Calculus 3 Solution Manual

Calculus III

Instructor’s Solution Manual

Tunc Geveci

Spring 2011

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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

11.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

12 Functions of Several Variables 19

12.1 Tangent Vectors and Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

12.2 Acceleration and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

12.3 Real-Valued Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . 27

12.4 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

12.5 Linear Approximations and the Differential . . . . . . . . . . . . . . . . . . . . . 39

12.6 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

12.7 Directional Derivatives and the Gradient . . . . . . . . . . . . . . . . . . . . . . . 51

12.8 Local Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

12.9 Absolute Extrema and Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . 63

13 Multiple Integrals 69

13.1 Double Integrals Over Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

13.2 Double Integrals Over Non-Rectangular Regions . . . . . . . . . . . . . . . . . . 72

13.3 Double Integrals in Polar Coordinates: . . . . . . . . . . . . . . . . . . . . . . . . 79

13.4 Applications of Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

13.5 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

13.6 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . 94

14 Vector Analysis 107

14.1 Vector Fields, Divergence and Curl . . . . . . . . . . . . . . . . . . . . . . . . . . 107

14.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

14.3 Line Integrals of Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . 118

14.4 Parametrized Surfaces and Tangent Planes . . . . . . . . . . . . . . . . . . . . . 124

14.5 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

14.6 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

14.7 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

14.8 Gauss’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

iii

iv CONTENTS

Chapter 11

Vectors

11.1 Cartesian Coordinates in 3D and Surfaces

1.

If x = c then2c+ y + 4z = 8⇒ y + 4z = 8− 2c.

These are lines.

If y = c then2x+ c+ 4z = 8⇒ 2x+ 4z = 8− c.

These are lines.

If z = c then2x+ y + 4c = 8⇒ 2x+ y = 8− 4c.

These are lines.

2.

1

2 CHAPTER 11. VECTORS

If x = c thenz = 4c− 3y.

These are lines.

If y = c thenz = 4x− 3c.

These are lines.

If z = c thenc = 4x− 3y.

These are lines.

3.

If x = c thenz = 4c2 + 9y2

These are parabolas.

If y = c thenz = 4x2 + 9c2


If z = c thenc = 4x2 + 9y2

These are ellipses if c > 0. If c = 0 the curve is reduced to a single point, the origin. The surfacedoes not have any points below the xy-plane corresponding to c < 0.

4.

If x = c thenc = y2 + 4z2

These are ellipses if c > 0. If c = 0 the curve is reduced to a single point, the origin. The surfacedoes not have any points corresponding to c < 0.

11.1. CARTESIAN COORDINATES IN 3D AND SURFACES 3

If y = c thenx = c2 + 4z2


If z = c thenx = y2 + 4c2


5.

If x = c thenc2 + 2y2 + 4z2 = 4⇒ 2y2 + 4z2 = 4− c2.

These are ellipses if −2 < c < 2. The curve is reduced to a single point if c = ±2. The surfacedoes not have any points corresponding to x < −2 or x > 2.If y = c then

x2 + 2c2 + 4z2 = 4⇒ x2 + 4z2 = 4− 2c2

These are ellipses if −√2 < c < √2. The curve is reduced to a single point if c = ±√2. Thesurface does not have any points corresponding to y >

√2 or y < −√2.

If z = c thenx2 + 2y2 + 4c2 = 4⇒ x2 + 2y2 = 4− 4c2

These are ellipses if −1 < c < 1. The curve is reduced to a single point if c = ±1. The surfacedoes not have any points corresponding to z > 1 or z < −1.6.

If x = c then4c2 + y2 + 9z2 = 4⇒ y2 + 9z2 = 4− 4c2

These are ellipses if −1 < c < 1. The curve is reduced to a single point if c = ±1. The surfacedoes not have any points corresponding to x < −1 or x > 1.If y = c then

4x2 + c2 + 9z2 = 4⇒ 4x2 + 9z2 = 4− c2


These are ellipses if −2 < c < 2. The curve is reduced a single point if c = ±2. The surfacedoes not have any points corresponding to y > 2 or y < −2.If z = c then

4x2 + y2 + 9c2 = 4⇒ 4x2 + y2 = 4− 9c2

These are ellipses if −2/3 < c < 2/3. The curve is reduced to a single point if c = ±2/3. Thesurface does not have any points corresponding to z > 2/3 or z < −2/3.7.

If x = c thenc2 − 9y2 − 4z2 = 1⇒ 9y2 + 4z2 = c2 − 1

These are ellipses if c < −1 or c > 1. The curve is reduced to a single point if c = ±1. Thesurface does not have any points corresponding to −1 < x < 1.If y = c then

x2 − 9c2 − 4z2 = 1⇒ x2 − 4z2 = 1 + 9c2

These are hyperbolas.

If z = c thenx2 − 9y2 − 4c2 = 1⇒ x2 − 9y2 = 1 + 4c2


8.

If x = c then4y2 − c2 − 2z2 = 1⇒ 4y2 − 2z2 = 1 + c2


If y = c then4c2 − x2 − 2z2 = 1⇒ x2 + 2z2 = 4c2 − 1

11.1. CARTESIAN COORDINATES IN 3D AND SURFACES 5

These are ellipses if c > 1/2 or c < −1/2. The surface does not have any points correspondingto −1/2 < y < 1/2.If z = c then

4y2 − x2 − 2c2 = 1⇒ 4y2 − x2 = 1 + 2c2These are hyperbolas.

9.

If x = c theny2 − c2 + z2 = 1⇒ y2 + z2 = 1 + c2

These are circles.

If y = c thenc2 − x2 + z2 = 1⇒ −x2 + z2 = 1− c2


If z = c theny2 − x2 + c2 = 1⇒ y2 − x2 = 1− c2


10.

If x = c thenc2 + y2 − z2 = 4⇒ y2 − z2 = 4− c2


If y = c thenx2 + c2 − z2 = 4⇒ x2 − z2 = 4− c2


If z = c thenx2 + y2 − c2 = 4⇒ x2 + y2 = 4 + c2

These are circles.


11.2 Vectors in Two and Three Dimensions

1.

P1 = (1, 2) , P2 = (3, 5) , Q1 = (2, 1)

a)

v = (2, 3)

b) −−→OQ2 =

−−→OQ1 + v = (2, 1) + (2, 3) = (4, 4) .

Therefore, Q2 = (4, 4) .

1 2 3 4x

1

2

3

4

5

y

P1

P2

Q1

Q2

2.

P1 = (−2, 3) , P2 = (−4, 2) , Q1 = (1, 3)a)

v = (−2,−1)b) −−→

OQ2 =−−→OQ1 + v =(1, 3) + (−2,−1) = (−1, 2) .

Therefore, Q2 = (−1, 2) .

-6 -4 -2 -1 1 2 3 5x

2

3

4

5

y

P1

P2

Q1

Q2

3.

P1 = (2,−3) , P2 = (4, 2) , Q1 = (3, 2)a)

v =(2, 5)

b) −−→OQ2 =

−−→OQ1 + v = (3, 2) + (2, 5) = (5, 7)

11.2. VECTORS IN TWO AND THREE DIMENSIONS 7

Therefore, Q2 = (5, 7).

2 3 4 5x

-3

2

7

y

P1

P2Q1

Q2

4.

P1 = (−2,−1) , P2 = (−4, 2) , Q1 = (−1, 2)a)

v = (−2, 3)b) −−→

OQ2 =−−→OQ1 + v =(−1, 2) + (−2, 3) = (−3, 5)

Therefore, Q2 = (−3, 5) .

-4 -3 -2 -1x

-1

2

5

y

P1

P2

Q1

Q2

5.

a)

v+w = (2, 1) + (3, 4) = (5, 5)

b)

2 3 5

1

4

5

v

wv+w


6.

a)

v +w = (2,−3) + (3, 2) = (5,−1)b)

2 3 5

-3

-1

2

3

5

v

w

v+w

7.

a)

v +w = (−2,−1) + (2, 4) = (0, 3)b)

-2 -1 2

-1

3

4

v

w

v+w

8.

a)

v+w = (2, 3) + (−1,−5) = (1,−2)b)

-1 1 2

-5

-3

-2

v

w

v+w

11.2. VECTORS IN TWO AND THREE DIMENSIONS 9

9.

a)

v +w = (2, 4) + (−1, 2) = (1, 6)b)

-1 2

2

4

6

v

w

v+w

10.

a)

v+w = (−4, 2) + (3, 1) = (−1, 3)b)

-4 3-1

2

1

3

v

w

v+w

11.

v −w = (2, 4)− (−2, 2) = (4, 2)

2-2 4

4

2 v

w

v- w

12.

a)

v−w = (−4, 2)− (3, 1) = (−7, 1)


b)

-4 3-7

2

1v

w

v- w

13.

2v− 3w = 2 (3,−1)− 3(−2, 5) = (6,−2) + (6,−15) = (12,−17)14.

−4v+ 5w = −4 (2, 4) + 5 (1,−4) = (−8,−16) + (5,−20) = (−3,−36)15.

a) We have

||v|| =p32 + 42 =

√25 = 5.

Therefore,

u =1

||v||v =1

5(3, 4) =

µ3

5,4

5

¶.

b)

3

4

v

u

16.

a) We have

||v|| =q(−2)2 + 22 =

√8.

Therefore,

u =1

||v||v ==1√8(−2, 2) =

µ− 2√

8,2√8

¶=

Ã−√8

4,

√8

4

!

b)

11.3. THE DOT PRODUCT 11

-2

2

v

u

17.

a)

v = 3i+ 2j, w = −2i+4jb)

2v− 3w = 2 (3i+ 2j)− 3 (−2i+4j) = 6i+ 4j+ 6i− 12j = 12i− 8j18.

a)

v = −4i+ j, w = 4i+ 3jb)

2v+w = 2 (−4i+ j) + 4i+ 3j = −8i+ 2j+ 4i+ 3j = −4i+ 5j19.

a)

v = −2i+ 3j+ 6k, w = 4i− 2j+ kb)

−v+ 4w = − (−2i+ 3j+ 6k) + 4 (4i− 2j+ k)= 2i− 3j− 6k+16i− 8j+ 4k = 18i− 11j− 2k

20.

a)

v = −i− 3j+ 5k, w = 7i+ 2j− 2kb)

3v − 2w = 3 (−i− 3j+ 5k)− 2 (7i+ 2j− 2k)= −3i− 9j+ 15k−14i− 4j+ 4k

11.3 The Dot Product

1.

a)

||v|| = 1, ||w|| =√2, v ·w = 1.

b)

θ = arccos

µv ·w

||v|| ||w||¶= arccos

µ1√2

¶=

π

4.

2.

a)

||v|| = 1, ||w|| = √1 + 3 =√4 = 2, v ·w = −1.


b)

θ = arccos

µv ·w


µ−12

¶= π − π

3=2π

3.

3.

a)

||v|| =√2, ||w|| =

vuutÃ√3 + 12

!2+

Ã√3− 12

!2=√2,

v ·w =√3 + 1

2+

√3− 12

=√3

b)

θ = arccos

µv ·w


Ã √3√2√2

!= arccos

Ã√3

2

!=

π

6

4.

a)

||v|| =√2, ||w|| =

√2, v ·w = 0.

b)

θ = arccos

µv ·w

||v|| ||w||¶= arccos (0) =

π

2.

Thus, v and w are orthogonal to each other.

5.

a)

||v|| =√2, ||w|| =

vuutÃ√3 + 12

!2+

Ã1−√32

!2=√2,

v ·w = 1b)

θ = arccos

µv ·w


µ1√2√2

¶= arccos

µ1

2

¶=

π

3.

6.

a)

||v|| =√2, ||w|| =

√2, v ·w = 0.

b) θ = arccos (0) = π/2. The vectors v and w are orthogonal to each other.

7.

a)

||v|| =√5, ||w|| =

√2, v ·w = −1

b)

cos (θ) =v ·w

||v|| ||w|| =−1√5√2= − 1√

10.

c)

θ = arccos

µ− 1√

10

¶∼= 1. 892 55

8.

a)

||v|| =√13, ||w|| =

√20, v ·w = (3) (−2) + (2) (4) = 2


b)

cos (θ) =v ·w

||v|| ||w|| =2√13√20=

2√260

c)

θ = arccos

µ2√260

¶= 1. 446 44

9.

a)

||v|| =√5, ||w|| =

√10, v ·w = (−2) (−1) + (−1) (3) = −1

b)

cos (θ) =v ·w

||v|| ||w|| =−1√5√10= − 1√

50.

c)

θ = arccos

µ− 1√

50

¶= 1. 712 69

10.

a)

||v|| =p1 + 32 + 1 =

√11, ||w|| =

p22 + 1 + 1 =

√6,

v ·w = (1) (2) + (3) (−1) + (−1) (1) = −2b)

cos (θ) =v ·w

||v|| ||w|| =−2√11√6= − 2√

66

c)

θ = arccos

µ− 2√

66

¶= 1. 819 54

11.

a)

||v|| = √4 + 9 =√13⇒ u =

1

||v||v =1√13(2, 3) =

µ2√13,3√13

¶b) The direction cosines of v are

2√13and

3√13

12.

a)

||v|| = √1 + 4 =√5⇒ u =

1

||v||v =1√5(−i+2j) = − 1√

5i+

2√5j

b) The direction cosines of v are

− 1√5and

2√5

13.

a)

||v|| = √9 + 16 = 5⇒ u =1

||v||v =1

5(−3, 4) =

µ−35,4

5

¶b) The direction cosines of v are

−35and

4

5


14.

a)

||v|| = √1 + 4 + 1 =√5⇒ u =

1

||v||v =1√5(i− 2j+ k) = 1√

5i− 2√

5j+

1√5k

b) The direction cosines of v are1√5, − 2√

5and

1√5

15.

a)

||w|| = √36 + 4 =√40⇒ u =

1

||w||w =1√40(6, 2) =

µ6√40,2√40

¶b)

compwv = v · u =(3, 4) ·µ

6√40,2√40

¶=

26√40

c)

Pwv =(compwv)u =26√40u =

26√40

µ6√40,2√40

¶=

µ39

10,13

10

¶d)

v2 = v−Pwv =(3, 4)−µ39

10,13

10

¶=

µ− 910,27

10

¶

-2 2 4 6 8

2

4

vw

Pwv

v2

16.

a)

||w|| = √4 + 1 =√5⇒ u =

1

||w||w =1√5(2, 1) =

µ2√5,1√5

¶b)

compwv = v · u =(−2, 1) ·1√5(2, 1) = − 3√

5

c)

Pwv =(compwv)u = −3√5

µ2√5,1√5

¶=

µ−65,−35

¶d)

v2 = v−Pwv =(−2, 1)−µ−65,3

5

¶=

µ−45,2

5

¶


-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

wv

Pwv

v2

17.

a)

||w|| = √4 + 1 =√5⇒ u =

1

||w||w =1√5(2,−1) =

µ2√5,− 1√

5

¶b)

compwv = v · u =(1,−2) ·µ2√5,− 1√

5

¶=

4√5

c)

Pwv =(compwv)u =4√5

µ2√5,− 1√

5

¶=

µ8

5,−45

¶d)

v2 = v −Pwv = (1,−2)−µ8,−4

5

¶=

µ−7,−6

5

¶

1 2 3

-1

-2

Pwv

v

w

v2

18.

a)

||w|| = √9 + 16 = 5⇒ u =1

||w||w =1

5(−3,−4) =

µ−35,−45

¶b)

compwv = v · u = (−2,−6) ·µ−35,−45

¶= 6

c)

Pwv =(compwv)u = 6

µ−35,−45

¶=

µ−185,−24

5

¶


d)

v2 = v −Pwv = (−2,−6)−µ−185,−24

5

¶=

µ8

5,−65

¶

-2-4-6

-2

-4

-6

w

v

Pwv

v2

11.4 The Cross Product

1.

v×w =¯¯ i j k

2 1 01 3 0

¯¯ = ¯ 1 0

3 0

¯i−

¯2 01 0

¯j+

¯2 11 3

¯k = 5k

2.

v ×w =¯¯ i j k

0 3 20 −2 1

¯¯ = ¯ 3 2

−2 1

¯i−

¯0 20 1

¯j+

¯0 30 −2

¯k = 7i

3.

v×w =¯¯ i j k

3 1 4−2 3 1

¯¯ =

¯1 43 1

¯i−

¯3 4−2 1

¯j+

¯3 1−2 3

¯k

= −11i− 11j+ 11k

4.

v ×w =¯¯ i j k

−2 1 41 −2 5

¯¯ =

¯1 4−2 5

¯i−

¯ −2 41 5

¯j+

¯ −2 11 −2

¯k

= 13i+ 14j+ 3k

5.

(−1, 3, 0)× (2, 6, 0) =¯¯ i j k

−1 3 02 6 0

¯¯ = ¯ −1 3

2 6

¯k = −12k

Therefore, the area of the parallelogram spanned by (−1, 3) and (2, 6) is

||(−1, 3, 0)× (2, 6, 0)|| = ||−12k|| = 12.

11.4. THE CROSS PRODUCT 17

6.

(−i+ 3j− 2k)× (4i− j− 6k) =

¯¯ i j k

−1 3 −24 −1 −6

¯¯

=

¯3 −2−1 −6

¯i−

¯ −1 −24 −6

¯j+

¯ −1 34 −1

¯k

= −20i− 14j+−11kTherefore, the area of the parallelogram spanned by −i+ 3j− 2k and 4i− j− 6k is

||−20i− 14j+−11k|| =p202 + 142 + 112 =

√717 (∼= 26. 776 9)

7.

u · (v ×w) =¯¯ 2 1 33 1 01 0 4

¯¯ = 2

¯1 00 4

¯−¯3 01 4

¯+ 3

¯3 11 0

¯= 2 (4)− 12 + 3 (−1) = −7

Therefore, the volume of the parallelepiped spanned by u,v and w is |−7| = 7.8.

u · (v×w) =¯¯ 2 2 53 −1 −41 −1 0

¯¯ = 2

¯ −1 −4−1 0

¯− 2

¯3 −41 0

¯+ 5

¯3 −11 −1

¯= 2 (−4)− 2 (4) + 5 (−2) = −26

Therefore, the volume of the parallelepiped spanned by u,v and w is |−26| = 26.9.

3 (x− 1) + 2 (y − 3) + (z − 4) = 0⇔ 3x+ 2y + z = 13

10.

(x− 2)− 2y + 4 (z − 5) = 0⇔ x− 2y + 4z = 2211.

− (x− 3) + (y − 1) + 2 (z + 3) = 0⇔ −x+ y + 2z = −812.

(x− 3)− 4 (y − 1)− (z − 2) = 0⇐⇒ x− 4y − z = −313. We have −−−→

P0P1 = (−1, 0,−1) and −−−→P0P2 = (0,−5, 2) .We set

N =−−−→P0P1 ×−−−→P0P2 =

¯¯ i j k

−1 0 −10 −5 2

¯¯

=

¯0 −1−5 2

¯i−

¯ −1 −10 2

¯j+

¯ −1 00 −5

¯k

= −5i+ 2j+ 5kThe vector N is orthogonal to the plane. Therefore, the equation of the plane is

(−5i+ 2j+ 5k)× ((x− 1) i+ (y − 2) j+ (z − 2)k) = 0


⇔−5 (x− 1) + 2 (y − 2) + 5 (z − 2) = 0⇔ −5 + 2y + 5z = 9

14. We have −−−→P0P1 = (−3, 6, 2) and −−−→P0P2 = (−2, 5,−1)

We set

N =−−−→P0P1 ×−−−→P0P2 =

¯¯ i j k

−3 6 2−2 5 −1

¯¯

=6 25 −1 i−

¯ −3 2−2 −1

¯j+

¯ −3 6−2 5

¯k

= −16i− 7j− 3k

The vector N is orthogonal to the plane. Therefore, the equation of the plane is

−16 (x− 3)− 7 (y + 2)− 3 (z + 1) = 0 ⇔ −16x− 7y − 3z = −31⇔ 16x+ 7y + 3z = 31

Chapter 12

Functions of Several Variables

12.1 Tangent Vectors and Velocity

1.

a)

σ0 (t) =d

dt

¡t, sin2 (4t)

¢=

µd

dt(t) ,

d

dtsin2 (4t)

¶= (1, 8 sin (4t) cos (4t))

(= i+ 8 sin (4t) cos (4t) j)

Therefore,

σ0³π3

´=

µ1 + 8 sin

µ4π

3

¶cos

µ4π

3

¶¶=

Ã1, 8

Ã−√3

2

!µ−12

¶!=

³1, 2√3´ ³

= i+ 2√3j´

Thus, ¯¯σ0³π3

´¯¯=√1 + 12 =

√13.

Therefore,

T³π3

´=

1√13

³1, 2√3´=

Ã1√13,2√3√13

! Ã=

1√13i+

2√3√13j

!b) We have

σ³π3

´=

µπ

3, sin2

µ4π

3

¶¶=

µπ

3,3

4

¶Therefore,

L (u) = σ³π3

´+ uσ0

³π3

´=

µπ

3,3

4

¶+ u

³1, 2√3´

=

µπ

3+ u,

3

4+ 2√3u

¶, −∞ < u ≤ +∞.

2.

19

20 CHAPTER 12. FUNCTIONS OF SEVERAL VARIABLES

a)

σ0 (t) =d

dt(2 cos (3t) , sin (t)) = (−6 sin (3t) , cos (t)) .

Therefore,

σ0³π4

´=

µ−6 sin

µ3π

4

¶, cos

³π4

´¶=

Ã−3√2,

√2

2

!Thus, ¯¯

σ0³π4

´¯¯=

r18 +

1

2=

r37

2

Therefore,

T³π4

´=

r2

37

Ã−3√2,

√2

2

!=

µ− 6√

37,1√37

¶b) We have

σ³π4

´=

µ2 cos

µ3π

4

¶, sin

³π4

´¶=

Ã−√2,

√2

2

!.

Therefore,

L (u) = σ³π4

´+ uσ0

³π4

´=

Ã−√2,

√2

2

!+ u

Ã−3√2,

√2

2

!

=

Ã−√2− 3

√2u,

√2

2+

√2

2u

!, u ∈ R

3.

a)

σ0 (t) =µd

dt

µ3t

t2 + 1

¶,d

dt

µ3t2

t2 + 1

¶¶=

Ã−3t2 + 3(t2 + 1)

2 ,6t

(t2 + 1)2

!Therefore,

σ0 (1) =µ0,3

2

¶Thus,

||σ0 (1)|| = 3

2.

Therefore,

T (1) =2

3σ0 (1) = (0, 1) (= j) .

b) We have

σ (1) =

µ3

2,3

2

¶.

Therefore,

L (u) = σ (1) + uσ0 (1) =µ3

2,3

2

¶+ u

µ0,3

2

¶=

µ3

2,3

2+3

2u

¶, u ∈ R.

4.

a)

σ0 (t) =µd

dt(4 cos (3t)) ,

d

dt(4 sin (3t)) ,

d

dt(2t)

¶= (−12 sin (3t) , 12 cos (3t) , 2) .

12.2. ACCELERATION AND CURVATURE 21

Therefore,

σ0³π2

´=

µ−12 sin

µ3π

2

¶, 12 cos

µ3π

2

¶, 2

¶= (12, 0, 2) .

Thus, ¯¯σ0³π2

´¯¯=√144 + 4 =

√148

Therefore,

T³π2

´=

1√148

σ0³π2

´=

1√148

(12, 0, 2) =

µ12√148

, 0,2√148

¶.

b) We have

σ³π2

´=

µ4 cos

µ3π

2

¶, 4 sin

µ3π

2

¶, 2³π2

´¶= (0,−4,π) .

Therefore,

L (u) = σ³π2

´+ uσ0

³π2

´= (0,−4,π) + u (12, 0, 2) = (12u,−4,π + 2u) , u ∈ R.

5.

v (t) = σ0 (t) =

µd

dt

¡e−t sin (t)

¢,d

dt

¡e−t cos (t)

¢¶=

¡e−t (cos (t)− sin (t)) ,−e−t (cos (t) + sin (t))¢

= e−t (cos (t)− sin (t)) i− e−t (cos (t) + sin (t)) j

Therefore,

v (π) =¡−e−π, e−π¢ = −e−πi+ e−πj

Thus, the speed at t = π is

||v (π)|| =pe−2π + e−2π =

√2e−π.

6.

v (t) = σ0 (t) =

µd

dt(t) ,

d

dtarctan (t)

¶=

µ1,

1

t2 + 1

¶= i+

1

t2 + 1j

Therefore,

v (1) =

µ1,1

2

¶= i+

1

2j

Thus, the speed at t = 1 is

||v (1)|| =r1 +

1

4=

√5

2

12.2 Acceleration and Curvature

1. We have

v (t) = σ0 (t) =

µd

dt(6 + 3 cos (4t)) ,

d

dt(6 + 3 sin (4t))

¶= (−12 sin (4t) , 12 cos (4t)) .


Therefore,

a (t) = v0 (t) = (−48 cos (4t) ,−48 sin (4t)) = −48 cos (4t) i− 48 sin (4t) j

Thus,

a (π/12) = −48 cos³π3

´i− 48 sin

³π3

´j = −48

µ1

2

¶i− 48

Ã√3

2

!j

= −24i− 24√3j

2. We have

v (t) = σ0 (t) =

µd

dt(6t− 6 sin (2t)) , d

dt(6t− 6 cos (2t))

¶= (6− 12 cos (2t) , 6 + 12 sin (2t)) .

Therefore,

a (t) = v0 (t) = (24 sin (2t) , 24 cos (2t)) .

Thus,

a (0) = (0, 24) .

3. We have

v (t) = σ0 (t) =µd

dt(t) ,

d

dtarctan (t)

¶=

µ1,

1

t2 + 1

¶= i+

1

t2 + 1j

Therefore,

a (t) = v0 (t) =d

dt

¡t2 + 1

¢−1j = −2t ¡t2 + 1¢−2 j = − 2t

(t2 + 1)2j.

Thus,

a (1) = −24j = −1

2j.

4. We have

v (t) = σ0 (t) =µd

dt

¡tet¢,d

dtet¶=¡et + tet, et

¢=¡et + tet

¢i+ etj

Therefore,

a (t) = v0 (t) =¡2et + tet

¢i+ etj

Thus,

a (2) = 4e2i+ e2j

5. We have

σ0 (t) =d

dt(4 + 2 cos (t) , 3 + 2 sin (t)) = (−2 sin (t) , 2 cos (t)) = 2 (− sin (t) , cos (t))

and

||σ0 (t)|| = 2qsin2 (t) + cos2 (t) = 2.

