∂M

∂y

¶dxdy


Since we can imagine that the width of the cut tends to 0, we have can replace the double

integral on D0 by the double integral on D. ThusZC1

Mdx+Ndy +

ZC2

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy

as claimed. ¥The above version of Green’s theorem can be generalized to "multiply connected regions"

with more than one hole by making cuts that enable us to use Green’s Theorem for a simply

connected region. For example, assume that the region D is bounded by the curves C1, C2 and

C3, with the orientations that are indicated the picture:

x

y

C1

C2C3

Figure 6

Then ZC1

Mdx+Ndy +

ZC2

Mdx+Ndy +

ZC3

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy.

Example 2 Let

F(x, y) = − y

x2 + y2i+

x

x2 + y2j

Assume that C1 is a simple closed curve that is the positively oriented boundary of the region

D and that the positively oriented (counterclockwise) circle C2 of radius a is in D.

x

y

C1

C2

Figure 7

a) Use Green’s Theorem to show thatZC1

F·dσ =ZC2

F·dσ

b) Use the result of part a) to evaluateRC1F · dσ.

Solution


a) We need to consider the negative orientation of the inner boundary C2 in order to apply

Green’s Theorem. ThusZC1

F ·dσ −ZC2

F ·dσ =Z Z

D

µ∂

∂x

µx

x2 + y2

¶− ∂

∂y

µ− y

x2 + y2

¶¶dxdy

=

Z ZD

Ã− x2 − y2(x2 + y2)

2 +x2 − y2(x2 + y2)

2

!dxdy = 0.

Therefore ZC1

F·dσ =ZC2

F·dσ,

as claimed.

b) We can parametrize C2 by σ (t) = (a cos (t) , a sin (t)), t ∈ [0, 2π]. By part a)ZC1

F·dσ =ZC2

F·dσ

=

ZC2

− y

x2 + y2dx+

x

x2 + y2dy

=

Z 2π

0

µ−a sin (t)

a2(−a sin (t)) + a cos (t)

a2(a cos (t))

¶dt

=

Z 2π

0


¢dt

=

Z 2π

0

1dt = 2π.

¤Green’s Theorem can be used to express the area of a region in terms of a line integral on its

boundary:

Proposition 1 Assume that D is the interior of C and that C is positively oriented. Then

Area of D =1

2

ZC

xdy − ydx.

Proof

1

2

ZC

xdy − ydx = 1

2

Z ZD

µ∂

∂x(x) +

∂

∂y(y)

¶dxdy

=1

2

Z ZD

(2) dxdy = Area of D.

¥

Interpretations of Green’s Theorem

In Section 14.1 we defined the curl of a two-dimensional vector field F (x, y) = M (x, y) i +N (x, y) j as

∇×F (x, y) =∇× (M (x, y) i+N (x, y) j+ 0k) =

µ∂N

∂x− ∂M

∂y

¶k.


Therefore, we can express Green’s Theorem,ZC

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy,

as ZC

F · dσ =Z Z

D

(∇×F) · k dxdy.

F

x

y

z

Figure 10

If F (x, y) is the velocity of a fluid particle at (x, y), the line integral is the circulation aroundC. Therefore, the circulation of the velocity field around C is equal to the integral

of the normal component of its curl over its interior D. We can think of the normal

component of curl of F at (x0, y0) as the average circulation of F at (x0, y0):

Proposition 2 Assume that Cr is a circle of radius r centered at (x0, y0) that is orientedcounterclockwise and Dr is the disk bounded by Cr. We have

limr→0

1

πr2

ZCr

F · dσ = (∇×F) (x0, y0) · k

(assuming that ∇×F is continuous).A plausibility argument for Proposition2:

By Green’s Theorem, ZCr

F · dσ =Z Z

Dr

∇×F · k dxdy.

By the continuity of ∇×F, if r is small,Z ZDr

∇×F · k dxdy ∼= (∇×F (x0, y0) · k) (area of Dr) = (∇×F (x0, y0) · k)¡πr2

¢.

Thus,

1

πr2

Z∂Dr

F ·dσ = 1

πr2

Z ZDr

∇×F ·k dS ∼= 1

πr2(∇×F (x0, y0) · k)

¡πr2

¢=∇×F (x0, y0) ·k.

This lends support to the assertion that

limr→0

1

πr2

Z∂Dr

F · dσ =∇×F (x0, y0) · k.