Therefore,

T (t) =σ0 (t)||σ0 (t)|| =

1

2(2 (− sin (t) , cos (t))) = (− sin (t) , cos (t))


Thus,

T (π/6) =³− sin

³π6

´, cos

³π6

´´=

Ã−12,

√3

2

!=1

2

³−1,√3´.

We have

dT

dt=d

dt(− sin (t) , cos (t)) = (− cos (t) ,− sin (t)) = − (cos (t) , sin (t))

and ¯¯dT

dt

¯¯= ||− (cos (t) , sin (t))|| =

qcos2 (t) + sin2 (t) = 1.

Therefore,

N (t) =T0 (t)||T0 (t)|| = T

0 (t) = − (cos (t) , sin (t)) .

Thus,

N (π/6) = −³cos³π6

´, sin

³π6

´´= −

Ã√3

2,1

2

!= −1

2

³√3, 1´.

6. We have

σ0 (t) =d

dt(cos (3t) , sin (3t) , 4t) = (−3 sin (3t) , 3 cos (3t) , 4)

and

||σ0 (t)|| =q9 sin2 (3t) + 9 cos2 (3t) + 16 =

√25 = 5.

Therefore,

T (t) =σ0 (t)||σ0 (t)|| =

1

5(−3 sin (3t) , 3 cos (3t) , 4) =

µ−35sin (3t) ,

3

5cos (3t) ,

4

5

¶.

Thus,

T (π/2) =

µ−35sin

µ3π

2

¶,3

5cos

µ3π

2

¶,4

5

¶=

µ3

5, 0,

4

5

¶=1

5(3, 0, 4) .

We have

dT

dt=

d

dt

µ−35sin (3t) ,

3

5cos (3t) ,

4

5

¶=

µ−95cos (3t) ,−9

5sin (3t) , 0

¶= −9

5(cos (3t) , sin 3t, 0)

and ¯¯dT

dt

¯¯=

¯¯−95(cos (3t) , sin 3t, 0)

¯¯=9

5

qcos2 (3t) + sin2 (3t) =

9

5.

Therefore,

N (t) =T0 (t)||T0 (t)|| =

5

9

µ−95(cos (3t) , sin (3t) , 0)

¶= − (cos (3t) , sin (3t) , 0)

Thus

N (π/2) = −µcos

µ3π

2

¶, sin

µ3π

2

¶, 0

¶= (0, 1, 0) .


Therefore,

B (π/2) = T (π/2)×N (π/2) =1

5(3, 0, 4)× (0, 1, 0) =

=1

5

¯¯ i j k

3 0 40 1 0

¯¯ = 1

5(−4i+ 3k) .

7.

a) We have

σ0 (t) =d

dt(2 cos (t) , 3 sin (t)) = (−2 sin (t) , 3 cos (t))

so that

||σ0 (t)|| =q4 sin2 (t) + 9 cos2 (t).

Therefore,

s (t) =

Z t

0

||σ0 (τ)|| dτ =Z t

0

q4 sin2 (τ) + 9 cos2 (τ)dτ .

b)

s (π) =

Z π

0

q4 sin2 (τ) + 9 cos2 (τ)dτ ∼= 7. 932 72

8.

a) We have

σ0 (t) =d

dt

¡tet, et

¢=¡et + tet, et

¢= et (1 + t, 1)

so that

||σ0 (t)|| = etq(1 + t)2 + 1 = et

pt2 + 2t+ 2.

Therefore,

s (t) =

Z t

0

||σ0 (τ)|| dτ =Z t

0

eτpτ2 + 2τ + 2dτ .

b)

s (1) =

Z 1

0

eτpτ2 + 2τ + 2dτ ∼= 3. 227 01

9. We have

σ0 (t) =d

dt(3 cos (t) , 2 sin (t) , 0) = (−3 sin (t) , 2 cos (t) , 0)

and

σ00 (t) =d

dt(−3 sin (t) , 2 cos (t) , 0) = (−3 cos (t) ,−2 sin (t) , 0) .

Thus,

σ0 (π/2) = (−3, 0, 0) and σ00 (π/2) = (0,−2, 0) .Therefore,

||σ0 (π/2)||3 = 23 = 8.

We have

σ00 (π/2)× σ0 (π/2) =

¯¯ i j k

0 −2 0−3 0 0

¯¯ = −6k


so that

||σ00 (π/2)× σ0 (π/2)|| = 6.Therefore,

κ (π/2) =||σ00 (π/2)× σ0 (π/2)||

||σ0 (π/2)||3 =6

8=3

4.

10. We have

σ0 (t) =d

dt

¡t, et, 0

¢=¡1, et, 0

¢and

σ00 (t) =d

dt

¡1, et, 0

¢=¡0, et, 0

¢.

Thus,

σ0 (2) =¡1, e2, 0

¢and σ00 (2) =

¡0, e2, 0

¢.

Therefore,

||σ0 (2)||3 =³p

1 + e4´3=¡1 + e4

¢3/2.

We have

σ00 (2)× σ0 (2) =

¯¯ i j k

0 e2 01 e2 0

¯¯ = −e2k

so that

||σ00 (2)× σ0 (2)|| = e2.Therefore,

κ (2) =||σ00 (2)× σ0 (2)||

||σ0 (2)||3 =e2

(1 + e4)3/2

11. We have

v (t) = σ0 (t) =d

dt(cos (2t) , sin (2t) , 0) = (−2 sin (2t) , 2 cos (2t) , 0)

so that the speed is

||v (t)|| =q4 sin2 (2t) + 4 cos2 (2t) = 2.

Therefored

dt||v (t)|| = d

dt(2) = 0.

Thus, the tangential component of the acceleration is always 0 (this is merely a circular motion).

We have

σ0 (π/6) = (−2 sin (π/3) , 2 cos (π/3) , 0) =Ã−2Ã√

3

2

!, 2

µ1

2

¶, 0

!=³−√3, 1, 0

´so that

||σ0 (π/6)|| = √3 + 1 = 2.We have

σ00 (t) =d

dt(−2 sin (2t) , 2 cos (2t) , 0) = (−4 cos (2t) ,−4 sin (2t) , 0) ,

so that

σ00 (π/6) = (−4 cos (π/3) ,−4 sin (π/3) , 0) =Ã−4µ1

2

¶,−4

Ã√3

2

!, 0

!= −2

³1,√3, 0´.


Therefore,

σ00 (π/6)× σ0 (π/6) =

¯¯ i j k

−2 −2√3 0

−√3 1 0

¯¯ = (−2− 6)k = −8k

so that

||σ00 (π/6)× σ0 (π/6)|| = 8.Thus,

κ (π/6) =||σ00 (π/6)× σ0 (π/6)||

||σ0 (π/6)||3 =8

23= 1.

Therefore, the normal component of the acceleration is

κ (π/6) ||v (π/6)||2 = (1) ¡23¢ = 8.12. We have

v (t) = σ0 (t) =d

dt

¡t, t2, 0

¢= (1, 2t, 0)

so that the speed is

||v (t)|| =p1 + 4t2.

Therefored

dt||v (t)|| = d

dt

p1 + 4t2 =

1

2√1 + 4t2

(8t) =4t√1 + 4t2

.

Thus, the tangential component of the acceleration at t = 1 is

d

dt||v (t)||

¯t=1

=4√5.

We have

σ0 (1) = (1, 2, 0)

so that

||σ0 (1)|| =√5.

We have

σ00 (t) =d

dt(1, 2t, 0) = (0, 2, 0)

Therefore,

σ00 (1)× σ0 (1) =

¯¯ i j k

0 2 01 2 0

¯¯ = −2k

so that

||σ00 (1)× σ0 (1)|| = 2.Thus,

κ (1) =||σ00 (1)× σ0 (1)||

||σ0 (1)||3 =2¡√5¢3 = 2

53/2.

Therefore, the normal component of the acceleration is

κ (1) ||v (1)||2 =µ2

53/2

¶(5) =

2

5

√5

12.3. REAL-VALUED FUNCTIONS OF SEVERAL VARIABLES 27

12.3 Real-Valued Functions of Several Variables

In the plots of the level curves for problems 1-6, the smaller values of f are indicated by thedarker color.

1.

a)

b)

2.

a)

b)


3.

a)

b)

4.

a)

b)

5.

a)

12.3. REAL-VALUED FUNCTIONS OF SEVERAL VARIABLES 29

b)

6.

a)

b)

7.


The level surfaces are spheres. In the pictures, the outer sphere is shown partially in order to

show the inner sphere.

8.

The level surfaces are ellipsoids. In the pictures, the outer ellipsoid is shown partially in order

to show the inner ellipsoid.

9. The pictures show two level surfaces of f .

12.4. PARTIAL DERIVATIVES 31

10. The pictures show two level surfaces of f .

12.4 Partial Derivatives

1.∂

∂x

¡4x2 + 9y2

¢= 8x,

∂

∂y

¡4x2 + 9y2

¢= 18y

2.∂

∂x

¡6x2 − 5y2¢ = 12x, ∂

∂y

¡6x2 − 5y2¢ = −10y

3.∂

∂x

p2x2 + y2 =

∂

∂x

¡2x2 + y2

¢1/2=1

2

¡2x2 + y2

¢−1/2(4x) =

2xp2x2 + y2

,

∂

∂y

p2x2 + y2 =

∂

∂y

¡2x2 + y2

¢1/2=1

2

¡2x2 + y2

¢−1/2(2y) =

yp2x2 + y2

4.∂

∂r

¡r2 cos (θ)

¢= 2r cos (θ) ,

∂

∂θ

¡r2 cos (θ)

¢= −r2 sin (θ)

5.∂

∂xe−x

2−y2 = −2xe−x2−y2 , ∂

∂ye−x

2−y2 = −2ye−x2−y2

7.

∂

∂xln¡x2 + y2

¢=

Ãd

duln (u)

¯u=x2+y2

!µ∂

∂x

¡x2 + y2

¢¶=

µ1

x2 + y2

¶(2x) =

2x

x2 + y2,


∂

∂yln¡x2 + y2

¢=

Ãd

duln (u)

¯u=x2+y2

!µ∂

∂y

¡x2 + y2

¢¶=

µ1

x2 + y2

¶(2y) =

2y

x2 + y2,

8.

∂

∂r

³er

2

tan (θ)´= 2rer

2

tan (θ) ,∂

∂θ

³er

2

tan (θ)´= er

2

µ1

cos2 (θ)

¶=

er2

cos2 (θ)

9.

∂

∂xarctan

Ãyp

x2 + y2

!=

Ãd

duarctan (u)

¯u=y/√x2+y2

!Ã∂

∂x

Ãyp

x2 + y2

!!

=

⎛⎜⎜⎝ 1

1 +y2

x2 + y2

⎞⎟⎟⎠µy ∂

∂x

¡x2 + y2

¢−1/2¶

=

µx2 + y2

x2 + 2y2

¶µy

µ−12

¡x2 + y2

¢−3/2(2x)

¶¶= − xy

¡x2 + y2

¢(x2 + 2y2) (x2 + y2)3/2

= − xy

(x2 + 2y2)px2 + y2

,

∂

∂yarctan

Ãyp

x2 + y2

!=

Ãd

duarctan (u)

¯u=y/√x2+y2

!Ã∂

∂y

Ãyp

x2 + y2

!!

=

⎛⎜⎜⎝ 1

1 +y2

x2 + y2

⎞⎟⎟⎠⎛⎝px2 + y2 − y

³∂∂y

¡x2 + y2

¢1/2´x2 + y2

⎞⎠

=

µx2 + y2

x2 + 2y2

¶⎛⎝px2 + y2 − y³12

¡x2 + y2

¢−1/2(2y)

´x2 + y2

⎞⎠=

µx2 + y2

x2 + 2y2

¶Ãx2 + y2 − y2(x2 + y2)

3/2

!=

x2

(x2 + 2y2)px2 + y2


10.

∂

∂xarccos

Ãxp

x2 + y2

!=

Ãd

duarccos (u)

¯u=x/√x2+y2

!Ã∂

∂x

Ãxp

x2 + y2

!!

=

Ã− 1√

1− u2¯u=x/√x2+y2

!⎛⎜⎜⎜⎜⎝px2 + y2 − x

Ã2x

2px2 + y2

!x2 + y2

⎞⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎝− 1s1− x2

x2 + y2

⎞⎟⎟⎟⎟⎠Ãx2 + y2 − x2(x2 + y2)

3//2

!

= − y2sy2

x2 + y2(x2 + y2)3//2

= − y2

|y| (x2 + y2)

11.

∂

∂x

Ã1p

x2 + y2 + z2

!=

∂

∂x

¡x2 + y2 + z2

¢−1/2=

Ãd

duu−1/2

¯u=x2+y2+z2

!µ∂

∂x

¡x2 + y2 + z2

¢¶

=

Ã−12u−3/2

¯u=x2+y2+z2

!(2x)

= − x

(x2 + y2 + z2)3/2

Similarly,

∂

∂y

Ã1p

x2 + y2 + z2

!= − y

(x2 + y2 + z2)3/2,

∂

∂z

Ã1p

x2 + y2 + z2

!= − z

(x2 + y2 + z2)3/2

12.

∂

∂x

¡xyzex−y+2z

¢= yzex−y+2z + xyzex−y+2z,

∂

∂y

¡xyzex−y+2z

¢= xzex−y+2z − xyzex−y+2z,

∂

∂z

¡xyzex−y+2z

¢= xyex−y+2z + 2xyzex−y+2z


13.

∂

∂xarcsin

Ã1p

x2 + y2 + z2

!=

Ãd

duarcsin (u)

¯u=1/√x2+y2+z2

!Ã∂

∂x

1px2 + y2 + z2

!

=

Ã1√1− u2

¯u=1/√x2+y2+z2

!Ã− x

(x2 + y2 + z2)3/2

!

=

⎛⎜⎜⎝ 1r1− 1

x2 + y2 + z2

⎞⎟⎟⎠Ã− x

(x2 + y2 + z2)3/2

!

=

Ã px2 + y2 + z2p

x2 + y2 + z2 − 1

!Ã− x

(x2 + y2 + z2)3/2

!= − xp

x2 + y2 + z2 − 1 (x2 + y2 + z2)14.

∂

∂ρ(ρ cos (ϕ) sin (θ)) = cos (ϕ) sin (θ) ,

∂

∂ϕ(ρ cos (ϕ) sin (θ)) = −ρ sin (ϕ) sin (θ) ,

∂

∂θ(ρ cos (ϕ) sin (θ)) = ρ cos (ϕ) cos (θ)

15.

a) We have

∂f

∂x(x, y) =

∂

∂x

px2 + y2 =

1

2px2 + y2

(2x) =xp

x2 + y2,

and∂f

∂y(x, y) =

∂

∂y

px2 + y2 =

1

2px2 + y2

(2y) =yp

x2 + y2.

Thus,∂f

∂x(3, 1) =

3√32 + 1

=3√10and

∂f

∂y(3, 1) =

1√32 + 1

=1√10

Therefore, the vector

i+∂f

∂x(3, 1)k = i+

3√10k

is tangential to C1 at (3, 1, f (3, 1)) =¡3, 1,√10¢.

The vector

j+∂f

∂y(3, 1)k = j+

1√10k

is tangential to C2 at¡3, 1,√10¢.

b) The line that is tangent to C1 at (3, 1, f (3, 1)) =¡3, 1,√10¢is parametrized by

L1 (u) =³3, 1,√10´+ u

µ1, 0,

3√10

¶=

µ3 + u, 1,

√10 +

3√10u

¶,


i.e.,

x (u) = 3 + u,

y (u) = 1,

z (u) =√10 +

3

10

√10u,

where u is an arbitrary real number.Similarly, The line that is tangent to C1 at (3, 1, f (3, 1)) =

¡3, 1,√10¢is parametrized by

L2 (u) =³3, 1,√10´+ u

µ0, 1,

1√10

¶=

µ3, 1 + u,

√10 +

1√10u

¶, u ∈ R.

16.

a) We have∂f

∂x(x, y) =

∂

∂x

¡10− x2 − y2¢ = −2x

and∂f

∂y(x, y) =

∂

∂x

¡10− x2 − y2¢ = −2y.

Thus,∂f

∂x(2, 1) = −4 and ∂f

∂y(2, 1) = −2.


i+∂f

∂x(2, 1)k = i− 4k

is tangential to C1 at (2, 1, f (2, 1)) = (2, 1, 51).The vector

j+∂f

∂y(2, 1)k = j− 2k

is tangential to C2 at (2, 1, 5) .

b) The line that is tangent to C1 at (2, 1, f (2, 1)) = (2, 1, 5) is parametrized by

L1 (u) = (2, 1, 5) + u (1, 0,−4) = (2, 1, 5) + (u, 0,−4u)= (2 + u, 1, 5− 4u) .

The line that is tangent to C2 at (2, 1, 5) is parametrized by

L2 (u) = (2, 1, 5) + u (0, 1,−2) = (2, 1, 5) + u (0, 1,−2)= (2, 1 + u, 5− 2u) .

17.

a) We have∂f

∂x(x, y) =

∂

∂x

³ex

2+y2´= 2xex

2+y2

and∂f

∂y(x, y) =

∂

∂x

³ex

2+y2´= 2yex

2+y2 .


Thus,∂f

∂x(1, 0) = 2e and

∂f

∂y(1, 0) = 0.


i+∂f

∂x(1, 0)k = i+ 2ek

is tangential to C1 at (1, 0, f (1, 0)) = (1, 0, e).The vector

j+∂f

∂y(1, 0)k = j

is tangential to C2 at (1, 0, e) .

b) The line that is tangent to C1 at (1, 0, f (1, 0)) = (1, 0, e) is parametrized by

L1 (u) = (1, 0, e) + u (1, 0, 2e) = (1, 0, e) + (u, 0, 2eu)

= (1 + u, 0, e+ 2eu) , u ∈ R.The line that is tangent to C2 at (1, 0, e) is parametrized by

L2 (u) = (1, 0, e) + u (0, 1, 0) = (1, 0, e) + (0, u, 0)

= (1, u, e) , u ∈ R.

18.

a) We have∂f

∂x(x, y) =

∂

∂x

¡x2 − y2¢ = 2x

and∂f

∂y(x, y) =

∂

∂x

¡x2 − y2¢ = −2y.

Thus,∂f

∂x(3, 2) = 6 and

∂f

∂y(3, 2) = −4.


i+∂f

∂x(3, 2)k = i+ 6k

is tangential to C1 at (3, 2, f (3, 2)) = (3, 2, 5).The vector

j+∂f

∂y(3, 2)k = j− 4k

is tangential to C2 at (3, 2, 5) .

b) The line that is tangent to C1 at (3, 2, f (3, 2)) = (3, 2, 5) is parametrized by

L1 (u) = (3, 2, 5) + u (1, 0, 6) = (3, 2, 5) + (u, 0, 6u)

= . (3 + u, 2, 5 + 6u) , u ∈ R.The line that is tangent to C2 at (2, 1, 5) is parametrized by

L2 (u) = (3, 2, 5) + u (0, 1,−4) = (3, 2, 5) + (0, u,−4u)= (3, 2 + u, 5− 4u) , u ∈ R.


19.

∂f

∂x(x, y) =

∂

∂x

¡4x2 + 9y2

¢= 8x,

∂f

∂y(x, y) =

∂

∂7

¡4x2 + 9y2

¢= 18y,

∂2f

∂x2(x, y) =

∂

∂x(8x) = 8,

∂2f

∂x∂y(x, y) =

∂

∂x(18y) = 0

20.

∂f

∂x(x, y) =

∂

∂x

¡x2 + y2

¢−1/2= −1

2

¡x2 + y2

¢−3/2(2x) = − x

(x2 + y2)3/2

∂2f

∂y∂x(x, y) =

∂

∂y

³−x ¡x2 + y2¢−3/2´ = −x ∂

∂y

¡x2 + y2

¢−3/2= −x

µ−32

¶¡x2 + y2

¢−5/2(2y) =

3xy

(x2 + y2)5/2,

∂f

∂y(x, y) =

∂

∂y

¡x2 + y2

¢−1/2= −1

2

¡x2 + y2

¢−3/2(2y) = − y

(x2 + y2)3/2,

∂2f

∂y2(x, y) =

∂

∂y

Ã− y

(x2 + y2)3/2

!= (−1)

Ã1

(x2 + y2)3/2

!− y

µ∂

∂y

¡x2 + y2

¢−3/2¶= − 1

(x2 + y2)3/2− y

µ−32

¡x2 + y2

¢−5/2(2y)

¶= − 1

(x2 + y2)3/2+

3y2

(x2 + y2)5/2

=− ¡x2 + y2¢+ 3y2(x2 + y2)

5/2=−x2 + 2y2(x2 + y2)

5/2

21.∂f

∂x(x, y) =

∂

∂xe−x

2+y2 = −2xe−x2+y2 ,

∂2f

∂x2(x, y) =

∂

∂x

³−2xe−x2+y2

´= −2e−x2+y2 + 4x2e−x2+y2 ,

∂2f

∂y∂x(x, y) =

∂

∂y

³−2xe−x2+y2

´= −2x (2y) e−x2+y2 = −4xye−x2+y2

22.

∂f

∂y(x, y) =

∂

∂ysin³p

x2 − y2´

= cos³p

x2 − y2´Ã 1

2px2 − y2 (−2y)

!= − yp

x2 − y2 cos³p

x2 − y2´


∂2f

∂x∂y(x, y) =

∂

∂x

Ã− yp

x2 − y2 cos³p

x2 − y2´!

= −yµ

∂

∂x

¡x2 − y2¢−1/2¶ cos³px2 − y2´

− ypx2 − y2

µ∂

∂xcos³p

x2 − y2´¶

= −yµ1

2

¡x2 − y2¢−3/2 2x¶ cos³px2 − y2´

+yp

x2 − y2 sin³p

x2 − y2´ÃÃ 1

2px2 − y2 (2x)

!!=

xy

(x2 − y2)3/2cos³p

x2 − y2´+

xy

x2 − y2 sin³p

x2 − y2´

23.∂f

∂z(x, y, z) =

∂

∂z

¡xyzex−y+2z

¢= xyex−y+2z + 2xyzex−y+2z,

∂2f

∂x∂z(x, y, z) =

∂

∂x

¡xyex−y+2z + 2xyzex−y+2z

¢= yex−y+2z + xyex−y+2z + 2yzex−y+2z + 2xyzex−y+2z,= ex−y+2z (y + xy + 2yz + 2xyz)

∂2f

∂y∂z(x, y, z) =

∂

∂y

¡xyex−y+2z + 2xyzex−y+2z

¢= xex−y+2z − xyex−y+2z + 2xzex−y+2z − 2xyzex−y+2z= ex−y+2z (x− xy + 2xz − 2xyz)

24.∂f

∂θ(ρ,ϕ, θ) =

∂

∂θ(ρ cos (ϕ) sin (θ)) = ρ cos (ϕ) cos (θ) ,

∂2f

∂ϕ∂θ(ρ,ϕ, θ) =

∂

∂ϕ(ρ cos (ϕ) cos (θ)) = −ρ sin (ϕ) cos (θ) ,

∂2f

∂ρ∂θ(ρ,ϕ, θ) =

∂

∂ρ(ρ cos (ϕ) cos (θ)) = cos (ϕ) cos (θ)

25.

fy (x, y) =∂

∂yln¡x2 + 4y2

¢=

1

x2 + 4y2(8y) =

8y

x2 + 4y2.

fxy (x, y) =∂

∂x

µ8y

x2 + 4y2

¶= 8y

∂

∂x

¡x2 + 4y2

¢−1= 8y

³− ¡x2 + 4y2¢−2 (2x)´ = − 16xy

(x2 + 4y2)2

fx (x, y) =∂

∂xln¡x2 + 4y2

¢=

1

x2 + 4y2(2x) =

2x

x2 + 4y2,

12.5. LINEAR APPROXIMATIONS AND THE DIFFERENTIAL 39

fyx (x, y) =∂

∂y

µ2x

x2 + 4y2

¶= 2x

∂

∂y

¡x2 + 4y2

¢−1= 2x

³− ¡x2 + 4y2¢−2 (8y)´ = − 16xy

(x2 + 4y2).

Thus, fyx = fxy.

26.

fy (x, y) =∂

∂y

³x2yex

2−2y2´

= x2ex2−2y2 + x2y

³−4yex2−2y2

´= x2ex

2−2y2 − 4x2y2ex2−2y2 ,

fxy (x, y) =∂

∂x

³x2ex

2−2y2 − 4x2y2ex2−2y2´

= 2xex2−2y2 + x2

³2xex

2−2y2´− 8xy2ex2−2y2 − 4x2y2

³2xex

2−2y2´

= ex2−2y2 ¡2x+ 2x3 − 8xy2 − 8x3y2¢

fx (x, y) =∂

∂x

³x2yex

2−2y2´

= 2xyex2−2y2 + x2y (2x) ex

2−2y2

= 2xyex2−2y2 + 2x3yex

2−2y2 ,

fyx (x, y) =∂

∂y

³2xyex

2−2y2 + 2x3yex2−2y2

´= 2xex

2−2y2 + 2xy³−4yex2−2y2

´+ 2x3ex

2−2y2 + 2x3y³−4yex2−2y2

´= ex

2−2y2 ¡2x− 8xy2 + 2x3 − 8x3y2¢Thus, fyx = fxy.

12.5 Linear Approximations and the Differential

1.

a) We have

∂f

∂x(x, y) =

∂

∂x

¡x2 + y2

¢3/2=3

2

¡x2 + y2

¢1/2(2x) = 3x

px2 + y2,

∂f

∂y(x, y) = 3y

px2 + y2,

Therefore,∂f

∂x(3, 4) = 45,

∂f

∂y(3, 4) = 60, and f (3, 4) = 125.

Thus,

L (x, y) = 125 + 45 (x− 3) + 60(y − 4).b)

f (3.1, 3.9) ∼= L (3.1, 3.9)= 125 + 45 (0.1) + 60 (−0.1) = 123.5


c) According to a calculator, f (3.1, 3.9) ∼= 123. 652. The absolute error is|f (3.1, 3.9)− 123.5| ∼= 0.152.

2.

a)∂f

∂x(x, y) = −e−x sin (y) , ∂f

∂y(x, y) = e−x cos (y) .

Therefore,∂f

∂x(0,π/2) = −1 and ∂f

∂y(0,π/2) = 0, and f (0,π/2) = 1.