¥


Example 3 Let

F (x, y) = −ωyi+ ωxj,

so that the vector field represents circulation around the origin.

In Section 14.1 we saw that

∇×F (x, y) = 2ωk⇒∇×F (x, y) · k = 2ω.Let Cr be the circle of radius r that is centered at the origin. We have

1

πr2

ZCr

F · dσ = ω

πr2

Z 2π

0

r2 (− sin (t) i+ cos (t) j) · (− sin (t) + cos (t)) dt

=ω

π

Z 2π

0


¢dt

=ω

π(2π) = 2ω,

as well. ¤We can also interpret Green’s Theorem in terms of the flux of a vector field across

a simple closed curve C. If F (x, y) =M (x, y) i+N (x, y) j, the divergence of F at (x, y) is

∇ · F (x, y) = ∂M

∂x(x, y) +

∂N

∂y(x, y)

By Green’s Theorem,ZC

−Ndx+Mdy =Z Z

D

µ∂N

∂y+

∂M

∂x

¶dxdy =

Z ZD

∇ · F (x, y) dA

Let us reexamine the expression on the left hand side. Assume that C is parametrized by

σ (t) = (x (t) , y (t)), where t ∈ [a, b]:ZC

−Ndx+Mdy =Z b

a

(−N (x (t) , y (t))) dxdt+M (x (t) , y (t))

dy

dt)dt

=

Z b

a

(M (x, y) i+N (x, y) j) ·µdy

dti− dx

dtj

¶dt

=

Z b

a

F (x (t) , y (t)) ·µdy

dti− dx

dtj

¶dt

=

Z b

a

F (x (t) , y (t)) ·

µdy

dti− dx

dtj

¶sµ

dx

dt

¶2+

µdy

dt

¶2sµ

dx

dt

¶2+

µdy

dt

¶2dt.

Note that

n (t) =

µdy

dti− dx

dtj

¶sµ

dx

dt

¶2+

µdy

dt

¶2is orthogonal to the tangent

dσ

dt=dx

dti+

dy

dtj,


||n (t)|| = 1, and n (t) points towards the exterior of C. We will refer to n (t) as the unitnormal at (x (t) , y (t)) that points towards the exterior of C. Thus,

ZC

−Ndx+Mdy =Z b

a

F (x (t) , y (t)) ·

µdy

dti− dx

dtj

¶sµ

dx

dt

¶2+

µdy

dt

¶2sµ

dx

dt

¶2+

µdy

dt

¶2dt

=

Z b

a

F (x (t) , y (t)) · n (t) ds

=

ZC

F · nds

The quantity ZC

F · nds.is the outward flux of F across C. By Green’s Theorem,Z

C

F · nds =Z Z

D

∇ · F (x, y) dxdy

Thus, .the outward flux of F across C is equal to the integral of the divergence of F

over the interior of C.

x

y

nT F

Figure 11

We can interpret the divergence of F at (x0, y0) as the flux per unit area of F at

(x0, y0).

Proposition 3 Assume that Cr is a circle of radius r centered at (x0, y0) that is orientedcounterclockwise and Dr is the disk bounded by Cr. We have

limr→0

1

πr2

ZCr

F · nds =∇ · F (x0, y0)

(assuming that ∇ · F is continuous).A plausibility argument for Proposition2:

We have ZCr

F · nds =Z Z

Dr

∇ · F (x, y) dxdy

By continuity, if r is smallZ ZDr

∇ · F (x, y) dxdy =∇ · F (x0, y0)× (area of Dr) =∇ · F (x0, y0)×¡πr2

¢.


Therefore1

πr2

ZCr

F · nds ∼= 1

πr2∇ · F (x0, y0)×

¡πr2

¢=∇ · F (x0, y0)

if r is small. This supports the claim that

limr→0

1

πr2

ZCr

F · nds =∇ · F (x0, y0) .

¥

Irrotational and Incompressible Flow

If F is the velocity field of a fluid in motion, we say that the flow is incompressible If

∇ · F (x, y) = 0 for each (x, y) in a region D that is relevant to the flow. The flow is irro-

tational if we have ∇×F (x, y) = 0 for each(x, y) ∈ D. Assume that C is simply connected.

If C is a closed curve that lies in D and F =M i+N j, thenZC

−Ndx+Mdy =ZC

F · nds =Z Z

D

∇ · F (x, y) dA = 0.