Thus,

L (x, y) = 1− x.b)

f (0.2,π/2 + 0.1) ∼= L (0.2,π/2 + 0.1) = 1− 0.2 = 0.8.c) According to a calculator, f (0.2,π/2 + 0.1) ∼= 0.814 641. The absolute error is

|f (0.2,π/2 + 0.1)− 0.8| ∼= 1.4× 10−23.

a)∂

∂xf (x, y) = − y

x2 + y2,

∂

∂yf (x, y) =

x

x2 + y2,

so that∂

∂xf³√3, 1´= −1

4, and

∂

∂yf³√3, 1´=

√3

4, and f

³√3, 1´=

π

6.

Therefore,

L (x, y) =π

6− 14

³x−√3´+

√3

4(y − 1) .

b)

f (1.8, 0.8) ∼= L (1.8, 0.8) = π

6− 14

³1.8−

√3´+

√3

4(0.8− 1) ∼= 0.420 009

c) According to a calculator f (1.8, 0.8) ∼= 0.418 224. The absolute error is|f (1.8, 0.8)− L (1.8, 0.8)| ∼= 1.8× 10−3

4.

a)∂

∂xf (x, y) = 2xex

2+y2 ,∂

∂yf (x, y) = 2yex

2+y2 .

Therefore,∂

∂xf (1, 0) = 2e,

∂

∂yf (1, 0) = 0, and f (1, 0) = e.

Thus,

L (x, y) = e+ 2e (x− 1) .b)

f (1.1,−0.2) ∼= L (1.1,−0.2) = e+ 2e (0.1) = 1.2e ∼= 3. 261 94c) According to a calculator f (1.1,−0.2) ∼= 3. 490 34. The absolute error is

|f (1.1,−0.2)− 1.2e| ∼= 0.23

12.5. LINEAR APPROXIMATIONS AND THE DIFFERENTIAL 41

5.

a)

∂f

∂x(x, y, z) =

xpx2 + y2 + z2

,∂f

∂y(x, y, z) =

ypx2 + y2 + z2

,

∂f

∂z(x, y, z) =

zpx2 + y2 + z2

.

Therefore,∂f

∂x(1, 2, 3) =

1√14,∂f

∂y(1, 2, 3) =

2√14,∂f

∂z(1, 2, 3) =

3√14.

and f (1, 2, 3) =√14.

Thus,

L (x, y, z) =√14 +

1√14(x− 1) + 2√

14(y − 2) + 3√

14(z − 3) .

b)

f (0.9, 2.2, 2.9) ∼= L (0.9, 2.2, 2.9)=√14 +

1√14(−0.1) + 2√

14(0.2) +

3√14(−0.1)

∼= 3. 741 66

c). According to a calculator f (0.9, 2.2, 2.9) ∼= f (0.9, 2.2, 2.9). The absolute error is

|f (0.9, 2.2, 2.9)− L (0.9, 2.2, 2.9)| ∼= 8× 10−3

6.

a)∂

∂xf (x, y) =

xpx2 + y2

,∂

∂yf (x, y) =

ypx2 + y2

Therefore,

df =xp

x2 + y2dx+

ypx2 + y2

dy.

b)

f (12.1, 4.9)− f (12, 5) ∼= df (12, 5, 0.1,−0.1)=12

13(0.1) +

5

13(−0.1) .

Therefore,

f (12.1, 4.9) ∼= f (12, 5) + 1213(0.1) +

5

13(−0.1)

= 13 +12

13(0.1) +

5

13(−0.1) ∼= 13. 053 8

c) According to a calculator f (12.1, 4.9) ∼= 13. 054 5. The absolute error is

|f (12.1, 4.9)− 13. 053 8| ∼= 1.5× 10−3

7.


a)∂

∂xf (x, y, z) =

yz

2√xyz

,∂

∂yf (x, y, z) =

xz

2√xyz

,∂

∂zf (x, y, z) =

xy

2√xyz

.

Therefore,

df =yz

2√xyz

dx+xz

2√xyz

dy +xy

2√xyz

dz.

b)

f (0.9, 2.9, 3.1)− f (1, 3, 3) ∼= df (1, 3, 3,−0.1,−0.1, 0.1)=

9

2 (3)(−0.1) + 3

2 (3)(−0.1) + 3

2 (3)(0.1) = −0.15

Therefore,

f (0.9, 2.9, 3.1) ∼= f (1, 3, 3)− 0.15 = 3− 0.15 = 2. 85c) According to a calculator f (0.9, 2.9, 3.1) ∼= 2. 844 47. The absolute error is

|f (0.9, 2.9, 3.1)− 2. 85| ∼= 5.5× 10−3

8.

a)∂f

∂x(x, y) =

2x

(x2 + y2)2+ 1

,∂f

∂y(x, y) =

2y

(x2 + y2)2+ 1

.

Therefore,

df =2x

(x2 + y2)2 + 1dx+

2y

(x2 + y2)2 + 1dy.

b)

f (0.1,−1.2)− f (0,−1) ∼= df (0,−1, 0.1,−0.2) = −22(−0.2) = 0.2.

Therefore,

f (0.1,−1.2) ∼= f (0,−1) + 0.2 = arctan (1) + 0.2 = π

4+ 0.2 ∼= 0.985 398

c) According to a calculator f (0.1,−1.2) ∼= 0.967 047. The absolute error i

|f (0.1,−1.2)− 0.985 398| ∼= 0.018

12.6 The Chain Rule

1.

a) If we set z = f (x, y), we have

∂z

∂x=

∂

∂x

px2 + y2 =

1

2px2 + y2

(2x) =xp

x2 + y2,

and∂z

∂y=

∂

∂y

px2 + y2 =

1

2px2 + y2

(2y) =yp

x2 + y2.

We havedx

dt=d

dt(sin (t)) = cos (t) and

dy

dt=d

dt(2 cos (t)) = −2 sin (t)

12.6. THE CHAIN RULE 43

Therefore,

d

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

dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt

=

Ãxp

x2 + y2

!cos (t) +

Ãyp

x2 + y2

!(−2 sin (t))

=sin (t) cos (t)− 4 cos (t) sin (t)q

sin2 (t) + 4 cos2 (t)= − 3 sin (t) cos (t)q

sin2 (t) + 4 cos2 (t)

b) We have

f (x (t) , y (t)) = f (sin (t) , 2 cos (t)) =

qsin2 (t) + 4 cos2 (t).

Therefore,

d

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

d

dt

qsin2 (t) + 4 cos2 (t)

=1

2qsin2 (t) + 4 cos2 (t)

(2 sin (t) cos (t)− 8 cos (t) sin (t))

= − 3 sin (t) cos (t)qsin2 (t) + 4 cos2 (t)

2.

a) If we set z = f (x, y), we have

∂z

∂x=

∂

∂xe−x

2+y2 = −2xe−x2+y2 ,

and∂z

∂y=

∂

∂ye−x

2+y2 = 2ye−x2+y2 .

We havedx

dt=d

dt(2t− 1) = 2 and dy

dt=d

dt(t+ 1) = 1.

Therefore,

d

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

dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt

=³−2xe−x2+y2

´(2) +

³2ye−x

2+y2´(1)

= e−x2+y2 (−4x+ 2y)

= e−(2t−1)2+(t+1)2 (−4 (2t− 1) + 2 (t+ 1))

= e6t−3t2

(6− 6t) .

b) We have

f (x (t) , y (t)) = f (2t− 1, t+ 1) = e−(2t−1)2+(t+1)2 = e6t−3t2 .Therefore,

d

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

d

dte6t−3t

2

= e6t−3t2

(6− 6t) .3.


a) If we set z = f (x), we havedz

dx=d

dxln (x) =

1

x.

We also have

∂x

∂u=

∂

∂u

µ1√

u2 + v2

¶=

Ã− 1

2 (u2 + v2)3/2

!(2u) = − u

(u2 + v2)3/2,

and

∂x

∂v=

∂

∂v

µ1√

u2 + v2

¶=

Ã− 1

2 (u2 + v2)3/2

!(2v) = − v

(u2 + v2)3/2.

Therefore,

∂

∂uf (x (u, v)) =

∂z

∂u=dz

dx

∂x

∂u=

µ1

x

¶Ã− u

(u2 + v2)3/2

!

=pu2 + v2

Ã− u

(u2 + v2)3/2

!= − u

u2 + v2

and

∂

∂vf (x (u, v)) =

∂z

∂v=dz

dx

∂x

∂v=

µ1

x

¶Ã− v

(u2 + v2)3/2

!

=pu2 + v2

Ã− v

(u2 + v2)3/2

!= − v

u2 + v2.

b) We have

f (x (u, v)) = f

µ1√

u2 + v2

¶= ln

³pu2 + v2

´.

Therefore.

∂

∂uf (x (u, v)) =

∂

∂uln

µ1√

u2 + v2

¶=

∂

∂u

³− ln

³pu2 + v2

´´= −

µ1√

u2 + v2

¶µu√

u2 + v2

¶= − u

u2 + v2,

and

∂

∂vf (x (u, v)) =

∂

∂vln

µ1√

u2 + v2

¶=

∂

∂v

³− ln

³pu2 + v2

´´= −

µ1√

u2 + v2

¶µv√

u2 + v2

¶= − v

u2 + v2.

4.


a) If we set z = f (x) we have

dz

dx=d

dxsin¡x2¢= 2x cos

¡x2¢.

We have∂x

∂u=

∂

∂u(u− 4v) = 1 and ∂x

∂v=

∂

∂v(u− 4v) = −4.

Therefore,

∂

∂uf (x (u, v)) =

∂z

∂u=

dz

dx

∂x

∂u

=¡2x cos

¡x2¢¢(1)

= 2x cos¡x2¢= 2 (u− 4v) cos

³(u− 4v)2

´and

∂

∂vf (x (u, v)) =

∂z

∂v=

dz

dx

∂x

∂v=¡2x cos

¡x2¢¢(−4)

= −8x cos ¡x2¢= −8 (u− 4v) cos

³(u− 4v)2

´b) We have

f (u, v) = sin³(u− 4v)2

´.

Therefore,

∂

∂uf (x (u, v)) = cos

³(u− 4v)2

´(2 (u− 4v)) = 2 (u− 4v) cos

³(u− 4v)2

´,

∂

∂vf (x (u, v)) = cos

³(u− 4v)2

´(2 (u− 4v)) (−4) = −8 (u− 4v) cos

³(u− 4v)2

´.

5.

a) If we set z = f (x, y) we have

∂z

∂x=

∂

∂xarcsin

Ãyp

x2 + y2

!=

1s1− y2

x2 + y2

µy∂

∂x

¡x2 + y2

¢−1/2¶

=

px2 + y2√x2

µy

µ−12

¡x2 + y2

¢−3/2¶(2x)

¶= − y

x2 + y2,

and

∂z

∂y=

∂

∂yarcsin

Ãyp

x2 + y2

!=

1s1− y2

x2 + y2

⎛⎜⎜⎝px2 + y2 − y

µy√x2+y2

¶x2 + y2

⎞⎟⎟⎠=

px2 + y2√x2

Ãx2

(x2 + y2)3/2

!=

x

x2 + y2.


We also have

∂x

∂u=

∂

∂u(u cos (v)) = cos (v) ,

∂x

∂v=

∂

∂v(u cos (v)) = −u sin (v) ,

and∂y

∂u=

∂

∂u(u sin (v)) = sin (v) ,

∂y

∂v=

∂

∂v(u sin (v)) = u cos (v) .

Therefore,

∂

∂uf (x (u, v) , y (u, v)) =

∂z

∂u

=dz

dx

∂x

∂u+dz

dy

∂y

∂u

=

µ− y

x2 + y2

¶cos (v) +

µx

x2 + y2

¶sin (v)

=− (u sin (v)) cos (v) + (u cos (v)) sin (v)

u2 cos2 (v) + u2 sin2 (v)

= 0,

and

∂

∂vf (x (u, v) , y (u, v)) =

∂z

∂v

=dz

dx

∂x

∂v+dz

dy

∂y

∂v

=

µ− y

x2 + y2

¶(−u sin (v)) +

µx

x2 + y2

¶(u cos (v))

=u2 sin2 (v) + u2 cos2 (v)

u2 cos2 (v) + u2 sin2 (v)= 1.

b) We have

f (x (u, v) , y (u, v)) = f (u cos (v) , u sin (v))

= arcsin

⎛⎝ u sin (v)qu2 cos2 (v) + u2 sin2 (v)

⎞⎠= arcsin (sin (v)) = v.

Therefore,∂

∂uf (x (u, v) , y (u, v)) = 0 and

∂

∂vf (x (u, v) , y (u, v)) = 1.

6.

a) If we set z = f (x, y) we have

∂z

∂x=

∂

∂xarctan

³yx

´=

1

1 +³yx

´2 ³− yx2´ = x2

x2 + y2

³− yx2

´= − y

x2 + y2,

and∂z

∂y=

∂

∂yarctan

³yx

´=

1

1 +³yx

´2 µ 1x¶=

x2

x2 + y2

µ1

x

¶=

x

x2 + y2


We also have

∂x

∂u=

∂

∂u(u+ 2v) = 1,

∂x

∂v=

∂

∂v(u+ 2v) = 2,

and

∂y

∂u=

∂

∂u(u− 2v) = 1, ∂y

∂v=

∂

∂v(u− 2v) = −2.

Therefore,

∂

∂uf (x (u, v) , y (u, v)) =

∂z

∂u

=dz

dx

∂x

∂u+dz

dy

∂y

∂u

=

µ− y

x2 + y2

¶(1) +

µx

x2 + y2

¶(1)

=x− yx2 + y2

=4v

2u2 + 8v2=

2v

u2 + 4v2

and

∂

∂vf (x (u, v) , y (u, v)) =

∂z

∂v

=dz

dx

∂x

∂v+dz

dy

∂y

∂v

=

µ− y

x2 + y2

¶(2) +

µx

x2 + y2

¶(−2)

= −2 (x+ y)x2 + y2

=−4u

2u2 + 8v2= − 2u

u2 + 4v2.

b) We have

f (x (u, v) , y (u, v)) = f (u+ 2v, u− 2v) = arctanµu− 2vu+ 2v

¶.

Therefore,

∂

∂uf (x (u, v) , y (u, v)) =

∂

∂uarctan

µu− 2vu+ 2v

¶=

1

1 +

µu− 2vu+ 2v

¶2Ã(u+ 2v)− (u− 2v)

(u+ 2v)2

!

=(u+ 2v)2

(u+ 2v)2+ (u− 2v)2

Ã4v

(u+ 2v)2

!=

4v

2u2 + 8v2=

2v

u2 + 4v2,


and

∂

∂vf (x (u, v) , y (u, v)) =

∂

∂varctan

µu− 2vu+ 2v

¶=

1

1 +

µu− 2vu+ 2v

¶2Ã−2 (u+ 2v)− 2 (u− 2v)

(u+ 2v)2

!

=(u+ 2v)2

(u+ 2v)2+ (u− 2v)2

Ã−4u

(u+ 2v)2

!=

−4u2u2 + 8v2

= − 2u

u2 + 4v2,

7. As in the text,

∂z

∂r= cos (θ)

∂z

∂x+ sin (θ)

∂z

∂y

∂z

∂θ= −r sin (θ) ∂z

∂x+ r cos (θ)

∂z

∂y.

Therefore,

µ∂z

∂r

¶2+1

r2

µ∂z

∂θ

¶2=

µcos (θ)

∂z

∂x+ sin (θ)

∂z

∂y

¶2+1

r2

µ−r sin (θ) ∂z

∂x+ r cos (θ)

∂z

∂y

¶2= cos2 (θ)

µ∂z

∂x

¶2+ 2 cos (θ) sin (θ)

µ∂z

∂x

¶µ∂z

∂y

¶+ sin2 (θ)

µ∂z

∂y

¶2+sin2 (θ)

µ∂z

∂x

¶2− 2 cos (θ) sin (θ)

µ∂z

∂x

¶µ∂z

∂y

¶+ cos2 (θ)

µ∂z

∂y

¶2=

¡cos2 (θ) + sin2 (θ)

¢µ∂z

∂x

¶2+¡cos2 (θ) + sin2 (θ)

¢µ∂z∂y

¶2=

µ∂z

∂x

¶2+

µ∂z

∂y

¶2.

8. We have

∂x

∂t= et cos (θ) ,

∂y

∂t= et sin (θ) ,

∂x

∂θ= −et sin (θ) , ∂y

∂θ= et cos (θ) .

Therefore,

∂z

∂t=

∂z

∂x

∂x

∂t+

∂z

∂y

∂y

∂t= et cos (θ)

∂z

∂x+ et sin (θ)

∂z

∂y,

∂z

∂θ=

∂z

∂x

∂x

∂θ+

∂z

∂y

∂y

∂θ= −et sin (θ) ∂z

∂x+ et cos (θ)

∂z

∂y.


Thus,µ∂z

∂t

¶2+

µ∂z

∂θ

¶2=

µet cos (θ)

∂z

∂x+ et sin (θ)

∂z

∂y

¶2+

µ−et sin (θ) ∂z

∂x+ et cos (θ)

∂z

∂y

¶2= e2t

Ãcos2 (θ)

µ∂z

∂x

¶2+ 2 cos (θ) sin (θ)

∂z

∂x

∂z

∂y+ sin2 (θ)

µ∂z

∂y

¶2!

+e2t

Ãsin2 (θ)

µ∂z

∂x

¶2− 2 cos (θ) sin (θ) ∂z

∂x

∂z

∂y+ cos2 (θ)

µ∂z

∂y

¶2!

= e2t

Ãµ∂z

∂x

¶2+

µ∂z

∂y

¶2!.

Therefore,

e−2t"µ

∂z

∂t

¶2+

µ∂z

∂θ

¶2#=

µ∂z

∂x

¶2+

µ∂z

∂y

¶2.

9.

a) Set w (x, t) = x+ at, so that u = f (w (x, t)). By the chain rule,

∂u

∂t=df

dw(w (x, t))

∂w

∂t=df

dw(w (x, t)) a = a

df

dw(w (x, t)) .

Therefore,

∂2u

∂t2= a

∂

∂t

µdf

dw(w (x, t))

¶= a

µd2f

dw2(w (x, t))

∂w

∂t

¶= a

d2f

dw2(w (x, t)) a = a2

d2f

dw2(x+ at)

Similarly,∂u

∂x=df

dw(w (x, t))

∂w

∂x=df

dw(w (x, t)) (1) =

df

dw(w (x, t))

and

∂2u

∂x2=

∂

∂x

µdf

dw(w (x, t))

¶=d2f

dw2(w (x, t))

∂w

∂x

=d2f

dw2(w (x, t)) (1) =

d2f

dw2(x− at)

Therefore,∂2u

∂t2= a2

∂2u

∂x2.

b) If f (w) = sin (w) and a = π/4 then

u (x, t) = f³x+

π

4t´= sin

³x+

π

4t´.

In particular,

u (x, 0) = sin (x) , u (x, 2) = sin³x+

π

2

´= cos (x) ,

u (x, 4) = sin (x+ π) = − cos (x)c)


sin(x)

-10 -5 5 10

-1.0

-0.5

0.5

1.0

x

y

cos(x)

-10 -5 5 10

-1

1

x

y

-cos(x)

-10 -5 5 10

-1

1

x

y

10.

a)

2z∂z

∂x− 2x = 0⇒ ∂z

∂x=x

z,

2z∂z

∂x− 2y = 0⇒ ∂z

∂y=y

z

b)

∂z

∂x

¯x=1,y=2,z=−3

= −13,∂z

∂y

¯x=1,y=2,z=−3

= −23.

Therefore, the tangent plane is the graph of the equation

z = −3− 13(x− 1)− 2

3(y − 2) .

1

yx

2

z

-3

11.

a)

2z∂z

∂x+ 2x = 0⇒ ∂z

∂x= −x

z,

2z∂z

∂x− 2y = 0⇒ ∂z

∂y=y

z

b)

∂z

∂x

¯x=3,y=6,z=6

= −36= −1

2,∂z

∂y

¯x=3,y=3,z=6

=6

6= 1.

Therefore, the tangent plane is the graph of the equation

z = 6− 12(x− 3) + (y − 6) .

12.7. DIRECTIONAL DERIVATIVES AND THE GRADIENT 51

z

6

3

y6 x

12.7 Directional Derivatives and the Gradient

1.

a)∂f

∂x(x, y) =

∂

∂x

¡4x2 + 9y2

¢= 8x and

∂f

∂y(x, y) =

∂

∂y

¡4x2 + 9y2

¢= 18y.

Therefore,

∇f (x, y) = 8xi+ 18yjb)

∇f (3, 4) = 24i+ 72j,and

||v|| = √4 + 1 =√5⇒ u =

1√5(−2, 1) =

µ− 2√

5,1√5

¶.

Therefore,

Duf (3, 4) = ∇f (3, 4) · u = (24i+ 72j) ·µ− 2√

5i+

1√5j

¶= − 48√

5+72√5=24√5

5

2.

a)∂f

∂x(x, y) =

∂

∂x

¡x2 − 4x− 3y2 + 6y + 1¢ = 2x− 4,

and∂f

∂y(x, y) =

∂

∂y

¡x2 − 4x− 3y2 + 6y + 1¢ = 6− 6y.

Therefore,

∇f (x, y) = (2x− 4) i+ (6− 6y) jb)

∇f (0, 0) = −4i+ 6j,and

||v|| = √1 + 1 =√2⇒ u =

1√2(1,−1) =

µ1√2,− 1√

2

¶.

Therefore,

Duf (0, 0) = ∇f (0, 0) · u = (−4i+ 6j) ·µ1√2i− 1√

2j

¶= − 4√

2− 6√

2= − 10√

2= −5

√2


3.

a)∂f

∂x(x, y) =

∂

∂xex

2−y2 = 2xex2−y2 and

∂f

∂y(x, y) =

∂

∂yex

2−y2 = −2yex2−y2 .

Therefore,

∇f (x, y) = 2xex2−y2i− 2yex2−y2jb)

∇f (2, 1) = 4e3i− 2e3j,and

||v|| =√10⇒ u =

1√10(−1, 3) =

µ− 1√

10,3√10

¶.

Therefore,

Duf (2, 1) = ∇f (2, 1) · u =¡4e3i− 2e3j¢ ·µ− 1√

10i+

3√10j

¶= − 4e

3

√10− 6e3√

10= −√10e3

4.

a)∂f

∂x(x, y) =

∂

∂x(sin (x) cos (y)) = cos (x) cos (y) ,

and∂f

∂y(x, y) =

∂

∂y(sin (x) cos (y)) = − sin (x) sin (y) .

Therefore,

∇f (x, y) = cos (x) cos (y) i− sin (x) sin (y) j.b)

∇f (π/3,π/4) = cos (π/3) cos (π/4) i− sin (π/3) sin (π/4) j

=

µ1

2

¶Ã√2

2

!i−

Ã√3

2

!Ã√2

2

!j =

√2

4i−√6

4j

and

||v|| = √4 + 9 =√13⇒ u =

1√13(2,−3) =

µ2√13,− 3√

13

¶.

Therefore,

Duf (π/3,π/4) = ∇f (π/3,π/4) · u =

Ã√2

4i−√6

4j

!·µ

2√13i− 3√

13j

¶=

√2

2√13+3√6

4√13=2√2 + 3

√6

4√13

5.

a)∂f

∂x(x, y, z) =

∂

∂x

¡x2 − y2 + 2z2¢ = 2x, ∂f

∂y(x, y, z) =

∂

∂y

¡x2 − y2 + 2z2¢ = −2y,

and∂f

∂z(x, y, z) =

∂

∂z

¡x2 − y2 + 2z2¢ = 4z.


Therefore,

∇f (x, y, z) = 2xi− 2yj+ 4zk.b)

∇f (1,−1, 2) = 2i+ 2j+ 8kand

||v|| =√3⇒ u =

1√3(1,−1, 1) =

µ1√3,− 1√

3,1√3

¶.

Therefore,

Duf (1,−1, 2) = ∇f (1,−1, 2) · u= (2i+ 2j+ 8k) ·

µ1√3i− 1√

3j+

1√3k

¶=

2√3− 2√

3+

8√3=

8√3

6.

a)

∂f

∂x(x, y, z) =

∂

∂x

¡x2 + y2 + z2

¢−1/2= −1

2

¡x2 + y2 + z2

¢−3/2(2x) = − x

(x2 + y2 + z2)3/2,

∂f

∂y(x, y, z) = − y

(x2 + y2 + z2)3/2,∂f

∂z(x, y, z) = − z

(x2 + y2 + z2)3/2

Therefore,

∇f (x, y, z) = − x

(x2 + y2 + z2)3/2i− y

(x2 + y2 + z2)3/2j− z

(x2 + y2 + z2)3/2k.

b)

∇f (2,−2, 1) = − 227i+

2

27j− 1

27k

and

||v|| =√26⇒ u =

1√26(3, 4, 1) =

µ3√26,4√26,1√26

¶.

Therefore,

Duf (2,−2, 1) = ∇f (2,−2, 1) · u=

µ− 227i+

2

27j− 1

27k

¶·µ

3√26i+

4√26j+

1√26k

¶= − 6

27√26+

8

27√26− 1

27√26=

1

27√26

7.

a) We have

∂f

∂x(x, y) =

∂

∂x

¡x2 + y2

¢−1/2= −1

2

¡x2 + y2

¢−3/2(2x) = − x

(x2 + y2)3/2,

∂f

∂y(x, y) =

∂

∂y

¡x2 + y2

¢−1/2= −1

2

¡x2 + y2

¢−3/2(2y) = − y

(x2 + y2)3/2.


Therefore,

∇f (x, y) = − x

(x2 + y2)3/2i− y

(x2 + y2)3/2j

Thus,

v = ∇f (2, 3) = − 2

133/2i− 3

133/2j

The corresponding rate of increase of f is

||v|| =sµ

2

133/2

¶2+

µ3

133/2

¶2=1

13

b)

w = −∇f (2, 3) = 2

133/2i+

3

133/2j

The corresponding rate of decrease of f is 1/13.

8.

a) We have

∂f

∂x(x, y) =

∂

∂xln³p

x2 + y2´=1

2

∂

∂xln¡x2 + y2

¢=1

2

µ1

x2 + y2

¶(2x) =

x

x2 + y2,

∂f

∂y(x, y) =

∂

∂yln³p

x2 + y2´=

y

x2 + y2.

Therefore,

∇f (x, y) = x

x2 + y2i+

y

x2 + y2j

Thus,

v = ∇f (3, 4) = 3

25i+

4

25j

The corresponding rate of increase of f is

||v|| =sµ

3

25

¶2+

µ4

25

¶2=1

5.

b)

w = −∇f (3, 4) = − 325i− 4

25j

The corresponding rate of decrease of f is 1/5.

9.

a) We have

∂f

∂x(x, y) =

∂

∂xex

2−y2 = 2xex2+y2 ,

∂f

∂y(x, y) =

∂

∂yex

2−y2 = −2yex2−y2 .

Therefore,

∇f (x, y) = 2xex2−y2i− 2yex2−y2j.Since

σ³π6

´=³2 cos

³π6

´, 2 sin

³π6

´´= 2

Ã√3

2,1

2

!=³√3, 1´,


and

∇f³√3, 1´= 2√3e2i− 2e2j = 2

√3e2i− 2e2j.

We havedσ

dt(t) = −2 sin (t) i+ 2 cos (t) j⇒ dσ

dt

³π6

´= −i+

√3j.

Therefore,

d

dtf (σ (t))

¯t=π/6

= ∇f³σ³π6

´´· dσdt

³π6

´=

³2√3e2i− 2e2j

´·³−i+

√3j´

= −2√3e2 − 2

√3e2 = −4

√3e.

b) We have ¯¯dσ

dt

³π6

´¯¯=¯¯−i+

√3j¯¯=√4 = 2,

so that

u =1

2

dσ

dt

³π6

´=1

2

³−i+

√3j´= −1

2i+

√3

2j

is the unit vector in the tangential direction. Therefore,

Ddσ/dtf (σ (π/6)) = Ddσ/dtf³√3, 1´

= ∇f³√3, 1´· u

=³2√3e2i− 2e2j

´·Ã−12i+

√3

2j

!= −

√3e2 −

√3e2 = −2

√3e2.

10.

a) We have

∂f

∂x(x, y) =

∂

∂xarctan

Ãyp

x2 + y2

!=

1

1 +y2

x2 + y2

µ∂

∂x

¡y¡x2 + y2

¢¢−1/2¶

=x2 + y2

x2 + 2y2y

µ−12

¡x2 + y2

¢−3/2¶(2x)

= − xy

(x2 + 2y2)px2 + y2

,

and

∂f

∂y(x, y) =

∂

∂yarctan

Ãyp

x2 + y2

!=

1

1 +y2

x2 + y2

⎛⎜⎜⎝px2 + y2 − y

µy√x2+y2

¶x2 + y2

⎞⎟⎟⎠=

x2 + y2

x2 + 2y2

Ãx2

(x2 + y2)3/2

!

=x2

(x2 + 2y2)px2 + y2

,


Therefore,

∇f (x, y) = − xy

(x2 + 2y2)px2 + y2

i+x2

(x2 + 2y2)px2 + y2

j.

Since

σ (2) =

µ−35,4

5

¶,

and

∇fµ−35,4

5

¶= 2√3e2i− 2e2j = 2

√3e2i− 2e2j.

We havedσ

dt(t) = −2 sin (t) i+ 2 cos (t) j⇒ dσ

dt

³π6

´= −i+

√3j.

Therefore,

d

dtf (σ (t))

¯t=π/6

= ∇f³σ³π6

´´· dσdt

³π6

´=

³2√3e2i− 2e2j

´·³−i+

√3j´

= −2√3e2 − 2

√3e2 = −4

√3e.

b) We have ¯¯dσ

dt

³π6

´¯¯=¯¯−i+

√3j¯¯=√4 = 2,

so that

u =1

2

dσ

dt

³π6

´=1

2

³−i+

√3j´= −1

2i+

√3

2j

is the unit vector in the tangential direction. Therefore,

Ddσ/dtf (σ (π/6)) = Ddσ/dtf³√3, 1´

= ∇f³√3, 1´· u

=³2√3e2i− 2e2j

´·Ã−12i+

√3

2j

!= −

√3e2 −

√3e2 = −2

√3e2.

11.

a) Let f (x, y) = 2x2 + 3y2. We have

∇f (x, y) = 4xi+ 6yj.Therefore,

∇f (2, 3) = 8i+ 18jis orthogonal to the curve f (x, y) = 35 at (2, 3).

b) The tangent line is the graph of the equation

8 (x− 2) + 18 (y − 3) = 0

12.

a) Let f (x, y) = x2 − y2. Then,∇f (x, y) = 2xi− 2yj.


Therefore,

∇f³3,√5´= 6i− 2

√5j

is orthogonal to the curve f (x, y) = 4 at¡3,√5¢.


6 (x− 3)− 2√5³y −√5´= 0.

13.

a) Let

f (x, y) = e25−x2−y2 .

Then,

∇f (x, y) = −2xe25−x2−y2i− 2ye25−x2−y2j,so that

∇f (3, 4) = −6i− 8jis orthogonal to the curve f (x, y) = 1 at (3, 4).


−6 (x− 3)− 8 (y − 4) = 0.14.

a) Let f (x, y, z) = z − x2 + y2. We have∇f (x, y, z) = −2xi+ 2yj+ k

Therefore

∇f (4, 3, 7) = −8i+ 6j+ kis orthogonal to the surface at (4, 3, 7).b) The plane that is tangent to the surface at (4, 3, 7) is the graph of the equation

−8 (x− 4) + 6 (y − 3) + (z − 7) = 0.15.

a) Let f (x, y, z) = x2 − y2 + z2. We have∇f (x, y, z) = 2xi− 2yj+ 2zk

Therefore

∇f (2, 2, 1) = 4i− 4j+ 2kis orthogonal to the surface at (2, 2, 1).b) The plane that is tangent to the surface at (2, 2, 1) is the graph of the equation

4 (x− 2)− 4 (y − 2) + 2 (z − 1) = 0.16.

a) Let f (x, y, z) = x2 − y2 − z2. We have∇f (x, y, z) = 2xi− 2yj− 2zk

Therefore

∇f (3, 2, 2) = 6i− 4j− 4k


is orthogonal to the surface at (3, 2, 2).b) The plane that is tangent to the surface at (3, 2, 2) is the graph of the equation

6 (x− 3)− 4 (y − 2)− 4 (z − 2) = 0.17.

Let f (x, y, z) = x− sin (y) cos (z). We have∇f (x, y, z) = i− cos (y) cos (z) j+ sin (y) sin (z)k

Therefore

∇f (1,π/2, 0) = iis orthogonal to the surface at (1,π/2, 0).b) The plane that is tangent to the surface at (1,π/2, 0) is the graph of the equation

x− 1 = 0⇔ x = 1.

12.8 Local Maxima and Minima

1.∂f

∂x(x, y) = 2x+ y + 1,

∂f

∂y(x, y) = −2y + x− 1.

Therefore, ∂xf (x, y) = 0 and ∂yf (x, y) = 0⇔2x+ y + 1 = 0,

−2y + x− 1 = 0

Thus, the critical point is (−1/5,−3/5).We have

∂2f

∂x2= 2,

∂2f

∂y2= −2 and ∂2f

∂y∂x= 1

Therefore,

D =∂2f

∂x2∂2f

∂y2−µ

∂2f

∂y∂x

¶2= (2) (−2)− 1 = −4− 1 < 0.

Thus, f has a saddle point at (−1/5,−3/5).

0

1

10

5

z

-3

-2

-1x

-1

0

1

-5

0

y -2

-3

2.∂f

∂x(x, y) = 2x− y, ∂f

∂y(x, y) = 2y − x+ 1.

Therefore, ∂xf (x, y) = 0 and ∂yf (x, y) = 0⇔2x− y = 0,

2y − x+ 1 = 0

12.8. LOCAL MAXIMA AND MINIMA 59


∂2f

∂x2= 2,

∂2f

∂y2= 2 and

∂2f

∂y∂x= −1

Therefore,

D =∂2f

∂x2∂2f

∂y2−µ

∂2f

∂y∂x

¶2= (2) (2) + 1 = 5 > 0,

and ∂2xxf = 2 > 0. Therefore, f has a local (and absolute) minimum at (−1/3,−2/3).

15

10

z5

0

10

-3-3

-1

y

01

x

-1-2 -2

3.

∂f

∂x(x, y) = 2x+ 2y,

∂f

∂y(x, y) = 2y + 2x.

Therefore, ∂xf (x, y) = 0 and ∂yf (x, y) = 0⇔2x+ 2y = 0,

2y + 2x = 0

Thus, any point on the line y = −x is a critical point.We have

∂2f

∂x2= 2,

∂2f

∂y2= 2 and

∂2f

∂y∂x= 2

Therefore,

D =∂2f

∂x2∂2f

∂y2−µ

∂2f

∂y∂x

¶2= (2) (2)− ¡22¢ = 0..

We expect that the quadratic function f attains its absolute maximum or minimum on the line

y = −x. Indeed,f (x, y) = x2 + y2 + 2xy − 10

= (x+ y)2 − 10,so that f (x, y) ≥ −10 for each (x, y), and f (x, y) = −10 iff y = −x.

-10

20

10

0

z

3

3

2

2

1

1

-2

0

0

yx

-1-1-2

-3 -3


4.∂f

∂x(x, y) = 2x+ 3y,

∂f

∂y(x, y) = 2y + 3x− 3.

Therefore, ∂xf (x, y) = 0 and ∂yf (x, y) = 0⇔2x+ 3y = 0,

2y + 3x− 3 = 0

Thus, the critical point is (9/5,−6/5).We have

∂2f

∂x2= 2,

∂2f

∂y2= 2 and

∂2f

∂y∂x= 3

Therefore,

D =∂2f

∂x2∂2f

∂y2−µ

∂2f

∂y∂x

¶2= (2) (2)− 32 = 4− 9 < 0.

Thus, f has a saddle point at (−9/5,−6/5).

10

30

z20

0

-3

-2-2

-1 0

0

1

1

y

x-1

5.

∂f

∂x(x, y) = 2x− 3y + 5, ∂f

∂y(x, y) = −3x− 2 + 12y.

Therefore, ∂xf (x, y) = 0 and ∂yf (x, y) = 0⇔2x− 3y + 5 = 0,

−3x+ 12y − 2 = 0


∂2f

∂x2= 2,

∂2f

∂y2= 12 and

∂2f

∂y∂x= −3

Therefore,

D =∂2f

∂x2∂2f

∂y2−µ

∂2f

∂y∂x

¶2= (2) (12)− 9 = 15 > 0,

and ∂2xxf = 2 > 0. Therefore, f has a local (and absolute) minimum at (−18/5,−11/15).

100

120

80

60z

40

20

02 1 0 -1

y

-2

x

-2 0-4-6-3 -4

12.8. LOCAL MAXIMA AND MINIMA 61

6.∂f

∂x(x, y) = 3− 3x2 − 3y2, ∂f

∂y(x, y) = −6xy

Thus, ∂xf (x, y) = 0 and ∂yf (x, y) = 0⇔3− 3x2 − 3y2 = 0,

−6xy = 0

From the second equation, x = 0 or y = 0. If x = 0, then

3− 3y2 = 0⇔ y2 = 1⇔ y = ±1.Therefore, (0, 1) and (0,−1) are critical points.If y = 0 then

3− 3x2 = 0⇔ x2 = 1⇔ x = ±1.Therefore, (1, 0) and (−1,−0) are critical points.We have

∂2f

∂x2= −6x, ∂2f

∂y2= −6x and ∂2f

∂y∂x= −6y

Therefore,

D (x, y) =∂2f

∂x2∂2f

∂y2−µ

∂2f

∂y∂x

¶2= (−6x) (−6x)− (−6y)2 = 36 ¡x2 − y2¢ .

(x, y) D (x, y) fxx (x, y) critical point

(1, 0) 36 −6 local maximum

(−1, 0) 36 6 local minimum

(0, 1) −36 0 saddle point

(0,−1) −36 0 saddle point

-10

z 0

10

20

-20

2

2

1

-1 x

1

0

y-1

0-2 -2

7.

∂f

∂x= −4x3 − 4y, ∂f

∂y(x, y) = −4y3 − 4x.

Therefore, ∂xf (x, y) = 0 and ∂yf (x, y) = 0⇔x3 + y = 0⇔ y = −x3y3 + x = 0⇔ x = −y3.

If we set y = −x3 in the second equation we obtain

x = − ¡−x3¢3 = x9 ⇔ x− x9 = 0⇔ x¡1− x8¢ = 0.


Therefore, x = 0, 1 or −1. Since y = −x3, we have y = 0, −1 and 1, respectively. Thus, thecritical points are (0, 0), (1,−1) and (−1, 1).We have

∂2f

∂x2= −12x2, ∂2f

∂y2= −12y2 and ∂2f

∂y∂x= −4

Therefore,

D (x, y) =∂2f

∂x2∂2f

∂y2−µ

∂2f

∂y∂x

¶2=¡−12x2¢ ¡−12y2¢− (−4)2 = 144x2y2 − 16



(1,−1) 128 −12 local maximum

(−1, 1) 128 −12 local maximum

1

y

0

-1-1

x0

1

-20

0

z-10

8.

f (x, y) = x3 − 12xy + 8y3

∂f

∂x= 3x2 − 12y, ∂f

∂y(x, y) = 24y2 − 12x

Therefore, ∂xf (x, y) = 0 and ∂yf (x, y) = 0⇔

x2 − 4y = 0,

2y2 − x = 0.

From the second equation x = 2y2. Substituting in the first equation,¡2y2¢2 − 4y = 0⇔ y4 − y = 0⇔ y

¡y3 − 1¢ = 0.

Thus, y = 0 or y = 1. The corresponding values of x are 0 and 1. Therefore, the critical pointsare (0, 0) and (2, 1).We have

∂2f

∂x2= 6x,

∂2f

∂y2= 48y and

∂2f

∂y∂x= −12

Therefore,

D (x, y) =∂2f

∂x2∂2f

∂y2−µ

∂2f

∂y∂x

¶2= (6x) (48y)− (−12)2 = 288xy − 144.



(2, 1) 432 12 local minimum

12.9. ABSOLUTE EXTREMA AND LAGRANGE MULTIPLIERS 63

0

4

2

xy

02

-100

-2-2

z 0

100

12.9 Absolute Extrema and Lagrange Multipliers

1. We have f (x, y) = x+ y and we set g (x, y) = x2+ y2, so that we will determine the extremaof f (x, y) subject to the constraint g (x, y) = 4 (i.e., the extrema of f on the circle of radius 2centered at the origin).

We have

∇f (x, y) = (1, 1) and ∇g (x, y) = (2x, 2y) .Therefore,

∇f (x, y) = λ∇g (x, y)⇔ 1 = 2λx and 1 = 2λy.

Thus, we need to solve the following system of equations:

1 = 2λx,

1 = 2λy,

x2 + y2 = 4

(the last equation is the constraint equation, g (x, y) = 4).From the first two equations,

x =1

2λand y =

1

2λ.

We substitute these expressions in the last equation:µ1

2λ

¶2+

µ1

2λ

¶2= 4 ⇒ 1

4λ2+

1

4λ2= 4

⇒ 2 = 16λ2

⇒ λ = ± 1√8= ± 1

2√2.

Therefore,

x = y = ± 1

2λ= ± 1

2

µ1

2√2

¶ = ±√2.

Thus, the necessary conditions for the extrema of f subject to the constraint x2 + y2 = 4 are

satisfied at the points¡√2,√2¢and

¡−√2,−√2¢. We havef³√2,√2´= 2√2 and f

³−√2,−√2´= −2

√2.

Therefore, the maximum value of f on the circle x2 + y2 = 4 is f¡√2,√2¢= 2√2 and the

minimum value is f¡−√2,−√2¢ = −2√2.


2. We have f (x, y) = 3x − 4y and we set g (x, y) = x2 + y2, so that we will determine theextrema of f (x, y) subject to the constraint g (x, y) = 9 (i.e., the extrema of f on the circle ofradius 3 centered at the origin).

We have

∇f (x, y) = (3,−4) and ∇g (x, y) = (2x, 2y) .Therefore,

∇f (x, y) = λ∇g (x, y)⇔ 3 = 2λx and − 4 = 2λy.Thus, we need to solve the following system of equations:

3 = 2λx,

−4 = 2λy,

x2 + y2 = 9

(the last equation is the constraint equation, g (x, y) = 9).From the first two equations,

x =3

2λand y = − 4

2λ.

We substitute these expressions in the last equation:µ3

2λ

¶2+

µ4

2λ

¶2= 9 ⇒ 9

4λ2+16

4λ2= 9

⇒ 25

9= 4λ2

⇒ λ = ±56

Therefore,

x =3

2λ=

3

2

µ5

6

¶ =9

5and y = − 4

2λ= − 4

2

µ5

6

¶ = −125,

or

x =3

2λ= − 3

2

µ5

6

¶ = −95and y = − 4

2λ=

4

2

µ5

6

¶ =12

5

Thus, the necessary conditions for the extrema of f subject to the constraint x2 + y2 = 9 are

satisfied at the points µ9

5,−12

5

¶and

µ−95,12

5

¶We have

f

µ9

5,−12

5

¶=7

5and f

µ−95,12

5

¶= −47

5.

Therefore, the maximum value of f on the circle x2 + y2 = 9 is

f

µ9

5,−12

5

¶=7

5,

and the mimimum value on x2 + y2 = 9 is

f

µ−95,12

5

¶= −47

5.


3. We have f (x, y) = xy and we set g (x, y) = 4x2 + y2, so that we will determine the extremaof f (x, y) subject to the constraint g (x, y) = 8 (i.e., the extrema of f on an elllipse that iscentered at the origin).

We have

∇f (x, y) = (y, x) and ∇g (x, y) = (8x, 2y) .Therefore,

∇f (x, y) = λ∇g (x, y)⇔ y = 8λx and x = 2λy.


y = 8λx,

x = 2λy,

4x2 + y2 = 8

(the last equation is the constraint equation, g (x, y) = 16).From the first equation, y = 8λx. We replace y in the second equation:

x = 16λ2x

We cannot have x = 0: Then y = 0, and (0, 0) does not satisfy the constraint equation.Therefore,

16λ2 = 1⇒ λ = ±14.

and we set y = 8λx in the last equation:

4x2 + 64λ2x2 = 8.

Since 16λ2 = 1,4x2 + 4x2 = 8⇒ 8x2 = 8⇒ x2 = 1⇒ x = ±1.

If λ = 1/4 and x = 1 then

y = 8λx = 8

µ1

4

¶(1) = 2,

If λ = −1/4 and x = 1 theny = 8λx = 8

µ−14

¶(1) = −2,

If λ = 1/4 and x = −1 theny = 8λx = 8

µ1

4

¶(−1) = −2,

If λ = −1/4 and x = −1 then

y = 8λx = 8

µ−14

¶(−1) = 2.

Therefore, the critical points are

(1, 2) , (1,−2) , (−1,−2) , (−1, 2) .

We have

f (1, 2) = 2, f (1,−2) = −2, f (−1,−2) = 2, f (−1, 2) = −2.Therefore, the minimum value of f (x, y) subject to 4x2 + y2 = 8 is -−2, the maximum value is

2.


4. We have f (x, y) = x + y2 and we set g (x, y) = 2x2 + y2, so that we will determine theextrema of f (x, y) subject to the constraint g (x, y) = 1 (i.e., the extrema of f on an elllipsethat is centered at the origin).

We have

∇f (x, y) = (1, 2y) and ∇g (x, y) = (4x, 2y) .Therefore,

∇f (x, y) = λ∇g (x, y)⇔ 1 = 4λx and 2y = 2λy.


4λx = 1,

2 (1− λ) y = 0,

2x2 + y2 = 1

(the last equation is the constraint equation, g (x, y) = 1).From the last equation, we have y = 0 or λ = 1. If y = 0 then

2x2 = 1⇒ x = ± 1√2.

If y 6= 0 then λ = 1. From the first equation

x =1

4λ=1

4.

Substituting in the last equation,

2

µ1

4

¶2+ y2 = 1⇒ y2 = 1− 1

8=7

8⇒ y = ±

r7

8.

Therefore, the critical points areµ1√2, 0

¶,

µ− 1√

2, 0

¶,

Ã1

4,

r7

8

!,

Ã1

4,−r7

8

!We have

f

µ1√2, 0

¶=

1√2, f

µ− 1√

2, 0

¶= − 1√

2, f

Ã1

4,±r7

8

!=9

8

Therefore, the minimum value of f (x, y) subject to 2x2+y2 = 1 is -−1/√2, the maximum valueis 9/8.

5. The critical point of f in the interior of D is (0, 0), and f (0, 0) = 0. In order to find thecritical point of f subject to x2 + y2 = 4, we set g (x, y) = x2 + y2, and apply the technique ofthe technique of Lagrange multipliers. We have

∇f (x, y) = (2x, 4y) and ∇g (x, y) = (2x, 2y) .Therefore,

∇f (x, y) = λ∇g (x, y)⇔ 2x = 2λx and 4y = 2λy.


2x = 2λx ⇔ (λ− 1)x = 04y = 2λy ⇔ (λ− 2) y = 0

x2 + y2 = 4


If λ = 1 then y = 0 and x = ±2. Thus, (2, 0) and (−2, 0) are a critical points.If x = 0 then y = ±2. Thus, (0, 2) and (0,−2) are critical point. We have

f (±2, 0) = 4, f (0,±2) = 8.Therefore, the minimum value of f in D is 0, and its maximum value in D is 8.

6. The critical point of f in the interior of D is (0, 0), and f (0, 0) = 0. In order to find thecritical point of f subject to x2 + 2y2 = 1, we set g (x, y) = x2 + 2y2, and apply the techniqueof the technique of Lagrange multipliers. We have

∇f (x, y) = (y, x) and ∇g (x, y) = (2x, 4y) .Therefore,

∇f (x, y) = λ∇g (x, y)⇔ y = 2λx and x = 4λy.


y = 2λx

x = 4λy

x2 + 2y2 = 1

We cannot have λ = 0: In that case x = 0 and y = 0, so that the constraint equation is notsatisfied.

We have

x = 4λy = 4λ (2λx) = 8λ2x,

so that ¡8λ2 − 1¢x = 0.

We cannot have x = 0. Therefore,

8λ2 − 1 = 0⇒ λ = ± 1√8.

Therefore

y = 2λx = ± 2√8x = ± 1√

2x.

Thus,

x2 + 2

µ1

2x2¶= 1⇒ 2x2 = 1⇒ x = ± 1√

2.

Therefore,

y = ±12.

Thus, the critical points on x2 + 2y2 = 1 areµ1√2,1

2

¶.

µ1√2,−12

¶,

µ− 1√

2,1

2

¶,

µ− 1√

2,−12

¶We have

f

µ1√2,1

2

¶=

√2

4, f

µ1√2,−12

¶= −√2

4

f

µ− 1√

2,1

2

¶= −

√2

4, f

µ− 1√

2,−12

¶=

√2

4

Therefore, the maximum value of f in D is −√2/4 and its maximum value in D is√2/4.

Chapter 13

Multiple Integrals

13.1 Double Integrals Over Rectangles

1. We have Z x=3

x=0

¡x2 + 3xy2

¢dx =

1

3x3 + 3

x2

2y2¯x=3x=0

= 9 +27

2y2

Therefore, Z y=2

y=1

µZ x=3

x=0

¡x2 + 3xy2

¢dx

¶dy =

Z y=2

y=1

µ9 +

27

2y2¶dy

= 9y +9

2y3¯21

=81

2

2. We have Z y=π/3

y=π/4

x sin (y) dy = −x cos (y)|y=π/3y=π/4

= −x cos³π3

´+ x cos

³π4

´=

Ã−12+

√2

2

!x =

x

2

³√2− 1

´Therefore,

Z x=1

x=0

ÃZ y=π/3

y=π/4

x sin (y) dy

!dx =

Z x=1

x=0

x

2

³√2− 1

´dx

=

√2− 14

x2

¯¯1

0

=

√2− 14

3. We have Z x=3

x==1

exyydx = exy|31 = e3y − ey.

69

70 CHAPTER 13. MULTIPLE INTEGRALS

Therefore, Z y=4

y=2

µZ x=3

x==1

exyydx

¶dy =

Z y=4

y=2

¡e3y − ey¢ dy

=1

3e3y − ey

¯42

=1

3e12 − e4 − 1

3e6 + e2

4. We set u = 1 + y2 so that du = 2ydy. Thus,Z y=3

y=1

xyp1 + y2dy = x

Z y=3

y=1

yp1 + y2dy

=x

2

Zu=10u=2 u

1/2du

=x

2

Ã2

3u3/2

¯102

!=x

3

³103/2 − 23/2

´.

Z x=2

x=0

µZ y=3

y=1

xyp1 + y2dy

¶dx =

Z x=2

x=0

x

3

³103/2 − 23/2

´dx

=103/2 − 23/2

6x2¯20

=2¡103/2 − 23/2¢

3

5. Z ZD

xy2

x2 + 1dA =

Z x=1

x=0

µZ y=2

y=−2

xy2

x2 + 1dy

¶dx

=

Z x=1

x=0

x

x2 + 1

µZ y=2

y=−2y2dy

¶dx

=

µZ x=1

x=0

x

x2 + 1dx

¶µZ y=2

y=−2y2dy

¶=

Ã1

2ln¡x2 + 1

¢¯10

!Ã1

3y3¯2−2

!

=1

2ln (2)

µ16

3

¶=8

3ln (2)

6. Z ZD

xyex2ydA =

Z y=2

y=1

µZ 1

x=0

xyex2ydx

¶dy

=

Z y=2

y=1

Ã1

2ex

2y

¯x=1x=0

!dy

=1

2

Z y=2

y=1

(ey − 1) dy

=1

2

³ey − y|21

´=1

2

¡e2 − 1− e¢

13.1. DOUBLE INTEGRALS OVER RECTANGLES 71

7. Z ZD

x2

1 + y2dA =

Z y=1

y=0

µZ x=1

x=0

x2

1 + y2dx

¶dy

=

Z y=1

y=0

1

1 + y2

Ãx3

3

¯10

!dy

=1

3

³arctan (y)|10

´=1

3arctan (1) =

π

12.