Therefore, the vector field −N i+Mj is the gradient of a scalar function, say g (x, y). Thus,

−N (x, y) = ∂g

∂x(x, y) and M (x, y) =

∂g

∂y(x, y) .

Since we also have ∇×F (x, y) = 0 for each(x, y) ∈ D, F is the gradient of scalar function f :

M (x, y) =∂f

∂x(x, y) and N (x, y) =

∂f

∂y(x, y) .

Therefore,∂f

∂x(x, y) =

∂g

∂y(x, y) and

∂f

∂y(x, y) = −∂g

∂x(x, y)

Also note that

∂2f

∂x2+

∂2f

∂y2=

∂M

∂x+

∂N

∂y=

∂2g

∂x∂y+

∂

∂y

µ−∂g∂x

¶= 0,

∂2g

∂x2+

∂2g

∂y2=

∂

∂x(−N) + ∂M

∂y= − ∂2f

∂x∂y+

∂

∂y

µ∂f

∂x

¶= 0.

Thus, if a vector field F is irrotational and conservative, a potential function f satisfies Laplace’s

equation. Such functions are said to be harmonic. The stream function, i.e., a potential g for

−N i+Mj also satisfies Laplace’s equation.

Sufficient Conditions for the Existence of a Potential Function

In Section 14.3 we saw that the integral of a gradient field is independent of path, and that a

necessary condition for F (x, y) = M (x, y) i + N (x, y) j to be a gradient field in a region D is

that∂M

∂y(x, y) =

∂N

∂x(x, y)

for each (x, y) ∈ D. Green’s Theorem enables us to discuss the sufficiency of this condition for

a vector field to be conservative:


Theorem 2 Assume that D is a simply connected region, F (x, y) =M (x, y) i+N (x, y) j and

∂M

∂y(x, y) =

∂N

∂x(x, y)

for each (x, y) ∈ D, and that these partial derivatives are continuous on D. Then the vectorfield F is conservative.

Proof

Step 1. To begin with, let us note that the line integral of a vector field is independent

of path if its integral around any simple closed curve is 0. We will not give a proof of

this statement under general conditions. For example, if C1 and C2 join P1 to P2 as in Figure

8, then C1 − C2 is a simple closed curve.

x

y

P1 P2

C2

C1

Figure 8

We have ZC1−C2

F · dσ = 0,

so that ZC1

F · dσ −ZC2

F · dσ = 0⇒ZC1

F · dσ =ZC2

F · dσ.

Step 2. If F (x, y) =M (x, y) i+N (x, y) j and

∂M

∂y(x, y) =

∂N

∂x(x, y)

for each (x, y) ∈ D, the line integrals of F are independent of path. By Step 1, it is

sufficient to show that ZC

F · dσ = 0

if C is a simple closed curve in D,. The interior of C is entirely in D since D is assumed to be

simply connected. By Green’s Theorem,ZC

F · dσ =ZC

Mdx+Ndy =

Z ZInt(C)

µ∂N

∂x− ∂M

∂y

¶dxdy = 0.

Step 3. Let’s fix a point P0 = (x0, y0) in D and define f (P ) = f (x, y) asZC(P0,P1)

F · dσ,

where C (P0, P1) is an arbitrary curve in D that joins P0 to P . The definition of f (x, y) makessense since the line integrals of F are independent of path, by Step 2.


Claim:

F (x, y) =∇f (x, y)for each (x, y) ∈ D.

x

y

P0

P x , yx x, y

Figure 9

Let |∆x| be small enough so that any point of the form (x+∆x, y) ∈ D. By independence ofpath, we can set

f (x+∆x, y)− f (x, y) =ZC(P0,P )+C∆x

F · dσ −ZC(P0,P )

F · d σ =ZC∆x

F · σ

where C∆x is the line segment from P = (x, y) to (x+∆x, y). We can parametrize C∆x byσ (t) = (x+ t∆x,y), where 0 ≤ t ≤ 1. Then,Z

C∆x

F · dσ =ZC∆x

Mdx+Ndy =

Z 1

0

µM (x+ t∆x, y)

dx

dt+N (x+ t∆x)

dy

dt

¶dt

=

Z 1

0

M (x+ t∆x, y)∆xdt

= ∆x

Z 1

0

M (x+ t∆x, y) dt

Therefore,

lim∆x→0

f (x+∆x, y)− f (x, y)∆x

= lim∆x→0

Z 1

0

M (x+ t∆x, y) dt

= lim∆x→0

µZ 1

0

M (x+ t∆x, y)−M (x, y)