8.Z ZD

y cos (x+ y) dA =

Z y=π/2

y=0

y

ÃZ x=π/3

x=0

cos (x+ y) dx

!dy

=

Z y=π/2

y=0

y³sin (x+ y)|x=π/3x=0

´dy

=

Z y=π/2

y=0

y³sin³π3+ y

´− sin (y)

´dy

=

Z y=π/2

y=0

y sin³π3+ y

´dy −

Z y=π/2

y=0

y sin (y) dy

=

Z u=5π/6

u=π/3

³u− π

3

´sin (u) du−

Z y=π/2

y=0

y sin (y) dy

=π

3cos (u)

¯u=5π/6u=π/3

+

Z y=5π/6

y=π/3

y sin (y) dy −Z y=π/2

y=0

y sin (y) dy

=π

3cos

µ5π

6

¶− π

3cos³π3

´+

Z y=5π/6

y=π/3


y=0

y sin (y) dy

= −π√3

6− π

6+

Z y=5π/6

y=π/3


y=0

y sin (y) dy

We set u = y and dv = sin (y) dy so that

du = dy and v = − cos (y) .

Therefore Zy sin (y) dy =

Zudv = uv −

Zvdu

= −y cos (y) +Zcos (y) dy

= −y cos (y) + sin (y)

Thus, Z y=5π/6

y=π/3

y sin (y) dy = −y cos (y) + sin (y)|5π/6π/3

= −5π6cos

µ5π

6

¶+ sin

µ5π

6

¶+

π

3cos³π3

´− sin

³π3

´=

1

6π +

5

12

√3π − 1

2

√3 +

1

2,


Z y=π/2

y=0

y sin (y) dy = −y cos (y) + sin (y)|π/20 = 1

Therefore,Z ZD

y cos (x+ y) dA = −π√3

6− π

6+

Z y=5π/6

y=π/3


y=0

y sin (y) dy

= −π√3

6− π

6+1

6π +

5

12

√3π − 1

2

√3 +

1

2− 1

=1

4

√3π − 1

2

√3− 1

2

13.2 Double Integrals Over Non-Rectangular Regions

1. We have Z √y0

xy2dx = y2Z √y0

xdx = y2

Ãx2

2

¯√y0

!

= y2³y2

´=y3

2.

Therefore, Z y=4

y=0

Z x=√y

x=0

xy2dxdy =

Z 4

0

y3

2dy =

y4

8

¯40

=44

8= 32

2. We have Z cos(θ)

0

esin(θ)dr = esin(θ)Z cos(θ)

0

dr = esin(θ) cos (θ) .

Therefore, Z θ=π/2

θ=0

Z r=cos(θ)

r=0

esin(θ)drdθ =

Z π/2

0

esin(θ) cos (θ) dθ.

We set u = sin (θ) so that du = cos (θ) dθ. Thus,Z π/2

0

esin(θ) cos (θ) dθ =

Z sin(π/2)

sin(0)

eudu =

Z 1

0

eudu

= eu|10 = e− 1.

3.

2

4

x

y

We have y = x2 and y = 2x if

x2 − 2x = 0⇔ x (x− 2) = 0⇔ x = 0 or x = 2.

13.2. DOUBLE INTEGRALS OVER NON-RECTANGULAR REGIONS 73

Thus, the graphs intersect at (0, 0) and (2, 4).Z ZD

4x3ydA =

Z x=2

x=0

Z y=2x

y=x24x3ydydx.

We have Z y=2x

y=x24x3ydy = 4x3

Z y=2x

y=x2ydy

= 4x3

Ã1

2y2¯2xx2

!= 2x3

¡4x2 − x4¢ .

Therefore, Z x=2

x=0

Z y=2x

y=x24x3ydydx =

Z 2

0

2x3¡4x2 − x4¢ dx

=

Z 2

0

¡8x5 − 2x7¢ dx

= 8

µx6

6

¶− 2

µx8

8

¶¯20

=4

3

¡26¢− 1

4

¡28¢=64

3

4.

4

16

x

y

We have Z ZD

y√xdA =

Z x=4

x=0

Z y=x2

y=0

y√xdydx,

and Z y=x2

y=0

y√xdy =

√x

Z y=x2

y=0

ydy

=√x

Ãy2

2

¯x20

!=

√x

2

¡x4¢=1

2x9/2.

Therefore, Z x=4

x=0

Z y=x2

y=0

y√xdydx =

Z 4

0

1

2x9/2dx

=1

2

µx11/2

11/2

¶¯40

=411/2

11=211

11.


5.

1

1

x

y

Z ZD

y2exydA =

Z y=1

y=0

Z x=y

x=0

y2exydxdy.

We have Z x=y

x=0

y2exydx = y

Z x=y

x=0

yexydx.

We set u = xy so that du = ydx. Therefore,Z x=y

x=0

yexydx =

Z y2

0

eudu = ey2 − 1.

Thus, Z y=1

y=0

Z x=y

x=0

y2exydxdy =

Z 1

0

y³ey

2 − 1´dy =

Z 1

0

yey2

dy −Z 1

0

ydy

=

Z 1

0

yey2

dy − 12

We set u = y2 so that du = 2ydy. Therefore,Z 1

0

yey2

dy =1

2

Z 1

0

eudu =1

2(e− 1) .

Thus, Z y=1

y=0

Z x=y

x=0

y2exydxdy =1

2(e− 1)− 1

2=1

2e− 1.

6.

x

y

Z ZD

y2 sin¡x2¢dA =

Z x=√π

x=0

Z y=x1/3

y=−x1/3y2 sin

¡x2¢dydx.


We have Z y=x1/3

y=−x1/3y2 sin

¡x2¢dy = sin

¡x2¢ Z y=x1/3

y=−x1/3y2dy

= sin¡x2¢⎛⎝ y3

3

¯x1/3−x1/3

⎞⎠= sin

¡x2¢µ23x

¶=2

3x sin

¡x2¢.

Thus, Z x=√π

x=0

Z y=x1/3

y=−x1/3y2 sin

¡x2¢dydx =

Z x=√π

x=0

2

3x sin

¡x2¢dx.

We set u = x2 so that du = 2xdx. Therefore,Z x=√π

x=0

2

3x sin

¡x2¢dx =

1

3

Z u=π

u=0

sin (u) du

=1

3(− cos (x)|π0 ) =

2

3

7.

y

-2

01

2

0

1

x

-1

0

2z

4

6

-1 0 1

1

x

y

Let R be the region in the xy-plane that is between y = x2 and y = 1. The volume of D isZ ZR

¡x2 + y2 + 1

¢dA =

Z x=1

x=−1

Z y=1

y=x2

¡x2 + y2 + 1

¢dydx

=

Z x=1

x=−1

Ãx2y +

y3

3+ y

¯y=1y=x2

!dx

=

Z x=1

x=−1

µx2 +

1

3+ 1− x4 − x

6

3− x2

¶dx

=

Z x=1

x=−1

µ4

3− x4 − x

6

3

¶dx

=4

3x− x

5

5− x

7

21

¯1−1=38

35+38

35=76

35


8.

1.0

0.5

1.0

0.5

y x0.0 0.0

-2

z 0

2

Let R be the square [0, 1]× [0, 1] in the xy-plane. The volume of D is

Z ZR

(ex + ey) dA =

Z x=1

x=0

Z y=1

y=0

(ex + ey) dydx

=

Z x=1

x=0

³exy + ey|y=1y=0

´dx

=

Z x=1

x=0

(ex + e− 1) dx

= ex + (e− 1)x|10 = e+ (e− 1)− 1 = 2e− 2.

9.

02

1

6

z

4

x

2

00

3

y

21

2

3

x

y


Let R be the shaded triangle. The volume of D isZ ZR

(6− 3x− 2y) dA =

Z x=2

x=0

Z y=3−3x/2

y=0

(6− 3x− 2y) dydx

=

Z x=2

x=0

³6y − 3xy − y2 ¯3−3x/2

0

´dx

=

Z x=2

x=0

Ã6

µ3− 3x

2

¶− 3x

µ3− 3x

2

¶−µ3− 3x

2

¶2!dx

=

Z x=2

x=0

µ9

4(x− 2)2

¶dx

=

Z u=0

u=−2

9

4u2du =

9

12u3¯0−2=9

12

¡23¢= 6

10.

a)

3

1

x

y

b) Z y=1

y=0

Z 3

x=3y

ex2

dxdy =

Z x=3

x=0

Z y=x/3

y=0

ex2

dydx

=

Z x=3

x=0

ex2

ÃZ y=x/3

y=0

dy

!dx

=

Z x=3

x=0

x

3ex

2

dx.

We set u = x2 so that du = 2xdx. Thus,Z x=3

x=0

x

3ex

2

dx =1

6

Z 9

u=0

eudu =1

6

¡e9 − 1¢ .

11.

a)


b) Z y=√π/2

y=0

Z x=√π/2

x=y

cos¡x2¢dxdy =

Z x=√π/2

x=0

Z y=x

y=0

cos¡x2¢dydx

=

Z x=√π/2

x=0

cos¡x2¢µZ y=x

y=0

dy

¶dx

=

Z x=√π/2

x=0

cos¡x2¢xdx.

We set u = x2 so that du = 2xdx. Thus,Z x=√π/2

x=0

cos¡x2¢xdx =

1

2

Z π/4

0

cos (u) du

=1

2

³sin (u)|π/40

´=1

2

³sin³π4

´− sin (0)

´=

√2

4

12.

b)Z y=1

y=0

Z π/2

x=arcsin(y)

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

Z x=π/2

x=0

Z y=sin(x)

y=0

cos (x)p1 + cos2 (x)dydx

=

Z x=π/2

x=0

cos (x)p1 + cos2 (x)

ÃZ y=sin(x)

y=0

dy

!dx

=

Z x=π/2

x=0

cos (x)p1 + cos2 (x) sin (x) dx

We set u = cos (x) so that du = − sin (x) dx. Thus,Z x=π/2

x=0

cos (x)p1 + cos2 (x) sin (x) dx = −

Z u=0

u=1

up1 + u2du

=

Z u=1

u=0

up1 + u2du.

If we set v = 1 + u2 then dv = 2udu. Thus,Z u=1

u=0

up1 + u2du =

1

2

Z v=2

v=1

v1/2dv

=1

2

µ2

3v3/2

¶¯21

=1

3

³√8− 1

´.

13.3. DOUBLE INTEGRALS IN POLAR COORDINATES: 79

13.3 Double Integrals in Polar Coordinates:

1.

a)

4x

4

y

b) Z ZD

xydA =

Z θ=π/2

θ=0

Z r=4

r=0

r cos (θ) r sin (θ) rdrdθ

=

Z θ=π/2

θ=0

cos (θ) sin (θ)

µZ r=4

r=0

r3dr

¶dθ

=

Z θ=π/2

θ=0

cos (θ) sin (θ)

Ãr4

4

¯r=4r=0

!dθ

= 64

Z θ=π/2

θ=0

cos (θ) sin (θ) dθ.

We set u = sin (θ) so that du = cos (θ) dθ. Thus,

64

Z θ=π/2

θ=0

cos (θ) sin (θ) dθ = 64

Z 1

0

udu = 64

Ãu2

2

¯10

!= 32.

2.

a)

-2 2

2

b) Z ZD

sin¡x2 + y2

¢dA =

Z θ=π

θ=0

Z 2

r=0

sin¡r2¢rdrdθ

=

ÃZ θ=π

θ=0

dθ

!µZ 2

r=0

sin¡r2¢rdr

¶= π

Z 2

r=0

sin¡r2¢rdr.


We set u = r2 so that du = 2rdr. Thus,

π

Z 2

r=0

sin¡r2¢rdr =

π

2

Z 4

u=0

sin (u) du =π

2

³− cos (u)|40

´=

π

2(1− cos (4)) .

3.

a)

3

-3

3

x

y

b) Z ZD

p9− x2 − y2dA =

Z θ=π/2

θ=−π/2

Z 3

r=0

p9− r2rdrdθ

=

ÃZ θ=π/2

θ=−π/2dθ

!µZ 3

r=0

p9− r2rdr

¶= π

Z 3

r=0

p9− r2rdr.

We set u = 9− r2 so that du = −2rdr. Thus,

π

Z r=3

r=0

p9− r2rdr = −π

2

Z 0

9

√udu

=π

2

Ã2

3u3/2

¯90

!= π

µ1

3

¶(27) = 9π.

4.

a)

-2 -1 1 2

1

2

x

y

b) Z ZD

e−x2−y2dA =

Z θ=π

θ=0

Z r=2

r=1

e−r2

rdrdθ

= π

Z r=2

r=1

e−r2

rdr

= π

Ã−12e−r

2

¯21

!=

π

2

¡e−1 − e−4¢

5.


a)

1 2

1

2

x

y

b) Z ZD

arctan³yx

´dA =

Z θ=π/4

θ=0

Z r=2

r=1

θrdrdθ

=

ÃZ θ=π/4

θ=0

θdθ

!µZ r=2

r=1

rdr

¶

=

Ãθ2

2

¯π/40

!Ãr2

2

¯21

!=1

4

µπ2

16

¶(3) =

3π2

64.

6.

a)

-0.5 0.5

-0.5

0.5

x

y

b) We have

r = sin (2θ) = 0 and 0 < θ ≤ π

2if 2θ = π ⇔ θ =

π

2.

The area inside one loop isZ θ=π/2

θ=0

Z sin(2θ)

r=0

rdrdθ =

Z θ=π/2

θ=0

Ã1

2r2¯sin(2θ)r=0

!dθ

=1

2

Z θ=π/2

θ=0

sin2 (2θ) dθ

=1

4

Z π/2

0

(1− cos (4θ)) dθ

=1

4

Ãθ − 1

4sin (4θ)

¯π/20

!=

1

4

³π2

´=

π

8.

7.

a)


1 2 3

-1

1

x

y

b) We have

1 + cos (θ) = 3 cos (θ) if 2 cos (θ) = 1⇔ cos (θ) =1

2.

The only such angle between 0 and π/2 is π/3. By symmetry, the area of D is

2

ÃZ π/2

θ=π/3

Z r=1+cos(θ)

r=3 cos(θ)

rdrdθ +

Z θ=π

θ=π/2

Z r=1+cos(θ)

r=0

rdrdθ

!.

We haveZ π/2

θ=π/3

Z r=1+cos(θ)

r=3cos(θ)

rdrdθ =

Z π/2

θ=π/3

Ãr2

2

¯1+cos(θ)3 cos(θ)

!dθ

=1

2

Z π/2

θ=π/3

¡1 + 2 cos (θ)− 8 cos2 (θ)¢ dθ

=1

2

Z π/2

θ=π/3

µ1 + 2 cos (θ)− 8

µ1 + cos (2θ)

2

¶¶dθ

= −32θ + sin (θ)− sin (2θ)

¯π//2π/3

=

µ−3π4+ 1

¶−Ãπ

2+

√3

2−√3

2

!= −π

4+ 1,

and Z θ=π

θ=π/2

Z r=1+cos(θ)

r=0

rdrdθ =

Z θ=π

θ=π/2

Ã1

2r2¯1+cos(θ)0

!dθ

=1

2

Z θ=π

θ=π/2

(1 + cos (θ))2 dθ

=1

2

Z θ=π

θ=π/2

¡1 + 2 cos (θ) + cos2 (θ)

¢dθ

=1

2

Z θ=π

θ=π/2

µ1 + 2 cos (θ) +

1 + cos (2θ)

2

¶dθ

=1

2

Ã3

2θ + 2 sin (θ) +

1

4sin (2θ)

¯ππ/2

!

=1

2

µ3π

2− 3π4− 2¶=3π

8− 1.


Therefore, the area of D is

2

ÃZ π/2

θ=π/3

Z r=1+cos(θ)

r=3 cos(θ)

rdrdθ +

Z θ=π

θ=π/2

Z r=1+cos(θ)

r=0

rdrdθ

!= 2

µ−π4+ 1 +

3π

8− 1¶

= 2³π8

´=

π

4.

8. The volume of D is Z ZR

px2 + y2dA,

where R is the disk that is inside the circle x2+y2 = 4, so that R is the disk of radius 2 centeredat the origin. Using polar coordinates,Z Z

R

px2 + y2dA =

Z θ=2π

θ=0

Z r=2

r=0

(r) rdrdθ

=

Z θ=2π

θ=0

Z r=2

r=0

r2drdθ

=

ÃZ θ=2π

θ=0

dθ

!µZ r=2

r=0

r2dr

¶

= 2π

Ãr3

3

¯20

!= 2π

µ8

3

¶=16π

3.

9. By symmetry, the volume of D is

2

Z ZR

p16− x2 − y2dA,

where R is the annular region between the circles x2 + y2 = 4 and x2 + y2 = 16. Thus,

2

Z ZR

p16− x2 − y2dA = 2

Z θ=2π

θ=0

Z r=4

r=2

p16− r2rdrdθ

= 2

ÃZ θ=2π

θ=0

dθ

!µZ r=4

r=2

p16− r2rdr

¶= 4π

Z r=4

r=2

p16− r2rdr.

We set u = 16− r2 so that du = −2rdr. Therefore,

4π

Z r=4

r=2

p16− r2rdr = −2π

Z u=0

u=12

u1/2du

= −2πÃ2

3u3/2

¯012

!=4π

3123/2.

10.

a)

-3

3

x

y


b) Z 3

−3

Z √9−x20

sin¡x2 + y2

¢dydx =

Z θ=π

θ=0

Z r=3

r=0

sin¡r2¢rdrdθ

=

ÃZ θ=π

θ=0

dθ

!µZ r=3

r=0

sin¡r2¢rdr

¶= π

Z r=3

r=0

sin¡r2¢rdr.

We set u = r2 so that du = 2rdr. Thus,

π

Z r=3

r=0

sin¡r2¢rdr =

π

2

Z 9

u=0

sin (u) du =π

2

³− cos (u)|90

´=

π

2(− cos (9) + 1)

11.

a) We have

x =p2− y2 ⇒ x2 + y2 = 2.

This is the equation of a circle of radius√2 centered at the origin. If y = 1 then x =

√2. The

polar coordinates of (1, 1) are r =√2 and θ = π/4

b) We have θ = π/4 on the line y = x. Thus,

Z y=1

y=0

Z x=√2−y2

x=y

xdxdy =

Z θ=π/4

θ=0

Z √2r=0

r cos (θ) rdrdθ

=

ÃZ θ=π/4

θ=0

cos (θ) dθ

!ÃZ √2r=0

r2dr

!

=³sin (θ)|π/40

´Ã 13r3¯√20

!

=

√2¡√2¢3

6=4

6=2

3.

13.4 Applications of Double Integrals

1. The mass of the plate is

13.4. APPLICATIONS OF DOUBLE INTEGRALS 85

m (D) =

Z ZD

ρ (x, y) dA =

Z x=4

x=0

Z y=2

y=0

(1 + x+ y) dydx

=

Z 4

0

(2x+ 4) dx = 32

The moment of the plate with respect to the y-axis is

My (D) =

Z ZD

xρ (x, y) dxdy =

Z x=4

x=0

Z y=2

y=0

x (1 + x+ y) dydx

=

Z 4

0

2x (x+ 2) dx =224

3

The moment of the plate with respect to the x-axis is

Mx (D) =

Z ZD

yρ (x, y) dxdy =

Z 4

0

Z 2

0

y (1 + x+ y) dydx

=

Z 4

0

µ2x+

14

3

¶dx =

104

3

Therefore, the x-coordinate of the center of mass is

x =My (D)

m (D)=

224

332

=7

3∼= 2.33

The y-coordinate of the center of mass is

y =Mx (D)

m (D)=

104

332

=13

12∼= 1. 083

2. The mass of the plate is

m (D) =

Z ZD

ρ (x, y) dA =

Z ZD

³81−

px2 + y2

´dA.

We transform to polar coordinates:Z ZD

³81−

px2 + y2

´dA =

Z θ=π

θ=0

Z r=3

r=0

(81− r) rdrdθ

= 2π

Z 3

r=0

¡81r − r2¢ dr = 711π

The moment of the plate with respect to the y-axis is

My (D) =

Z ZD

xρ (x, y) dA =

Z ZD

x³81−

px2 + y2

´dA.


x³81−

px2 + y2

´dA =

Z θ=π

θ=0

Z r=3

r=0

r cos (θ) (81− r) rdrdθ

=

ÃZ θ=π

θ=0

cos (θ) dθ

!µZ r=3

r=0

¡81r2 − r3¢ dr¶

= (0)

µ2835

4

¶= 0.


The moment of the plate with respect to the x-axis is

Mx (D) =

Z ZD

yρ (x, y) =

Z ZD

y³81−

px2 + y2

´dA.


y³81−

px2 + y2

´dA =

Z θ=π

θ=0

Z r=3

r=0

r sin (θ) (81− r) rdrdθ

=

ÃZ θ=π

θ=0

sin (θ) dθ

!µZ r=3

r=0

¡81r2 − r3¢ dr¶

= (2)

µ2835

4

¶=2835

2.

Therefore, the x-coordinate of the center of mass is

x =My (D)

m (D)= 0.

The y-coordinate of the center of mass is

y =Mx (D)

m (D)=

2835

2711π

=315

158π∼= 0.634.

3. The moment of inertia of the plate about the x-axis is

Ix =

Z ZD

y2ρ (x, y) dxdy =

Z ZD

y2ρ0dxdy = ρ0

Z ZD

y2dxdy.

We transform to polar coordinates:

ρ0

Z ZD

y2dxdy = ρ0

Z θ=2π

θ=0

Z r=r0

r=0

r2 sin2 (θ) rdrdθ

= ρ0

ÃZ θ=2π

θ=0

sin2 (θ) dθ

!µZ r=r0

r=0

r3dr

¶= ρ0 (π)

µ1

4r40

¶=

π

4ρ0r

40

The moment of inertia of the plate about the y-axis is

Iy =

Z ZD

x2ρ (x, y) dxdy =

Z ZD

x2ρ0dxdy = ρ0

Z ZD

x2dxdy.


ρ0

Z ZD

x2dxdy = ρ0

Z θ=2π

θ=0

Z r=r0

r=0

r2 cos2 (θ) rdrdθ

= ρ0

ÃZ θ=2π

θ=0

cos2 (θ) dθ

!µZ r=r0

r=0

r3dr

¶= ρ0 (π)

µ1

4r40

¶=

π

4ρ0r

40.

13.4. APPLICATIONS OF DOUBLE INTEGRALS 87

The moment of inertia of the plate about the origin is

I0 = Ix + Iy =π

4ρ0r

40 +

π

4ρ0r

40 =

π

2ρ0r

40.

4. The moment of inertia of the plate about the x-axis is

Ix =

Z ZD

y2ρ (x, y) dxdy =

Z ZD

y2³81−

px2 + y2

´dxdy.


Z ZD

y2³4−

px2 + y2

´dxdy =

Z θ=π

θ=0

Z r=3

r=0

r2 sin2 (θ) (4− r) rdrdθ

=

ÃZ θ=π

θ=0

sin2 (θ) dθ

!µZ r=3

r=0

¡4r3 − r4¢ dr¶

=

µ1

2π

¶µ162

5

¶= 16.2π

The moment of inertia of the plate about the y-axis is

Iy =

Z ZD

x2ρ (x, y) dxdy =

Z ZD

x2³81−

px2 + y2

´dxdy


Z ZD

x2³81−

px2 + y2

´dxdy =

Z θ=π

θ=0

Z r=3

r=0

r2 cos2 (θ) (4− r) rdrdθ

=

ÃZ θ=π

θ=0

cos2 (θ) dθ

!µZ r=3

r=0

¡4r3 − r4¢ dr¶

=

µ1

2π

¶µ162

5

¶= 16.2π

The moment of inertia of the plate about the origin is

I0 = Ix + Iy = 16.2π + 16.2π = 32.4π

5. Since X and Y are independent, their joint density function is

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

½112e−−x/6e−y/2 if x ≥ 0 and y ≥ 0,0 otherwise.

Therefore, the probability that X + Y < 8 isZ ZD

1

12e−x/6e−y/2dxdy,


where D is the triangular region in the first quadrant below the line x+ y = 8. We have

Z ZD

1

12e−−x/6e−y/2dxdy =

Z x=8

x=0

Z 8−x

y=0

1

12e−−x/6e−y/2ydx

=1

12

Z x=8

x=0

e−x/6µZ 8−x

y=0

e−y/2dy¶dx

=1

12

Z x=8

x=0

e−x/6³2− 2ex/2−4

´dx

=1

12

Z x=8

x=0

³2e−x/6 − 2ex/3−4

´dx

=1

12

³6e−4 − 18e−4/3 + 12

´=

1

2e−4 − 3

2e−4/3 + 1 ∼= 0.613 762.

6. The density functions of X and Y are

f1 (x) =1

0.2√2πe−(x−3)

2/0.08 and f2 (y) =1

0.1√2πe−(y−5)

2/0.02,

respectively. Therefore,

P (2.5 < X < 3.5 and 4.5 < Y < 5.5)

=

Z y=5.5

y=4.5

Z x=3.5

x=2.5

f1 (x) f2 (y) dxdy

=

µZ y=5.5

y=4.5

f2 (y) dy

¶µZ x=3.5

x=2.5

f1 (x) dx

¶=

µZ y=5.5

y=4.5

1

0.1√2πe−(y−5)

2/0.02dy

¶µZ x=3.5

x=2.5

1

0.2√2πe−(x−3)

2/0.08dx

¶∼= (0.999 999 ) (0.987 581 ) ∼= 0.987 58.

13.5 Triple Integrals

1. The surfaces intersect when

x2 + y2 = 18− x2 − y2 ⇔ x2 + y2 = 9.

This is a circle of radius 3 centered at the origin. The volume in question is

Z ZR

ÃZ z=18−x2−y2

z=x2+y2dz

!dxdy

13.5. TRIPLE INTEGRALS 89

where R is the disk that is enclosed by the circle x2 + y2 = 9. Let’s use polar coordinates:

Z ZR

ÃZ z=18−x2−y2

z=x2+y2dz

!dxdy =

Z θ−2π

θ=0

Z r=3

r=0

ÃZ z=18−r2

z=r2dz

!rdrdθ

=

Z θ−2π

θ=0

Z r=3

r=0

¡18− 2r2¢ rdrdθ

= 2π

Z r=3

r=0

¡18r − 2r3¢ dr

= 2π

Ã9r2 − 1

2r4¯30

!

= 2π

µ81

2

¶= .81π.

2. The volume in question is

Z ZR

µZ z=1−x−y

z=−10dz

¶dxdy,

where R is the disk that is enclosed by the circle of radius 2 centered at the origin. Let’s use

polar coordinates:

Z ZR

µZ z=1−x−y

z=−10dz

¶dxdy =

Z θ=2π

θ=0

Z r−2

r−0

ÃZ z=1−r(cos(θ)+sin(θ))

z=−10dz

!rdrdθ

=

Z θ=2π

θ=0

Z r−2

r−0

¡11r − r2 cos (θ)− r2 sin (θ)¢ drdθ

=

Z θ=2π

θ=0

Ã11

2r2 − 1

3r3 cos (θ)− 1

3r3 sin (θ)

¯r=2r=0

!dθ

=

Z θ=2π

θ=0

µ22− 8

3cos (θ)− 8

3sin (θ)

¶dθ

= 22θ − 83sin (θ) +

8

3cos (θ)

¯2π0

= 44π.

3.

Z Z ZD

x2dxdydz =

Z z=1

z=0

Z y=1

y=0

Z x=1

x=0

x2dxdydz

=

Z z=1

z=0

Z y=1

y=0

Ã1

3x3¯10

!dydz

=1

3

Z z=1

z=0

Z y=1

y=0

dydz =1

3.


4.

Z Z ZD

e−xyydxdydz =Z z=1

z=0

Z y=1

y=0

Z x=1

x=0

e−xyydxdydz

=

Z z=1

z=0

Z y=1

y=0

³−e−xy ¯x=1

x=0

´dydz

=

Z z=1

z=0

Z y=1

y=0

¡−e−y + 1¢ dydz=

Z z=1

z=0

³e−y + y

¯y=1y=0

´dz

=

Z z=1

z=0

¡e−1 + 1− 1¢ dz = 1

e.

5.

Z Z ZD

zex+ydxdydz =

Z z=2

z=0

Z y=1

y=0

Z x=1

x=0

zex+ydxdydz

=

Z z=2

z=0

Z y=1

y=0

zeyµZ x=1

x=0

exdx

¶dydz

=

Z z=2

z=0

Z y=1

y=0

zey (e− 1) dydz

= (e− 1)Z z=2

z=0

Z y=1

y=0

zeydydz

= (e− 1)Z z=2

z=0

z

Z y=1

y=0

eydydz

= (e− 1)Z z=2

z=0

z (e− 1) dz

= (e− 1)2Ã1

2z2¯z=2z=0

!= 2 (e− 1)2 .

Remark: One can also note that

Z z=2

z=0

Z y=1

y=0

Z x=1

x=0

zex+ydxdydz =

Z z=2

z=0

Z y=1

y=0

Z x=1

x=0

zexeydxdydz

=

µZ z=2

z=0

zdz

¶µZ y=1

y=0

eydy

¶µZ 1

x=0

ex¶dx

at the beginning.


6. Z Z ZD

x2 cos (z) dxdydz =

Z z=π/2

z=0

µZ 1

y=0

µZ x=1−y

x=0

x2 cos (z) dx

¶dy

¶dz

=

Z z=π/2

z=0

cos (z)

ÃZ y=1

y=0

Ã1

3x3¯1−yx=0

!dy

!

=

Z z=π/2

z=0

cos (z)

µZ y=1

y=0

1

3(1− y)3 dy

¶dz

=

Z z=π/2

z=0

cos (z)

µ1

12

¶dz

=1

12

³sin (z)|π/20

´=1

12.

7. Z Z ZD

zdxdydz =

Z z=1

z=0

z

µZ ZR

dxdy

¶dz =

1

2

Z ZR

dxdy,

where

R = (x, y) : x2 + y2 ≤ 1, x ≥ 0 and y ≥ 0.Therefore, Z Z

R

dxdy =

Z θ=π/2

θ=0

Z 1

r=0

rdrdθ =π

2

Ã1

2r2¯10

!=

π

4

(merely the area of the part of the unit disk in the first quadrant).

Thus, Z Z ZD

zdxdydz =1

2

Z ZR

dxdy =π

8.

8. Part of the boundary of the tetrahedron is the plane that passes trough the points (1, 0, 0) , (0, 1, 0) , (0, 0, 1).That plane is the graph of an equation of the form is

ax+ by + cz = d.

You can confirm that a = b = c = d = 1, so that the equation is

x+ y + z = 1,

so that z = 1− x− y = 1− y − x. The projection of that plane onto the xy-plane is the linex+ y = 1.

Thus, Z Z ZD

x2dxdydz =

Z ZR

µZ z=1−x−y

z=0

x2dz

¶dxdy,

where R is the triangular region bounded by the x and y axes and the line x+ y = 1, as in thepicture:

1

1

x

y


Therefore,

Z ZR

µZ z=1−x−y

z=0

x2dz

¶dxdy =

Z x=1

x=0

Z 1−x

y=0

x2µZ z=1−x−y

z=0

dz

¶dydx

=

Z x=1

x=0

Z 1−x

y=0

x2 (1− x− y) dydx

=

Z x=1

x=0

Z 1−x

y=0

¡x2 − x3 − x2y¢ dydx

=

Z x=1

x=0

Ãx2y − x3y − 1

2x2y2

¯1−x0

!dx

=

Z x=1

x=0

µ1

2x4 − x3 + 1

2x2¶dx

=1

10x5 − 1

4x4 +

1

6x3¯10

=1

60

9.

Z Z ZD

xydxdxydz =

Z ZR

µZ z=x+y

z=0

xydz

¶dxdy,

where R is the region in the xy-plane that is bounded by y = x2 and x = y2.


Thus,

Z ZR

µZ z=x+y

z=0

xydz

¶dxdy =

Z ZR

xy

µZ z=x+y

z=0

dz

¶dxdy

=

Z ZR

xy (x+ y) dxdy

=

Z ZR

¡x2y + xy2

¢dxdy

=

Z x=1

x=0

Z y=√x

y=x2

¡x2y + xy2

¢dydx

=

Z x=1

x=0

Ãx2y2

2+xy3

3

¯y=√xy=x2

!dx

=

Z x=1

x=0

µx3

2+x5/2

3− x

6

2− x

7

3

¶dx

=x4

8+2

21x7/2 − x

7

14− x

8

24

¯10

=1

8+2

21− 1

14− 1

24=3

28

10.

Z Z ZD

x2dydxdz =

Z ZR

x2µZ z=5−y

z=1

dz

¶dxdy,

where R is the region inside the circle x2 + y2 = 9. Thus,

Z ZR

x2µZ z=5−y

z=1

dz

¶dxdy =

Z ZR

x2 (4− y) dxdy

=

Z θ=2π

θ=0

Z 3

r=0

r3 cos2 (θ) (4− r sin (θ)) drdθ

=

Z θ=2π

θ=0

cos2 (θ)

Ãr4 − r

5

5sin (θ)

¯3r=0

!dθ

=

Z θ=2π

θ=0

cos2 (θ)

µ81− 243

5sin (θ)

¶dθ

=

Z θ=2π

θ=0

µ81 cos2 (θ)− 243

5cos2 (θ) sin (θ)

¶dθ

=81

2

Z θ=2π

θ=0

(1 + cos (2θ)) dθ − 2435

Z 2π

0

cos2 (θ) sin (θ) dθ

=81

2

Ãθ +

1

2sin (2θ)

¯2π0

!+243

5

Ãcos3 (θ)

3

¯2π0

!= 81π


13.6 Triple Integrals in Cylindrical and Spherical Coordi-

nates

1.

x = 2 cos³π4

´= 2

Ã√2

2

!=√2,

y = 2 sin³π4

´= 2

Ã√2

2

!=√2,

z = 1.

2.

x = 4cos³π6

´= 4

Ã√3

2

!= 2√3,

y = 4 sin³π6

´= 4

µ1

2

¶= 2,

z = 2.

3.

x = 3 cos

µ2π

3

¶= 3

µ−12

¶= −3

2,

y = 3 sin

µ2π

3

¶= 3

Ã√3

2

!=3√3

2,

z = 4.

4.

x = 2cos³−π6

´= 2

Ã√3

2

!=√3,

y = 2 sin³−π6

´= 2

µ−12

¶= −1,

z = 4.

5.

r =√1 + 1 =

√2,

θ = arccos

µ− 1√

2

¶= π − π

4=3π

4,

z = 4.

6.

r =√4 + 12 =

√16 = 4,

θ = − arccosµ2

4

¶= − arccos

µ1

2

¶= −π

3,

z = −1

13.6. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 95

7.

r =√1 + 1 =

√2

θ = − arccosµ1√2

¶= −π

4,

z = 2

8.

r =√1 + 3 =

√4 = 2,

θ = arccos

µ1

2

¶=

π

3,

z = 2

9. We have

x2 + y2 = 16⇔ r2 = 16⇒ r = 4.

5-5

0z

5

5

y

0

x0

-5-5

Thus, Z Z ZD

px2 + y2dxdydz =

Z ZR

µZ z=4

z=−5

px2 + y2dz

¶dxdy,

where R is the disk of radius 4 centered at the origin in the xy-plane. Thus,Z ZR

µZ z=4

z=−5

px2 + y2dz

¶dxdy =

Z ZR

px2 + y2

µZ z=4

z=−5dz

¶dxdy

= 9

Z ZR

px2 + y2dxdy

= 9

Z θ=2π

θ=0

Z r−4

r=0

r2drdθ

= 9 (2π)

Ãr3

3

¯40

!= 9 (2π)

µ64

3

¶= 384π

10. We have

z = 1− x2 − y2 = z = 1− r2,and

z = 0⇒ 1− r2 = 0⇒ r = 1.

Thus, the intersection of the paraboloid with the xy-plane is the unit circle.

00

1

z

1

x y1 0


Therefore, Z Z ZD

¡x3 + xy2

¢dxdydz =

Z ZR

ÃZ 1−x2−y2

z=0

¡x3 + xy2

¢dz

!dxdy,

where R is the part of the unit disk in the first quadrant. Thus,Z ZR

ÃZ 1−x2−y2

z=0

¡x3 + xy2

¢dz

!dxdy

=

Z θ=π/2

θ=0

Z r=1

r=0

ÃZ 1−r2

z=0

¡r3 cos3 (θ) + r3 cos (θ) sin2 (θ)

¢dz

!rdrdθ

We have

r3 cos3 (θ) + r3 cos (θ) sin2 (θ) = r3 cos (θ)¡cos2 (θ) + sin2 (θ)

¢= r3 cos (θ) .

Therefore, Z θ=π/2

θ=0

Z r=1

r=0

ÃZ 1−r2

z=0

¡r3 cos3 (θ) + r3 cos (θ) sin2 (θ)

¢dz

!rdrdθ

=

Z θ=π/2

θ=0

Z r=1

r=0

ÃZ 1−r2

z=0

r4 cos (θ) dz

!drdθ

=

Z θ=π/2

θ=0

Z r=1

r=0

r4 cos (θ)

ÃZ 1−r2

z=0

dz

!drdθ

=

Z θ=π/2

θ=0

Z r=1

r=0

r4 cos (θ)¡1− r2¢ drdθ

=

ÃZ θ=π/2

θ=0

cos (θ) dθ

!µZ r=1

r=0

¡r4 − r6¢ dr¶

=³sin (θ)|π/20

´Ã 15r5 − 1

7r7¯1r=0

!

= (1)

µ2

35

¶=2

35.

11.

0

1

2

z

6

y

0

2

x

-2 0-2

We have Z Z ZD

ezdxdydz =

Z ZR

ÃZ z=1+x2+y2

z=0

ezdz

!dxdy,


where R is the disk that is bounded by the circle x2 + y2 = 5 in the xy-plane. This is the circleof radius

√5 centered at the origin. Thus,Z θ=2π

θ=0

Z r=√5

r=0

ÃZ z=1+r2

z=0

ezdz

!rdrdθ =

Z θ=2π

θ=0

Z r=√5

r=0

³ez|1+r20

´rdrdθ

=

Z θ=2π

θ=0

Z r=√5

r=0

³e1+r

2 − 1´rdrdθ

= 2π

Z r=√5

r=0

³e1+r

2

r − r´dr.

We set u = 1 + r2 so that du = 2rdr. Therefore,Ze1+r

2

rdr =1

2

Zeudu =

1

2eu =

1

2e1+r

2

.

Thus,

2π

Z r=√5

r=0

³e1+r

2

r − r´dr = 2π

Ã1

2e1+r

2 − 12r2¯√50

!= π

¡e6 − 5− e¢

12. In cylindrical coordinates,

x2 + y2 = 1⇔ r2 = 1⇔ r = 1,

and

z =p4x2 + 4y2 ⇔ z = 2

px2 + y2 ⇔ z = 2r.

01

z

2

1

y

0

x0

-1 -1

We have Z Z ZD

x2dxdydz =

Z ZR

µZ z=2r

z=0

r2 cos2 (θ) dz

¶dxdy

=

Z θ=2π

θ=0

Z r=1

r=0

µZ z=2r

z=0

r2 cos2 (θ) dz

¶rdrdθ

=

Z θ=2π

θ=0

Z r=1

r=0

r3 cos2 (θ)

µZ z=2r

z=0

dz

¶drdθ

=

Z θ=2π

θ=0

Z r=1

r=0

2r4 cos2 (θ) drdθ

=

ÃZ θ=2π

θ=0

cos2 (θ) dθ

!µZ r=1

r=0

2r4dr

¶

=

ÃZ θ=2π

θ=0

1 + cos (2θ)

2dθ

!Ã2

5r5¯r=1r=0

!

=2

5

Ãθ

2+1

4sin (2θ)

¯θ=2πθ=0

!=2

5π.


13.

2

2

0

-1

-2

1

z

2

1

0

y -1

-2

1

x0

-1-2

The volume of the region is Z ZR

ÃZ z=√4−x2−y2

z=−√4−x2−y2

dz

!dxdy,

where R is the disk in the xy-plane that is bounded by the circle x2 + y2 = 1. Thus,Z ZR

ÃZ z=√4−x2−y2

z=−√4−x2−y2

dz

!dxdy =

Z θ=2π

θ=0

Z r=1

r=0

ÃZ z=√4−r2

z=−√4−r2dz

!rdrdθ

=

Z θ=2π

θ=0

Z r=1

r=0

³2p4− r2

´rdrdθ

= 4π

Z r=1

r=0

p4− r2rdr.

We set u = 4− r2 so that du = −2rdr. Therefore,

4π

Z r=1

r=0

p4− r2rdr = −2π

Z u=3

u=4

u1/2du

= −2πÃ2

3u3/2

¯34

!=

4π

3

³8− 33/2

´14.

x = ρ sin (φ) cos (θ) = 1 sin (0) cos (0) = 0,

y = ρ sin (φ) sin (θ) = 1 sin (0) sin (0) = 0,

z = ρ cos (φ) = 1 cos (0) = 1.

Thus, the point is (0, 0, 1) in Cartesian coordinates.

15.

x = ρ sin (φ) cos (θ) = 2 sin³π4

´cos³π3

´= 2

Ã√2

2

!µ1

2

¶=

√2

2,

y = ρ sin (φ) sin (θ) = 2 sin³π4

´sin³π3

´= 2

Ã√2

2

!Ã√3

2

!=

√6

2,

z = ρ cos (φ) = 2 cos³π4

´= 2

Ã√2

2

!=√2.


Thus, the point is Ã√2

2,

√6

2,√2

!in Cartesian coordinates.

16.


ćos (π) = 5 (1) (−1) = −5,


´sin (π) = 5 (1) (0) = 0,


´= 5 (0) = 0.

Thus, the point is (−5, 0, 0) in Cartesian coordinates.17.


ćos

µ3π

4

¶= 4

Ã√3

2

!Ã−√2

2

!= −√6,


´sin

µ3π

4

¶= 4

Ã√3

2

!Ã√2

2

!=√6,


´= 4

µ1

2

¶= 2.

Thus, the point is¡−√6,√6, 2¢ in Cartesian coordinates.

18.

ρ =√1 + 3 + 12 =

√16 = 4,

φ = arccos

µz

ρ

¶= arccos

Ã2√3

4

!= arccos

Ã√3

2

!=

π

6,

θ = arccos

Ãxp

x2 + y2

!= arccos

µ1√1 + 3

¶= arccos

µ1

2

¶=

π

3

(since y =√3 > 0).

Thus, the point is ³4,π

3,π

6

ín spherical coordinates.

19.

ρ =√0 + 1 + 1 =

√2,

φ = arccos

µz

ρ

¶= arccos

µ− 1√

2

¶= π − π

4=3π

4,

θ = − arccosÃ

xpx2 + y2

!= − arccos (0) = −π

2

(since y = −1 < 0).Thus, the point is µ√

2,−π2,3π

4

¶


in spherical coordinates.

20.

ρ =√0 + 3 + 1 =

√4 = 2,

φ = arccos

µz

ρ

¶= arccos

µ1

2

¶=

π

3,

θ = arccos

Ã0p

x2 + y2

!= arccos (0) =

π

2

(since y = 1 > 0).

Thus, the point is ³2,π

2,π

3

´in spherical coordinates.

21.

ρ =√1 + 1 + 6 =

√8 = 2

√2,

φ = arccos

µz

ρ

¶= arccos

Ã √6

2√2

!= arccos

Ã√3

2

!=

π

6,

θ = arccos

Ãxp

x2 + y2

!= arccos

µ− 1√

1 + 1

¶= arccos

µ− 1√

2

¶=3π

4

(since y = 1 > 0).

Thus, the point is µ2√2,3π

4,π

6

¶

in spherical coordinates.

22.

1

x0

0

y

2

2

-2-2

0

z

2

3

Since ρ = 3 on the sphere x2 + y2 + z2 = 9, and 0 ≤ φ < π/2 above the xy-plane,


Z Z ZD

¡9− x2 − y2¢ dxdydz

=

Z θ=2π

θ=0

Z π/2

φ=0

Z ρ=3

ρ=0

¡9− ρ2 sin2 (φ) cos2 (θ)− ρ2 sin2 (φ) sin2 (θ)

¢ρ2 sin (φ) dρdφdθ

=

Z θ=2π

θ=0

Z π/2

φ=0

Z ρ=3

ρ=0

¡9− ρ2 sin2 (φ)

¢ρ2 sin (φ) dρdφdθ

=

Z θ=2π

θ=0

Z π/2

φ=0

Ã3ρ3 − 1

5ρ5 sin2 (φ)

¯ρ=3ρ=0

!sin (φ) dφdθ

=

Z θ=2π

θ=0

Z π/2

φ=0

µ81− 243

5sin2 (φ)

¶sin (φ) dφdθ

= 81

Z θ=2π

θ=0

Z π/2

φ=0

sin (φ) dφdθ − 2435

Z θ=2π

θ=0

Z π/2

φ=0

sin3 (φ) dφdθ

We have

81

Z θ=2π

θ=0

Z π/2

φ=0

sin (φ) dφdθ = 81

Z θ=2π

θ=0

³− cos (φ)|π/20

´dθ

= 81

Z θ=2π

θ=0

dθ = 81 (2π) = 162π

We also have Zsin3 (φ) dφ =

Zsin2 (φ) sin (φ) dφ =

Z ¡1− cos2 (φ)¢ sin (φ) dφ.

If we set u = cos (φ) then du = − sin (φ) dφ so thatZ ¡1− cos2 (φ)¢ sin (φ) dφ = −

Z ¡1− u2¢ du

= −u+ 13u3 = − cos (φ) + 1

3cos3 (φ) .

Thus, Z π/2

φ=0

sin3 (φ) dφ = − cos (φ) + 13cos3 (φ)

¯π/20

= 1− 13=2

3.

Therefore,Z Z ZD

¡9− x2 − y2¢ dxdydz = 81

Z θ=2π

θ=0

Z π/2

φ=0

sin (φ) dφdθ − 2435

Z θ=2π

θ=0

Z π/2

φ=0

sin3 (φ) dφdθ

= 162π − 2435

Z θ=2π

θ=0

µ2

3

¶dθ

= 162π − 1625(2π) =

486

5π

23.


2

y

1

0

00

x

1

2

z 1

2

Z Z ZD

zdxdydz =

Z θ=π/2

θ=0

Z φ=π/2

φ=0

Z ρ=2

ρ=1

ρ cos (φ) ρ2 sin (φ) dρdφdθ

=

Z θ=π/2

θ=0

Z φ=π/2

φ=0

Z ρ=2

ρ=1

ρ3 cos (φ) sin (φ) dρdφdθ

=

ÃZ θ=π/2

θ=0

dθ

!ÃZ φ=π/2

φ=0

cos (φ) sin (φ) dφ

!µZ ρ=2

ρ=1

ρ3dρ

¶

=³π2

´Ã 12sin2 (φ)

¯π/20

!Ã1

4ρ4¯21

!

=³π2

´µ12

¶µ24

4− 14

¶=15

16π

24.

3

2x

z

2

1

1

00

0

1 y

2

3

3

Z Z ZD

e√x2+y2+z2dxdydz =

Z θ=π/2

θ=0

Z φ=π/2

φ=0

Z ρ=3

ρ=0

eρρ2 sin (φ) dρdφdθ

=

ÃZ θ=π/2

θ=0

dθ

!ÃZ φ=π/2

φ=0

sin (φ) dφ

!µZ ρ=3

ρ=0

eρρ2dρ

¶=

³π2

´³− cos (φ)|π/20

´Z ρ=3

ρ=0

eρρ2dρ

=π

2

Z ρ=3

ρ=0

eρρ2dρ.

In order to evaluate the last integral, let’s apply integration by parts by setting u = ρ2 anddv = eρdρ. Thus,

du = 2ρdρ and v =

Zeρdρ = eρ.


Therefore, Zρ2eρdρ =

Zudv = uv −

Zvdu

= ρ2eρ − 2Zeρρdρ.

Now we apply integration by parts by setting u = ρ and dv = eρ. Therefore,Zρeρdρ =

Zudv = uv −

Zvdu

= ρeρ −Zeρdρ = ρeρ − eρ.

Thus, Zρ2eρdρ = ρ2eρ − 2

Zeρρdρ = ρ2eρ − 2 (ρeρ − eρ)

= ρ2eρ − 2ρeρ + 2eρ.

Therefore,

π

2

Z ρ=3

ρ=0

eρρ2dρ =π

2

³ρ2eρ − 2ρeρ + 2eρ¯ρ=3

ρ=0

´=

π

2

¡5e3 − 2¢ = π

2

¡5e3 − 2¢

25.

2

0z

-2

-4-4

-2

x

0

2

4 02

y

4

4

The spheres have radius 3 and 4, respectively, and y > 0 in the region. Therefore,Z Z ZD

x2dxdydz =

Z θ=π

θ=0

Z φ=π

φ=0

Z ρ=4

ρ=3

ρ2 sin2 (φ) cos2 (θ) ρ2 sin3 (θ) dρdφdθ

=

ÃZ θ=π

θ=0

cos2 (θ) dθ

!ÃZ φ=π

φ=0

sin3 (φ) dφ

!µZ ρ=4

ρ=3

ρ4dρ

¶

=

µ1

2θ +

1

4sin (2θ)

¯π0

¶ÃZ φ=π

φ=0

sin3 (φ) dφ

!Ã1

5ρ5¯43

!

=³π2

´ÃZ φ=π

φ=0

sin3 (φ) dφ

!µ781

5

¶.

We have Zsin3 (φ) dφ =




We set u = cos (φ):Z ¡1− cos2 (φ)¢ sin (φ) dφ = −

Z ¡1− u2¢ du

= −u+ 13u3 = − cos (φ) + 1

3cos3 (φ) .

Therefore, Z φ=π

φ=0

sin3 (φ) dφ = − cos (φ) + 13cos3 (φ)

¯π0

= 1− 13+ 1− 1

3=4

3

Thus, ³π2

´ÃZ φ=π

φ=0

sin3 (φ) dφ

!µ781

5

¶=³π2

´µ43

¶µ781

5

¶=1562

15π

26.

0

2

4

z

y

0

42

x0-2-4

Z Z ZD

x2zdxdydz =

Z θ=2π

θ=0

Z φ=π/3

φ=0

Z ρ=4

ρ=2

ρ2 sin2 (φ) cos2 (θ) ρ cos (φ) ρ2 sin (φ) dρdφdθ

=

Z θ=2π

θ=0

Z φ=π/3

φ=0

Z ρ=4

ρ=2

ρ5 sin3 (φ) cos (φ) cos2 (θ) dρdφdθ

=

ÃZ θ=2π

θ=0

cos2 (θ) dθ

!ÃZ φ=π/3

φ=0

sin3 (φ) cos (φ) dφ

!µZ ρ=4

ρ=2

ρ5dρ

¶

=

Ã1

2θ +

1

4sin (2θ)

¯2π0

!Ã1

4sin4 (θ)

¯π/30

!Ã1

5ρ5¯42

!

= (π)

⎛⎝14

Ã√3

2

!4⎞⎠µ15

¡45 − 25¢¶ = 279

10π

27.

0

2

4

4

z

2

y

0

-2 42

x0

-4 -2-4


Z Z ZD

dxdydz =

Z θ=2π

θ=0

Z φ=π/3

φ=π/6

Z ρ=3

ρ=0

ρ2 sin (φ) dρdφdθ

=

ÃZ θ=2π

θ=0

dθ

!ÃZ φ=π/3

φ=π/6

sin (φ) dφ

!µZ ρ=3

ρ=0

ρ2dρ

¶

= (2π)³− cos (φ)|π/3π/6

´Ã 13ρ3¯30

!

= (2π)

Ã−12+

√3

2

!µ1

3

¡33¢¶= 9

³√3− 1

´π.

28.

0

2

4

z

y

0

4

x

20-2-4

Z Z ZD

zdxdydz =

Z θ=2π

θ=0

Z φ=π/3

φ=0

Z ρ=4 cos(φ)

ρ=0

ρ cos (φ) ρ2 sin (φ) dρdφdθ

=

Z θ=2π

θ=0

Z φ=π/3

φ=0

cos (φ) sin (φ)

ÃZ ρ=4 cos(φ)

ρ=0

ρ3dρ

!dφdθ

= 2π

Z φ=π/3

φ=0

cos (φ) sin (φ)

Ãρ4

4

¯4 cos(φ)0

!dφ

= 2π

Z φ=π/3

φ=0

cos (φ) sin (φ)¡64 cos4 (θ)

¢dφ

= 128π

Z φ=π/3

φ=0

cos5 (φ) sin (φ) dφ

= 128π

Ã−16cos6 (φ)

¯π/30

!

=128π

6

Ã1−

µ1

2

¶6!= 21π

29.