¶+M (x, y)

= lim∆x→0

µZ 1

0

M (x+ t∆x, y)−Z 1

0

M (x, y) dt

¶+M (x, y)

= lim∆x→0

Z 1

0

(M (x+ t∆x)−M (x, y)) dt+M (x, y)

=M (x, y) ,

since

lim∆x→0

Z 1

0

(M (x+ t∆x)−M (x, y)) dt =

Z 1

0

lim∆x→0

(M (x+ t∆x)−M (x, y)) dt = 0,

by the continuity of M at (x, y).


Thus, we have shown that

∂f

∂x(x, y) = lim

∆x→0f (x+∆x, y)− f (x, y)

∆x=M (x, y) .

Similarly, by considering the vertical line that connects (x, y) to (x, y +∆y), we can show that

∂f

∂y(x, y) = N (x, y) .

Therefore„

F (x, y) =M (x, y) i+N (x, y) j =∂f

∂x(x, y) i+

∂f

∂y(x, y) j =∇f (x, y) ,

as claimed. ¥

Remark 1 The assumption that D is simply connected is essential. If

F(x, y) = − y

x2 + y2i+

x

x2 + y2j,

as in Example 2, we have∂

∂y

µ− y

x2 + y2

¶=

∂

∂x

µx

x2 + y2

¶for each (x, y) 6= (0, 0). But F is not conservative in D =

©(x, y) ∈ R2: (x, y) 6= (0, 0)ª, since

we showed that ZC

F·dσ = 2π

if C is any simple closed curve that contains the origin in its interior (if F were conservative

in D such a line integral would have been 0). There is no contradiction here, since F does not

satisfy the condition∂

∂y

µ− y

x2 + y2

¶=

∂

∂x

µx

x2 + y2

¶at the origin. D has a hole at the origin, so that it is not simply connected. ♦

Problems

In problems 1-4

a) Use Green’s Theorem to express the given line integral as a double integral,

b) Evaluate the double integral.

1. ZC

y3dx− x3dy,

where C is the positively oriented boundary of the disk bounded by the circle x2 + y2 = 4.

2. ZC

xy2dx+ 2x2ydy,

where C is the positively oriented boundary of the triangular region with vertices (0, 0) , (2, 0)and (2, 4).

3. ZC

cos (y) dx+ x2 sin (y) dy,


where C is the positively oriented boundary of the rectangular region with vertices (0, 0) , (5, 0) , (5,π)and (0,π).

4. ZC

xe−2xdx+¡x4 + 2x2y2

¢dy,

where C is the positively oriented boundary of the annular region bounded by the circles x2+y2 =1 an x2 + y2 = 4.


a) Use Green’s Theorem to express ZC

F · dσ

as a double integral,

b) Evaluate the double integral.

5.

F (x, y) = y2 cos (x) i+¡x2 + 2y sin (x)

¢j

and C = C1+C2+C3, where C1 is the line segment from (0, 0) to (2, 6), C2 is the line segmentfrom (2, 6) to (2, 0), and C3 is the line segment from (2, 0) to (0, 0) (pay attention to orientation).

6.

F (x, y) =¡ex + x2y

¢i+

¡ey − xy2¢ j

and C is the positively oriented boundary of the disk bounded by the circle x2 + y2 = 25.

7. Let

F(x, y) =x

x2 + y2i+

y

x2 + y2j

Assume that C1 is a positively oriented simple closed curve such that the circle of radius 1centered at the origin is contained in its interior. CalculateZ

C1

F · dσ

Caution: Green’s Theorem is not applicable in the interior of C1 since F is not defined at the

origin.

In problems 8 and 9, make use of the appropriate form of Green’s Theorem to calculateZC

F · nds,

where n is the unit normal that points towards the exterior of C:

8.

F (x, y) = x2y3i− xyjand C is the positively oriented boundary of the triangular region with vertices (0, 0) , (1, 0)and (1, 2).

9.

F (x, y) = x3i+ y3j

and C is the positively oriented boundary of the disk bounded by the circle x2 + y2 = 4.

calculus iii - cognella academic publishing · preface this is the third volume of my calculus...

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