022

1z

2

xy

00

-2-2


The sphere is the graph of ρ = 2. We have

z =px2 + y2 ⇒ ρ cos (φ) =

qρ2 sin2 (φ) cos2 (θ) + ρ2 sin2 (φ) sin2 (θ)

⇒ ρ cos (φ) = ρ sin (φ)⇒ cos (φ) = sin (φ) ,

so that the cone is the graph of φ = π/4. Therefore,Z Z Zdxdydz =

Z θ=2π

θ=0

Z φ=π/2

φ=π/4

Z ρ=2

ρ=0

ρ2 sin (φ) dφdθ

=

ÃZ θ=2π

θ=0

dθ

!ÃZ φ=π/2

φ=π/4

sin (φ) dφ

!µZ ρ=2

ρ=0

ρ2dρ

¶

= (2π)³− cos (θ)|π/2π/4

´Ã 13ρ3¯20

!

= 2π

Ã√2

2

!µ8

3

¶=8√2

3π

Chapter 14

Vector Analysis

14.1 Vector Fields, Divergence and Curl

1.

a)

∇ · F (x, y, z) = ∂

∂x(xyz)− ∂

∂z

¡x2y

¢= yz

b)

∇×F (x, y, z) =

¯¯ i j k

∂x ∂y ∂zxyz 0 −x2y

¯¯

=

¯∂y ∂z0 −x2y

¯i−

¯∂x ∂zxyz −x2y

¯j+

¯∂x ∂yxyz 0

¯k

=

µ∂

∂y

¡−x2y¢¶ i−µ ∂

∂x

¡−x2y¢− ∂

∂z(xyz)

¶j+

µ− ∂

∂y(xyz)

¶k

= −x2i− (−2xy − xy) i− xzk = −x2i+ 3xyi− xzk2.

a)

∇ · F (x, y, z) =∂

∂x

¡e−xxy2

¢+

∂

∂y

¡e−xx2y

¢=

¡−e−xxy2 + e−xy2¢+ e−xx2 = e−x ¡−xy2 + y2 + x2¢b)

∇×F (x, y, z) =

¯¯ i j k

∂x ∂y ∂ze−xxy2 e−xx2y 0

¯¯

=

¯∂y ∂z

e−xx2y 0

¯i−

¯∂x ∂z

e−xxy2 0

¯j+

¯∂x ∂y

e−xxy2 e−xx2y

¯k

=∂

∂x

¡e−xx2y

¢− ∂

∂y

¡e−xxy2

¢= −e−xx2y + 2xe−xy − 2ye−xx = −e−xx2y

3.

107

108 CHAPTER 14. VECTOR ANALYSIS

a)

∇ · F (x, y, z) = ∂

∂ycos (xz)− ∂

∂zsin (xy) = 0.

b)

∇×F (x, y, z) =

¯¯ i j k

∂x ∂y ∂z0 cos (xz) − sin (xy)

¯¯

=

¯∂y ∂z

cos (xz) − sin (xy)¯i−

¯∂x ∂z0 − sin (xy)

¯j+

¯∂x ∂y0 cos (xz)

¯k

=

µ− ∂

∂ysin (xy)− ∂

∂zcos (xz)

¶i+

∂

∂xsin (xy) j+

∂

∂xcos (xz)k

= (−x cos (xy) + x sin (xz)) i+ y cos (xy) j− z sin (xz)k

4.

a)

∇ · F (x, y, z) =∂

∂xarctan

³yx

´+

∂

∂yarctan

µx

y

¶

=

⎛⎜⎝ 1

1 +³yx

´2⎞⎟⎠³− y

x2

´+

⎛⎜⎜⎜⎝ 1

1 +

µx

y

¶2⎞⎟⎟⎟⎠µ− xy2

¶

=

µx2

x2 + y2

¶³− yx2

´+

µy2

x2 + y2

¶µ− xy2

¶= − y

x2 + y2− x

x2 + y2= − x+ y

x2 + y2

b)

∇×F (x, y, z)

=

¯¯ i j k

∂x ∂y ∂zarctan (y/x) arctan (x/y) 0

¯¯

=

¯∂y ∂z

arctan (x/y) 0

¯i−

¯∂x ∂z

arctan (y/x) 0

¯i+

¯∂x ∂y

arctan (y/x) arctan (x/y)

¯k

=

µ∂

∂xarctan

µx

y

¶− ∂

∂yarctan

³yx

´¶k

=

⎛⎜⎜⎜⎝⎛⎜⎜⎜⎝ 1

1 +

µx

y

¶2⎞⎟⎟⎟⎠µ1

y

¶+

⎛⎜⎝ 1

1 +³yx

´2⎞⎟⎠µ 1

x

¶⎞⎟⎟⎟⎠k=

µµy2

x2 + y2

¶µ1

y

¶+

µx2

x2 + y2

¶µ1

x

¶¶k

=

µx+ y

x2 + y2

¶k

5.

14.2. LINE INTEGRALS 109

a)

∇ · F (x, y, z) = ∂

∂x(2xy) +

∂

∂y

¡x2 + 2yz

¢+

∂

∂zy2 = 2y + 2z

b)

∇×F (x, y, z) =

¯¯ i j k

∂x ∂y ∂z2xy x2 + 2yz y2

¯¯

=

¯∂y ∂z

x2 + 2yz y2

¯i−

¯∂x ∂z2xy y2

¯j+

¯∂x ∂y2xy x2 + 2yz

¯k

= (2y − 2y) i− (0) j+ (2x− 2x)k = 06.

a)

∇ · F (x, y, z) =∂

∂xln (x) +

∂

∂yln (xy) +

∂

∂zln (xyz)

=1

x+1

y+1

z

b)

×F (x, y, z) =

¯¯ i j k

∂x ∂y ∂zln (x) ln (xy) ln (xyz)

¯¯

=

¯∂y ∂z

ln (xy) ln (xyz)

¯i−

¯∂x ∂zln (x) ln (xyz)

¯j+

¯∂x ∂yln (x) ln (xy)

¯k

=1

yi− 1

xj+

1

xk

14.2 Line Integrals

1. We have σ0 (t) =¡3t2, 1

¢so that

ds = ||σ0 (t)|| dt =p9t4 + 1dt

Therefore, ZC

y3ds =

Z 2

0

t3p9t4 + 1dt.

We set u = 9t4 + 1 so that du = 36t3dt. Thus,Z 2

0

t3p9t4 + 1dt =

1

36

Z 145

1

u1/2du

=1

36

Ã2

3u3/2

¯1451

!=1

54

³1453/2 − 1

´2. We have

σ0 (t) = (−2 sin (t) , 2 cos (t))so that

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


Therefore, ZC

xy2ds =

Z π/2

−π/2(2 cos (t)) (2 sin (t))

2(2) dt = 16

Z π/2

−π/2sin2 (t) cos (t) dt.

We set u = sin (t) so that du = cos (t) dt. Thus,

16

Z π/2

−π/2sin2 (t) cos (t) dt = 16

Z 1

−1u2du

= 16

Ã1

3u3¯1−1

!=16

3(1 + 1) =

32

3

3. We can paramterize C by

σ (t) = (2, 1) + t (4− 2, 5− 1) = (2, 1) + t (2, 4) = (2 + 2t, 1 + 4t) ,where 0 ≤ t ≤ 1. Thus,

σ0 (t) = (2, 4)⇒ ||σ0 (t)|| = √4 + 16 =√20 = 2

√5

Therefore, ZC

xeyds =

Z 1

0

(2 + 2t) e1+4tdt = 2e

Z 1

0

e4tdt+ 2e

Z 1

0

te4tdt.

We have

2e

Z 1

0

e4tdt = 2e

Ã1

4e4t¯10

!=e

2

¡e4 − 1¢ = e5

2− e2

We evaluate the other integral by setting u = t and dv = e4tdt. Thus

du = dt and v =1

4e4t.

Therefore, Zte4tdt =

Zudv = uv −

Zvdu

= t

µ1

4e4t¶− 14

Ze4tdt

=t

4e4t − 1

16e4t.

Thus,

2e

Z 1

0

te4tdt = 2e

Ãt

4e4t − 1

16e4t¯10

!

= 2e

µ1

4e4 − 1

16e4 +

1

16

¶=1

8e+

3

8e5

Therefore, ZC

xeyds = 2e

Z 1

0

e4tdt+ 2e

Z 1

0

te4tdt

=e5

2− e2+1

8e+

3

8e5 =

7

8e5 − 3

8e


4. We can parametrize C by

σ (θ) = (4, 3) + 2 (cos (θ) , sin (θ))

= (4 + 2 cos (θ) , 3 + 2 sin (θ)) , 0 ≤ θ ≤ π.

Thus,

σ0 (θ) = 2 (− sin θ) , cos (θ) ,so that

ds = ||σ0 (θ)|| = 2.Therefore, Z

C

y − 3(x− 4)2 + (y − 3)2 ds =

Z π

0

2 sin (θ)

4 cos2 (θ) + 4 sin2 (θ)(2) dθ

=

Z π

0

sin (θ) dθ = − cos (θ)|π0 = 2.


σ (t) = (2, 3, 4) + t (1, 2, 3) = (2 + t, 3 + 2t, 4 + 3t) , 0 ≤ t ≤ 1.Thus,

ds = ||σ0 (t)|| dt = ||(1, 2, 3)|| dt =√14dt.

Therefore, ZC

(x− 2) e(y−3)(z−4)ds =Z 1

0

te(2t)(3t)√14dt =

√14

Z 1

0

te6t2

dt

=√14

Ã1

12e6t

2

¯10

!

=

√14

12

¡e6 − 1¢ .


σ (t) = (1, 2) + t (3− 1, 4− 2) = (1, 2) + t (2, 2) = (1 + 2t, 2 + 2t) ,where 0 ≤ t ≤ 1. Thus,

σ0 (t) = 2i+ 2j,

and

F (σ (t)) = F (1 + 2t, 2 + 2t) = (1 + 2t)2 i+ (2 + 2t)2 j

Therefore,

F (σ (t)) · σ0 (t) =³(1 + 2t)2 i+ (2 + 2t)2 j

´· (2i+ 2j)

= 2 (1 + 2t)2+ 2 (2 + 2t)

2

= 16t2 + 24t+ 10.

Thus, ZC

F·d σ =Z 1

0

F (σ (t)) · σ0 (t) dt =

Z 1

0

¡16t2 + 24t+ 10

¢dt

=16

3t3 + 12t2 + 10t

¯10

=16

3+ 12 + 10 =

82

3



σ (θ) = (cos (θ) , sin (θ)) ,π

2≤ θ ≤ π.

Thus,

σ0 (θ) = − sin (θ) i+ cos (θ) jand

F (σ (θ)) = F (cos (θ) , sin (θ)) = sin (θ) i− cos (θ) jTherefore,

F (σ (θ)) · σ0 (θ) = (sin (θ) i− cos (θ) j) · (− sin (θ) i+ cos (θ) j)= − sin2 (θ)− cos2 (θ) = −1.

Thus ZC

F·dσ =

Z π

π/2

F (σ (θ)) · σ0 (θ) dθ

=

Z π

π/2

(−1) dθ = −³π − π

2

´= −π

2.

8. We have

σ0 (t) =1

ti+ 3t2j

and

F (σ (t)) = F¡ln (t) , t3

¢= ln

¡t3¢i− eln(t)j = 3 ln (t) i− tj

Therefore,

F (σ (t)) · σ0 (t) = (3 ln (t) i− tj) ·µ1

ti+ 3t2j

¶= 3

ln (t)

t− 3t3

Thus, ZC

F·d σ =Z e

1

F (σ (t)) · σ0 (t) dt =

Z e

1

µ3ln (t)

t− 3t3

¶dt

= 3ln2 (t)

2− 34t4¯e1

=3

2− 34e4 +

3

4=9

4− 34e4

9. We have

σ0 (t) = − sin (t) i+ j+ kand

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

Therefore,

F (σ (t)) · σ0 (t) = (cos (t) i+ sin (t) j+ cos (t)k) · (− sin (t) i+ j+ k)= − cos (t) sin (t) + sin (t) + cos (t)


Thus ZC

F·dσ =

Z π/2

π/4

F (σ (t)) · σ0 (t) dt

=

Z π/2

π/4

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

= −12sin2 (t)− cos (t) + sin (t)

¯π/2π/4

= −12+ 1 +

1

2

µ1√2

¶2+

√2

2−√2

2=3

4

10. We have ZC

F·d σ =ZC1+C2

F·d σ =ZC1

F·dσ +ZC2

F·dσ.

We can parametrize C1 byσ1 (x) = (x, 0) , 0 ≤ x ≤ 2

and C2 byσ2 (t) = (2, 0) + t (3− 2, 2− 0) = (2, 0) + t (1, 2) = (2 + t, 2t) ,

where 0 ≤ t ≤ 1.On C1,

σ01 (x) = i

and

F (σ1 (x)) = xj

Thus, ZC1

F·d σ =Z 2

0

F (σ1 (x)) · σ01 (x) dx =Z 2

0

(xj) · idx = 0

On C2,σ02 (t) = i+ 2j

and

F (σ2 (t)) = F (2 + t, 2t) = (2 + t) 2ti+ (2 + t− 2t) j=

¡4t+ 2t2

¢i+ (2− t) j

Thus, ZC2

F·d σ =Z 1

0

F (σ2 (t)) · σ02 (t) dt =

Z 1

0

¡¡4t+ 2t2

¢i+ (2− t) j¢ · (i+ 2j) dt

=

Z 1

0

¡4t+ 2t2 + 4− 2t¢ dt

=

Z 1

0

¡2t+ 2t2 + 4

¢dt

= t2 +2

3t3 + 4t

¯10

=17

3

Therefore, ZC

F·d σ =ZC1

F·dσ +ZC2

F·dσ = 0 + 173=17

3.


11. We have ZC

F ·Tds = −ZC2

F ·Tds,

where C2is the part of the circle of radius 2 centered at the origin traversed from (0, 2) to (−2, 0).We can parametrize C2 by

σ (θ) = (2 cos (θ) , 2 sin (θ)) ,π

2≤ θ ≤ π.

Thus,

σ0 (θ) = −2 sin (θ) i+ 2 cos (θ) j,so that ||σ0 (θ)|| = 2,

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

and

ds = ||σ0 (θ)|| dθ =q4 sin2 (θ) + 4 cos2 (θ)dθ = 2dθ.

We have

F (σ (θ)) = −2 (2 cos (θ)) (2 sin (θ)) i+ (2 sin (θ) + 1) j= −8 cos (θ) sin (θ) i+ (2 sin (θ) + 1) j

Thus,

(F ·T) (θ) = (−8 cos (θ) sin (θ) i+ (2 sin (θ) + 1) j) · (− sin (θ) i+ cos (θ) j)= 8 cos (θ) sin2 (θ) + 2 sin (θ) cos (θ) + cos (θ) .

Therefore,

−ZC2

F ·Tds = −Z π

θ=π/2

¡8 cos (θ) sin2 (θ) + 2 sin (θ) cos (θ) + cos (θ)

¢2dθ

= −Z π

θ=π/2

¡16 cos (θ) sin2 (θ) + 4 sin (θ) cos (θ) + 2 cos (θ)

¢dθ

= −Ã16

3sin3 (θ) + 2 sin2 (θ) + 2 sin (θ)

¯ππ/2

!=

16

3+ 2 + 2 =

28

3


σ (t) = (1, 2, 3) + t (−1− 1,−2− 2,−3− 3)= (1, 2, 3)− t (2, 4, 6) = (1− 2t, 2− 4t, 3− 6t) ,

where 0 ≤ t ≤ 1.Thus,

σ0 (t) = −2i− 4j− 6kso that

||σ0 (t)|| = √4 + 16 + 36 = 2√14

Thus,

T (t) =σ0 (t)||σ0 (t)|| =

1

2√14(−2i− 4j− 6k)


and

ds = ||σ0 (t)|| dt = 2√14dt.

We have

F (σ (t)) = F (1− 2t, 2− 4t, 3− 6t)= e1−2t+2−4ti+ (1− 2t) (3− 6t) j+ (2− 4t)k= e3−6ti+

¡12t2 − 12t+ 3¢ j+ (2− 4t)k

Therefore,

F (σ (t)) ·T (t)=

¡e3−6ti+

¡12t2 − 12t+ 3¢ j+ (2− 4t)k¢ ·µ 1

2√14(−2i− 4j− 6k)

¶=

1

2√14

¡−2e3−6t − 4 ¡12t2 − 12t+ 3¢− 6 (2− 4t)¢=

1

2√14

¡−2e3−6t − 48t2 + 72t− 24¢Thus, Z

C

F ·Tds =

Z 1

0

F (σ (t)) ·T (t) ds

=

Z 1

0

1

2√14

¡−2e3−6t − 48t2 + 72t− 24¢ 2√14dt=

Z 1

0

¡−2e3−6t − 48t2 + 72t− 24¢ dt=

1

3e3e−6t − 16t3 + 36t2 − 24t

¯10

=1

3e3e−6 − 16 + 36− 24− 1

3e3 =

1

3e−3 − 1

3e3 − 4

13.

a) ZC

F·d σ =ZC

−xydx+ 1

x2 + 1dy.

b) ZC

−xydx+ 1

x2 + 1dy =

Z −1−4

µ−xydx

dt+

1

x2 + 1

dy

dt

¶dt

=

Z −1−4

µ−t ¡t2¢ d

dt(t) +

1

t2 + 1

d

dt

¡t2¢¶dt

=

Z −1−4

µ−t3 + 2t

t2 + 1

¶dt

= −14t4 + ln

¡t2 + 1

¢¯−1−4

= −14+ ln (2) + 43 − ln (17) = ln

µ2

17

¶+255

4

14.


a) ZC

F·d σ =ZC

ydx− xdy

b) We can parametrize C by

σ (θ) = (cos (θ) , sin (θ)) , −π2≤ θ ≤ π

2.

Thus, ZC

ydx− xdy =

Z π/2

−π/2

µy (θ)

dx

dθ− x (θ) dy

dθ

¶dθ

=

Z π/2

−π/2(sin (θ) (− sin (θ))− cos (θ) cos (θ)) dθ

= −Z π/2

π/2

¡sin2 (θ) + cos2 (θ)

¢dθ

= −Z π/2

π/2

dθ = −³θ|π/2−π/2

´= −π


σ (θ) = (cos (θ) , sin (θ)) , 0 ≤ θ ≤ π

2.

Thus, ZC

3x2dx− 2y3dy =

Z π/2

0

¡3 cos2 (θ) (− sin (θ))− 2 sin3 (θ) cos (θ)¢ dθ

= −Z π/2

0

¡3 cos2 (θ) sin (θ) + 2 sin3 (θ) cos (θ)

¢dθ

= −Ã− cos3 (θ) + 1

2sin4 (θ)

¯π/20

!

= −µ1

2+ 1

¶= −3

2

16. The equation of the ellipse can be written as

x2

4+ y2 = 1.

This suggests that we set

σ (θ) = (2 cos (θ) , sin (θ)) .

Indeed(2 cos (θ))2

4+ sin2 (θ) = cos2 (θ) + sin2 (θ) = 1.

Thus, −C is parametrized by

σ (θ) = (2 cos (θ) , sin (θ)) , 0 ≤ θ ≤ π

2

(since C is from (0, 1) to (2, 0).


Therefore, ZC

exdx+ eydy = −Z π/2

0

³e2 cos(θ) (−2 sin (θ)) + esin(θ) cos (θ)

´dθ

=

Z π/2

0

e2 cos(θ)2 sin (θ) dθ −Z π/2

0

esin(θ) cos (θ) dθ

=

µ−e2 cos(θ)

¯π/20

¶−µesin(θ)

¯π/20

¶=

¡−1 + e2¢− (e− 1) = e2 − e17. We can parametrize C by

σ (x) =¡x, x2

¢, 0 ≤ x ≤ 2.

Thus, ZC

− sin (x) dx+ cos (x) dy =

Z π

0

(− sin (x) + cos (x) (2x)) dx

= −Z π

0

sin (x) dx+ 2

Z π

0

x cos (x) dx

= cos (x)|π0 + 2Z π

0

x cos (x) dx

= −2 + 2Z π

0

x cos (x) dx

In order to evaluate the last integral, we set u = x and dv = cos (x) dx. Thus,

du = dx and v =

Zcos (x) dx = sin (x) .

Therefore, Z π

0

x cos (x) dx = x sin (x)|π0 −Z π

0

sin (x) dx = cos (x)|π0 = −2.

Thus, ZC

− sin (x) dx+ cos (x) dy = −2 + 2Z π

0

x cos (x) dx

= −2 + 2 (−2) = −6.


σ (t) = t (2, 3, 4) = (2t, 3t, 4t) , 0 ≤ t ≤ 1.

Thus, ZC

x3dx+ y2dy + zdz =

Z 1

0

³(2t)

3(2) + (3t)

2(3) + (4t) (4)

´dt

=

Z 1

0

¡16t3 + 27t2 + 16t

¢dt

= 4t4 + 9t3 + 8t2¯10= 4 + 9 + 8 = 21.


14.3 Line Integrals of Conservative Vector Fields

1.

a)∂

∂y(2x− 3y) = −3 and ∂

∂x(−3x+ 4y − 8) = −3.

Since the above partial derivatives are equal F may be conservative.

b) We need to have

∂f

∂x(x, y) = 2x− 3y,

∂f

∂y(x, y) = −3x+ 4y − 8

From the first equation,

f (x, y) =

Z(2x− 3y) dx = x2 − 3yx+ g (y) .

Therefore,

−3x+ 4y − 8 = ∂f

∂y(x, y) = −3x+ dg (y)

dy

Thus,dg (y)

dy= 4y − 8⇒ g (y) =

Z(4y − 8) dy = 2y2 − 8y +K,

where K is an arbitrary constant.

Therefore,

f (x, y) = x2 − 3yx+ g (y) = x2 − 3yx+ 2y2 − 8y +Kis a potential for F (you can check that ∇f = F).2.

a) We have∂

∂y(ex cos (y)) = −ex sin (y)

and∂

∂x(ex sin (y)) = ex sin (y) .

We have

−ex sin (y) = ex sin (y)⇔ sin (y) = 0⇔ y = nπ, n = 0,±1,±2, . . .Since the necessary conditions for the existence of a potential function are satisfied only at

isolated points, F is not conservative.

3.

a) We have∂

∂y(ex sin (y)) = ex cos (y)

and∂

∂x(ex cos (y)) = ex cos (y) .

Since the abve partial derivatives are equal everywhere, F may be conservative.

14.3. LINE INTEGRALS OF CONSERVATIVE VECTOR FIELDS 119

b) We need to have

∂f

∂x(x, y) = ex sin (y) ,

∂f

∂y(x, y) = ex cos (y) .


f (x, y) =

Zex sin (y) dx = ex sin (y) + g (y)

⇒ex cos (y) =

∂f

∂y(x, y) = ex cos (y) +

dg (y)

dy⇒

dg (y)

dy= 0⇒ g (y) = K,

where K is an arbitrary constant. Thus,

f (x, y) = ex sin (y) +K

is a is a potential for F (confirm).

4.

a)

∇×F (x, y, z) =

¯¯ i j k

∂x ∂y ∂z− cos (z) e−x−y − cos (z) e−x−y − sin (z) e−x−y

¯¯

=

¯∂y ∂z

− cos (z) e−x−y − sin (z) e−x−y¯i

−¯

∂x ∂z− cos (z) e−x−y − sin (z) e−x−y

¯j

+

¯∂x ∂y

− cos (z) e−x−y − cos (z) e−x−y¯k

=¡−∂y sin (z) e−x−y + ∂z cos (z) e

−x−y¢ i− ¡−∂x sin (z) e−x−y + ∂z cos (z) e

−x−y¢ j+¡−∂x cos (z) e−x−y + ∂y cos (z) e

−x−y¢k=

¡sin (z) e−x−y − sin (z) e−x−y¢ i− ¡sin (z) e−x−y − sin (z) e−x−y¢ j+¡cos (z) e−x−y − cos (z) e−x−y¢k

= 0

Thus, F may be conservative.

b) We need to have

∂f

∂x(x, y, z) = − cos (z) e−x−y,

∂f

∂y(x, y, z) = − cos (z) e−x−y,

∂f

∂z(x, y, z) = − sin (z) e−x−y.



f (x, y, z) = −Zcos (z) e−x−ydx = − cos (z) e−y

Ze−xdx

= cos (z) e−ye−x + g (y, z)= cos (z) e−x−y + g (y, z) .

From the second equation,

− cos (z) e−x−y = ∂f

∂y(x, y, z) = − cos (z) e−x−y + ∂g

∂y(y, z)

⇒∂g

∂y(y, z) = 0

⇒g (y, z) = h (z) .

Therefore,

f (x, y, z) = cos (z) e−x−y + h (z) .

From the third equuation,

− sin (z) e−x−y = ∂f

∂z(x, y, z) = − sin (z) e−x−y + dh (z)

dz

⇒dh (z)

dz= 0⇒ h (z) = K,

where K is an arbitrary constant. Thus,

f (x, y, z) = cos (z) e−x−y

is a is a potential for F (confirm).

5.

a) We need to have∂f

∂x(x, y) = xy2 and

∂f

∂y(x, y) = x2y.


f (x, y) =

Zxy2dx =

1

2x2y2 + g (y) .

Therefore,∂f

∂y(x, y) = x2y +

dg (y)

dy.


x2y =∂f

∂y(x, y) = x2y +

dg (y)

dy

⇒dg (y)

dy= 0⇒ g (y) = K,

where K is an arbitrary constant. We can set K = 0. Thus,

f (x, y) =1

2x2y2


is a potential for F (confirm by differentiation).

b) ZC

F · dσ = f (2, 1)− f (0, 1) = 2− 0 = 2.

6.

a) We need to have∂f

∂x(x, y) = yexy − 1 and ∂f

∂y(x, y) = xexy


f (x, y) =

Z(yexy − 1) dx = exy − x+ g (y)

Therefore,∂f

∂y(x, y) = xexy +

dg (y)

dy


xexy =∂f

∂y(x, y) = xexy +

dg (y)

dy

⇒dg (y)

dy= 0⇒ g (y) = K.

We can set K = 0. Thus,f (x, y) = exy − x

is a potential for F.

b) ZC

F · dσ = f (4, ln (2))− f (0, 1)

=³e4 ln(2) − 4

´− 1 = 24 − 5 = 11

7.

a) We need to have

∂f

∂x(x, y, z) = yz,

∂f

∂y(x, y, z) = xz,

∂f

∂z(x, y, z) = xy + 2z.


f (x, y, z) =

Zyzdx = xyz + g (y, z) .

Thus,∂f

∂y(x, y, z) = xz +

∂g

∂y(y, z) .


xz =∂f

∂y(x, y, z) = xz +

∂g

∂y(y, z)

⇒∂g

∂y(y, z) = 0⇒ g (y, z) = h (z) .


Thus,

f (x, y, z) = xyz + h (z) .

Therefore,∂f

∂z(x, y, z) = xy +

dh (z)

dz.

From the third equation,

xy + 2z =∂f

∂z(x, y, z) = xy +

dh (z)

dz

⇒dh (z)

dz= 2z ⇒ h (z) = z2 +K.

Thus, we can set

f (x, y, z) = xyz + z2.

b) ZC

F · dσ = f (4, 6, 3)− f (1, 0,−2)=

¡(4) (6) (3) + 32

¢− (4) = 778. We have

df =∂f

∂x(x, y) dx+

∂f

∂y(x, y) dy =

y

x2 + y2dx− x

x2 + y2dy

⇔∂f

∂x(x, y) =

y

x2 + y2and

∂f

∂y(x, y) = − x

x2 + y2.


f (x, y) =

Zy

x2 + y2dx =

Z1

(x/y)2 + 1

µ1

y

¶dx

We set u = x/y so that du = dx/y. Thus,

f (x, y) =

Z1

(x/y)2 + 1

µ1

y

¶dx =

Z1

u2 + 1du

= arctan (u) + g (y) = arctan (x/y) + g (y) .

Therefore,∂f

∂y(x, y) =

1

1 +x2

y2

µ− xy2

¶+dg (y)

dy= − x

x2 + y2+dg (y)

dy.


− x

x2 + y2=

∂f

∂y(x, y) = − x

x2 + y2+dg (y)

dy

⇒dg (y)

dy= 0⇒ g (y) = K.

We can set K = 0 and set

f (x, y) = arctan

µx

y

¶.


Since we have y > 0 on C and C joins (0, 1) to (1, 1) we haveZC

y

x2 + y2dx− x

x2 + y2dy =

ZC

d arctan

µx

y

¶= arctan (1)− arctan (0) = π

4.

9. We have

df =∂f

∂x(x, y) dx+

∂f

∂y(x, y) dy =

y2

1 + x2dx+ 2y arctan (x) dy

⇔∂f

∂x(x, y) =

y2

1 + x2and

∂f

∂y(x, y) = 2y arctan (x) .


f (x, y) =

Zy2

1 + x2dx = y2 arctan (x) + g (y) .

Therefore,∂f

∂y(x, y) = 2y arctan (x) +

dg (y)

dy.


2y arctan (x) =∂f

∂y(x, y) = 2y arctan (x) +

dg (y)

dy

⇒dg (y)

dy= 0⇒ g (y) = K.

Thus, we can set K = 0 andf (x, y) = y2 arctan (x) .

Therefore, ZC

y2

1 + x2dx+ 2y arctan (x) dy =

ZC

df

= f (1, 2)− f (0, 0)= 4 arctan (1) = 4

³π4

´= π.

10. We have

df =∂f

∂xdx+

∂f

∂ydy +

∂f

∂z= y2 cos (z) dx+ 2xy cos (z) dy − xy2 sin (z) dz

⇔∂f

∂x= y2 cos (z) ,

∂f

∂y= 2xy cos (z) ,

∂f

∂z= −xy2 sin (z)

From the first equuation,

f (x, y, z) =

Zy2 cos (z) dx = xy2 cos (z) + g (y, z) .

Therefore,∂f

∂y(x, y, z) = 2xy cos (z) +

∂g (y, z)

∂y.



2xy cos (z) =∂f

∂y(x, y, z) = 2xy cos (z) +

∂g (y, z)

∂y

⇒∂g (y, z)

∂y= 0⇒ g (y, z) = h (z) .

Thus,

f (x, y, z) = xy2 cos (z) + h (z) .

Therefore,∂f

∂z(x, y, z) = −xy2 sin (z) + dh (z)

dz.

From the third equation,

−xy2 sin (z) = −xy2 sin (z) + dh (z)dz

⇒ dh (z)

dz= 0.

Therefore, we can set h (z) = 0 and

f (x, y, z) = xy2 cos (z)

Then ZC

y2 cos (z) dx+ 2xy cos (z) dy − xy2 sin (z) dz =

ZC

df

= f (1, 1,π)− f (0, 0, 0)= cos (π) = −1.

14.4 Parametrized Surfaces and Tangent Planes

1.

a) We have

∂Φ

∂u(u, v) = −3 sin (u) j+ 3 cos (u)k,

∂Φ

∂v(u, v) = i

Therefore,

∂Φ

∂u

³π6, 1´= −3 sin

³π6

´j+ 3 cos

³π6

´k = −3

2j+

3√3

2k,

and∂Φ

∂v

³π6, 1´= i.

Thus,

N³π6, 1´

=∂Φ

∂u

³π6, 1´× ∂Φ

∂v

³π6, 1´

=

Ã−32j+

3√3

2k

!× i

=

¯¯ i j k

0 −32 3√3

21 0 0

¯¯

=

¯−32 3

√32

0 0

¯i−

¯0 3

√32

1 0

¯j+

¯0 −321 0

¯k =

3√3

2j+

3

2k

14.4. PARAMETRIZED SURFACES AND TANGENT PLANES 125

b) We have

P0 = Φ³π6, 1´=³1, 3 cos

³π6

´, 3 sin

³π6

´´=

Ã1,3√3

2,3

2

!.

Therefore, the required tangent plane is the set of points P = (x, y, z) such that

N³π6, 1´·−−→P0P = 0

⇔ Ã3√3

2j+

3

2k

!·Ã(x− 1) i+

Ãy − 3

√3

2

!j+

µz − 3

2

¶k

!= 0

⇔3√3

2

Ãy − 3

√3

2

!+3

2

µz − 3

2

¶= 0

2.

a) We have∂Φ

∂u(u, v) = 2ui+ cos (v) j+ sin (v)k

and∂Φ

∂v(u, v) = −u sin (v) j+ u cos (v)k.

Therefore,

∂Φ

∂u

³3,π

3

´= 6i+

1

2j+

√3

2k

and

∂Φ

∂v

³3,π

3

´= −3

Ã√3

2

!j+ 3

µ1

2

¶k = −3

√3

2j+

3

2k

Thus,

N³3,π

3

´=

∂Φ

∂u

³3,π

3

´× ∂Φ

∂v

³3,π

3

´=

Ã6i+

1

2j+

√3

2k

!×Ã−3√3

2j+

3

2k

!

=

¯¯ i j k

6 12

√32

0 −3√32

32

¯¯

=

¯¯ 1

2

√32

−3√32

32

¯¯ i−

¯6

√32

0 32

¯j+

¯6 1

2

0 −3√3

2

¯k

= 3i−9j− 9√3k.

b) We have

P0 = Φ³3,π

3

´=³9, 3 cos

³π3

´, 3 sin

³π3

´´=

Ã9,3

2,3√3

2

!Therefore, the required tangent plane is the set of points P = (x, y, z) such that

N³3,π

3

´·−−→P0P = 0


⇐⇒ ³3i−9j− 9

√3k´·Ã(x− 9) i+

µy − 3

2

¶j+

Ãz − 3

√3

2

!k

!= 0

⇐⇒3 (x− 8)− 9

µy − 3

2

¶− 9√3

Ãz − 3

√3

2

!= 0.

3.

a) We have∂Φ

∂r(r, θ) = 3 cos (θ) i+ 2rj+ 2 sin (θ)k

and∂Φ

∂θ(r, θ) = −3r sin (θ) i+ 2r cos (θ)k.

Therefore,

∂Φ

∂r

³2,π

4

´=3√2

2i+ 4j+

√2k

and∂Φ

∂θ

³2,π

4

´= −3

√2i+ 2

√2k.

Thus,

N³2,π

4

´=

∂Φ

∂r

³2,π

4

´× ∂Φ

∂θ

³2,π

4

´=

Ã3√2

2i+ 4j+

√2k

!×³−3√2i+ 2

√2k´

=

¯¯ i j k3√22 4

√2

−3 0 2√2

¯¯

=

¯4√2

0 2√2

¯i−

¯¯ 3√22

√2

−3 2√2

¯¯ j+

¯3√2

2 4−3 0

¯k

= 8√2i−

³3√2 + 6

´j+ 12k

b) We have

Φ³2,π

4

´=³3√2, 4, 2

√2´

Therefore, he required tangent plane is the set of points (x, y, z) such that³8√2i−

³3√2 + 6

´j+ 12k

´·³³x− 3

√2´i+ (y − 4) j+

³z − 2

√2´k´= 0

⇔8√2³x− 3

√2´−³3√2 + 6

´(y − 4) + 12

³z − 2

√2´= 0.

4.

a) We have∂Φ

∂u(u, v) = sinh (u) cos (v) i+ cosh (u) j+ sinh (u) sin (v)k

and∂Φ

∂v(u, v) = − cosh (u) sin (v) i+ cosh (u) cos (v)k.


Therefore,∂Φ

∂u

³1,π

2

´= cosh (1) j+ sinh (1)k

and∂Φ

∂v

³1,π

2

´= − cosh (1) i.

Thus,

N³1,π

2

´=

∂Φ

∂u

³1,π

2

´× ∂Φ

∂v

³1,π

2

´= (cosh (1) j+ sinh (1)k)× (− cosh (1) i)

=

¯¯ i j k

0 cosh (1) sinh (1)− cosh (1) 0 0

¯¯

=

¯cosh (1) sinh (1)0 0

¯i−

¯0 sinh (1)

− cosh (1) 0

¯j+

¯0 cosh (1)

− cosh (1) 0

¯k

= − sinh (1) cosh (1) j+cosh2 (1)k.b) We have

P0 = Φ³1,π

2

´= (0, sinh (1) , cosh (1)) .

Therefore, he required tangent plane is the set of points (x, y, z) such that¡− sinh (1) cosh (1) j+cosh2 (1)k¢ · (xi+ (y − sinh (1)) j+ (z − cosh (1))k) = 0⇔

− sinh (1) cosh (1) (y − sinh (1)) + cosh2 (1) (z − cosh (1)) = 0.5.

a) We have

∂Φ

∂ρ(ρ, θ) =

1

2cos (θ) i+

1

2sin (θ) j+

√3

2k

and∂Φ

∂θ(ρ, θ) = −1

2ρ sin (θ) i+

1

2ρ cos (θ) j.

Therefore,

∂Φ

∂ρ

³2,π

3

´=1

4i+

√3

4j+

√3

2k

and∂Φ

∂θ

³2,π

3

´= −√3

2i+1

2j

Thus,

N³2,π

3

´=

Ã1

4i+

√3

4j+

√3

2k

!×Ã−√3

2i+1

2j

!

=

¯¯ i j k

14

√34

√32

−√32

12 0

¯¯

=

¯ √34

√32

12 0

¯i−

¯¯ 1

4

√32

−√32 0

¯¯ j+

¯¯ 1

4

√34

−√32

12

¯¯k

= −√3

4i−34j+

1

2k


b) We have

Φ³2,π

3

´=

Ã1

2,

√3

2,√3

!Therefore, he required tangent plane is the set of points (x, y, z) such thatÃ

−√3

4i−34j+

1

2k

!·Ãµ

x− 12

¶i+

Ãy −√3

2

!j+

³z −√3´k

!= 0

⇐⇒−√3

4

µx− 1

2

¶− 34

Ãy −√3

2

!+1

2

³z −√3´= 0.

6. We have∂Φ

∂φ(φ, θ) = 2 cos (φ) cos (θ) i+ cos (φ) sin (θ) j− 1

3sin (φ)k

and∂Φ

∂θ(φ, θ) = −2 sin (φ) sin (θ) i+ sin (φ) cos (θ) j

Therefore,

∂Φ

∂φ

³π4,π

2

´=

√2

2j− 1

3k

and∂Φ

∂θ

³π4,π

2

´= −√2i

Thus,

N³π4,π

2

´=

Ã√2

2j− 1

3k

!×³−√2i´

=

¯¯ i j k

0√22 − 13

−√2 0 0

¯¯

=

¯ √22 −130 0

¯i−

¯0 −13−√2 0

¯j+

¯¯ 0

√22

−√2 0

¯¯k

=

√2

3j+ k

b) We have

Φ³π4,π

2

´=

Ã0,

√2

2, 0

!Therefore, he required tangent plane is the set of points (x, y, z) such thatÃ√

2

3j+ k

!·Ãxi+

Ãy −√2

2

!j+ zk

!= 0

⇐⇒ √2

3

Ãy −√2

2

!+ z = 0

7.


a) We have∂Ω

∂x(x, θ) = i+ ex sin (θ) j+ ex cos (θ)k

and∂Ω

∂θ(x, θ) = ex cos (θ) j− ex sin (θ)k.

Therefore,

∂Ω

∂x

³1,π

6

´= i+

1

2ej+

√3

2ek

and∂Ω

∂θ

³1,π

6

´=

√3

2ej− 1

2ek

Thus,

N³1,π

6

´=

Ãi+

1

2ej+

√3

2ek

!×Ã√

3

2ej− 1

2ek

!

=

¯¯ i j k

1 12e

√32 e

0√32 e −12e

¯¯

=

¯¯ 1

2e√32 e√

32 e −12e

¯¯ i−

¯1

√32 e

0 −12e¯j+

¯1 1

2e

0√32 e

¯k

= −e2i+ 12ej+

√3

2k

b) We have

Ω³1,π

6

´=

Ã1,1

2e,

√3

2e

!.

Therefore, he required tangent plane is the set of points (x, y, z) such thatÃ−e2i+ 1

2ej+

√3

2k

!·Ã(x− 1) i+

µy − 1

2e

¶j+

Ãz −√3

2e

!k

!= 0

⇐⇒−e2 (x− 1) + 1

2e

µy − 1

2e

¶+

√3

2

Ãz −√3

2e

!= 0.

8.

a) We have∂Ω

∂x(x, θ) = cos (θ) i+ sin (θ) j+ exk

and∂Ω

∂θ(x, θ) = −x sin (θ) i+ x cos (θ) j.

Therefore,∂Ω

∂x

³1,π

2

´= j+ ek

and∂Ω

∂θ

³1,π

2

´= −i


Thus,

N³1,π

2

´= (j+ ek)× (−i)

=

¯¯ i j k

0 1 e−1 0 0

¯¯

=

¯1 e0 0

¯i−

¯0 e−1 0

¯j+

¯0 1−1 0

¯k

= ej+ k

b) We have

Ω³1,π

2

´= (0, 1, e) .

Therefore, he required tangent plane is the set of points (x, y, z) such that

(ej+ k) · (xi+ (y − 1) j+ (z − e)k) = 0

⇐⇒e (y − 1) + (z − e) = 0

14.5 Surface Integrals

1 (a cylinder)

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

\We have∂Φ

∂u(u, v) = −3 sin (u) j+ 3 cos (u)k,

∂Φ

∂v(u, v) = i

Therefore,

N (u, v) =∂Φ

∂u(u, v)× ∂Φ

∂v(u, v)

= (−3 sin (u) j+ 3cos (u)k)× i

=

¯¯ i j k

0 −3 sin (u) 3 cos (u)1 0 0

¯¯

=

¯ −3 sin (u) 3 cos (u)0 0

¯i−

¯0 3 cos (u)1 0

¯j+

¯0 −3 sin (u)1 0

¯k

= −3 cos (u) j+ 3 sin (u)k

Thus,

||N (u, v)|| =q9 cos2 (u) + 9 sin2 (u) = 3.

Therefore the area of the surface isZ 2π

u=0

Z 4

v=−43dvdu = 3 (8) (2π) = 48π.

14.5. SURFACE INTEGRALS 131

2 (a paraboloid)

Φ (u, v) =¡u2, u cos (v) , u sin (v)

¢, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2π.

We have∂Φ

∂u(u, v) = 2ui+ cos (v) j+ sin (v)k

and∂Φ

∂v(u, v) = −u sin (v) j+ u cos (v)k.

Therefore,

N (u, v) =∂Φ

∂u(u, v)× ∂Φ

∂v(u, v)

= (2ui+ cos (v) j+ sin (v)k)× (−u sin (v) j+ u cos (v)k)

=

¯¯ i j k

2u cos (v) sin (v)0 −u sin (v) u cos (v)

¯¯

=

¯cos (v) sin (v)−u sin (v) u cos (v)

¯i−

¯2u sin (v)0 u cos (v)

¯j+

¯2u cos (v)0 −u sin (v)

¯k

=¡u cos2 (v) + u sin2 (v)

¢i−2u2 cos (v) j− 2u2 sin (v)k

= ui−2u2 cos (v) j− 2u2 sin (v)k.

Thus,

||N (u, v)|| =qu2 + 4u4 cos2 (v) + 4u4 sin2 (v) =

pu2 + 4u4

Therefore, the area of the surface isZ 3

u=0

Z 2π

v=0

pu2 + 4u4dudv =

Z 3

u=0

Z 2π

v=0

up1 + 4u2dudv

= 2π

Z 3

u=0

up1 + 4u2du.

We set w = 1 + 4u2 so that dw = 8udu. Thus,

2π

Z 3

u=0

up1 + 4u2du =

π

4

Z w=37

w=1

w1/2dw

=π

4

Ã2

3w3/2

¯371

!=

π

6

³373/2 − 1

´.

3 (a cone)

Φ (ρ, θ) =

Ã1

2ρ cos (θ) ,

1

2ρ sin (θ) ,

√3

2ρ

!where 0 ≤ ρ ≤ 4 and 0 ≤ θ ≤ 2π.We have

∂Φ

∂ρ(ρ, θ) =

1

2cos (θ) i+

1

2sin (θ) j+

√3

2k

and∂Φ

∂θ(ρ, θ) = −1

2ρ sin (θ) i+

1

2ρ cos (θ) j.


Therefore,

N (ρ, θ) =

Ã1

2cos (θ) i+

1

2sin (θ) j+

√3

2k

!×µ−12ρ sin (θ) i+

1

2ρ cos (θ) j

¶

=

¯¯ i j k

12 cos (θ)

12 sin (θ)

√32−12ρ sin (θ) 1

2ρ cos (θ) 0

¯¯

=

¯12 sin (θ)

√32

12ρ cos (θ) 0

¯i−

¯12 cos (θ)

√32−12ρ sin (θ) 0

¯j+

¯12 cos (θ)

12 sin (θ)−12ρ sin (θ) 12ρ cos (θ)

¯k

= −√3

4ρ cos (θ) i+

√3

4ρ sin (θ) j+

µ1

4ρ sin2 (θ) +

1

4ρ cos2 (θ)

¶k

= −√3

4ρ cos (θ) i+

√3

4ρ sin (θ) j+

1

4ρk.

Thus,

||N (ρ, θ)|| =r3

16ρ2 cos2 (θ) +

3

16ρ2 sin2 (θ) +

1

16ρ2 =

1

4

p4ρ2 =

1

2ρ.

Therefore, the area of the surface isZ θ=2π

θ=0

Z ρ=4

ρ=0

1

2ρdρdθ = 2π

µZ ρ=4

ρ=0

1

2ρdρ

¶= 2π

Ã1

4ρ2¯40

!= 8π.

4 (an ellipsoid)

Φ (φ, θ) =

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

1

3cos (φ)

¶,

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π.We have

∂Φ

∂φ(φ, θ) = 2 cos (φ) cos (θ) i+ cos (φ) sin (θ) j− 1

3sin (φ)k

and∂Φ

∂θ(φ, θ) = −2 sin (φ) sin (θ) i+ sin (φ) cos (θ) j

Therefore,

N (φ, θ) =

µ2 cos (φ) cos (θ) i+ cos (φ) sin (θ) j− 1

3sin (φ)k

¶× (−2 sin (φ) sin (θ) i+ sin (φ) cos (θ) j)

=

¯¯ i j k

2 cos (φ) cos (θ) cos (φ) sin (θ) −13 sin (φ)−2 sin (φ) sin (θ) sin (φ) cos (θ) j 0

¯¯

=

¯cos (φ) sin (θ) −13 sin (φ)sin (φ) cos (θ) j 0

¯i−

¯2 cos (φ) cos (θ) −13 sin (φ)−2 sin (φ) sin (θ) 0

¯j

+

¯2 cos (φ) cos (θ) cos (φ) sin (θ)−2 sin (φ) sin (θ) sin (φ) cos (θ) j

¯k

=1

3sin2 (φ) cos (θ) i− 2

3sin2 (φ) sin (θ) j+

¡2 sin (φ) cos (φ) cos2 (θ) + 2 sin (φ) cos (φ) sin2 (θ)

¢k

=1

3sin2 (φ) cos (θ) i− 2

3sin2 (φ) sin (θ) j+2 sin (φ) cos (φ)k.


Thus,

||N (φ, θ)|| =

r1

9sin4 (φ) cos2 (θ) +

4

9sin4 (φ) sin2 (θ) + 4 sin2 (φ) cos2 φ

=

ssin4 (φ)

µ1

9cos2 (θ) +

4

9sin2 (θ)

¶+ 4 sin2 (φ) cos2 φ

= sin (φ)

ssin2 (φ)

µ1

9cos2 (θ) +

4

9sin2 (θ)

¶+ 4cos2 (φ)

Therefore, the area of the surface is

Z θ=2π

θ=0

Z φ=π

φ=0

sin (φ)

ssin2 (φ)

µ1

9cos2 (θ) +

4

9sin2 (θ)

¶+ 4 cos2 (φ)dφdθ

5 (a surface of revolution)

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

where 0 ≤ x ≤ 2, 0 ≤ θ ≤ 2π.We have

∂Ω

∂x(x, θ) = i+ ex sin (θ) j+ ex cos (θ)k

and∂Ω

∂θ(x, θ) = ex cos (θ) j− ex sin (θ)k.

Therefore,

N (x, θ) = (i+ ex sin (θ) j+ ex cos (θ)k)× (ex cos (θ) j− ex sin (θ)k)

=

¯¯ i j k

1 ex sin (θ) ex cos (θ)0 ex cos (θ) −ex sin (θ)

¯¯

=

¯ex sin (θ) ex cos (θ)ex cos (θ) −ex sin (θ)

¯i−

¯1 ex cos (θ)0 −ex sin (θ)

¯j+

¯1 ex sin (θ)0 ex cos (θ)

¯k

=¡−e2x sin2 (θ)− e2x cos2 (θ)¢ i+ ex sin (θ) j+ ex cos (θ)k

= −e2xi+ ex sin (θ) j+ ex cos (θ)k

Thus,

||N (x, θ)|| =qe4x + e2x sin2 (θ) + e2x cos2 (θ) =

pe4x + e2x = ex

pe2x + 1.

Therefore, the area of the surface isZ θ=2π

θ=0

Z x=2

x=0

expe2x + 1dxdθ = 2π

Z x=2

x=0

expe2x + 1dx

We have Zexpe2x + 1dx =

1

2expe2x + 1 +

1

2arcsinh (ex)


(you can set u = ex so that the integrand takes the form√1 + u2du: Then set u = sinh (v)).

Therefore,

2π

Z x=2

x=0

expe2x + 1dx = 2π

Ã1

2expe2x + 1 +

1

2arcsinh (ex)

¯20

!

= 2π

µ1

2e2p(e4 + 1) +

1

2arcsinh

¡e2¢− 1

2

√2− 1

2arcsinh (1)

¶6. The radius of the sphere is 3. As in the lecture notes,

N (φ, θ) = 9 sin (φ) (sin (φ) cos (θ) i+ sin (φ) sin (θ) j+ cos (φ)k)

and

||N (φ, θ)|| = 9 sin (φ) .Therefore,Z Z

S

¡x2 + y2

¢dS =

Z π

θ−0

Z π

φ=0

¡9 sin2 (φ) cos2 (θ) + 9 sin2 (φ) sin2 (θ)

¢ ||N (φ, θ)|| dφdθ

=

Z π

θ−0

Z π

φ=0

9 sin2 (φ)¡cos2 (θ) + sin2 (θ)

¢9 sin (φ) dφdθ

= 81

Z π

θ−0

Z π

φ=0

sin3 (φ) dφdθ

= 81π

Z π

φ=0

sin3 (φ) dφ

Now, Zsin3 (φ) dφ =



We set u = cos (φ) so that du = − sin (φ) dφ. Thus,Z ¡1− cos2 (φ)¢ sin (φ) = −Z ¡

1− u2¢ du = −u+ 13u3 = − cos (φ) + 1

3cos3 (φ) .

Therefore, Z ZS

¡x2 + y2

¢dS = 81π

Z π

φ=0

sin3 (φ) dφ

= 81π

Ã− cos (φ) + 1

3cos3 (φ)

¯φ=πφ=0

!

= 81π

µ1− 1

3+ 1− 1

3

¶= 81π

µ4

3

¶= 108π.

7. We have

N (u, v) =

¯¯ i j k

0 −2 sin (u) 2 cos (u)1 0 0

¯¯

=

¯ −2 sin (u) 2 cos (u)0 0

¯i−

¯0 2 cos (u)1 0

¯j+

¯0 −2 sin (u)1 0

¯k

= 2 cos (u) j+ 2 sin (u)k.


Therefore,

||N (u, v)|| =q4 cos2 (u) + 4 sin2 (u) = 2.

Thus,

Z ZS

y2dS =

Z v=4

v=−4

Z u=π

u=0

cos2 (u) 2dudv = 16

Z u=π

u=0

cos2 (u) du

= 16

Z u=π

u=0

1 + cos (2u)

2du

= 8π + 8

Z π

0

cos (2u) du

= 8π + 8

µ1

2sin (2u)

¯π0

¶= 8π

8. We have

N (r, θ) =

¯¯ i j k

cos (θ) sin (θ) 0−r sin (θ) r cos (θ) 1

¯¯

=

¯sin (θ) 0r cos (θ) 1

¯i−

¯cos (θ) 0−r sin (θ) 1

¯j+

¯cos (θ) sin (θ)−r sin (θ) r cos (θ)

¯k

= sin (θ) i− cos (θ) j+ ¡r cos2 (θ) + r sin2 (θ)¢k= sin (θ) i− cos (θ) j+rk.

Thus,

||N (r, θ)|| =qsin2 (θ) + cos2 (θ) + r2 =

p1 + r2.

Therefore,

Z ZS

px2 + y2dS =

Z θ=2π

θ=0

Z r=4

r=0

qr2 cos2 (θ) + r2 sin2 (θ) ||N (r, θ)|| drdθ

=

Z θ=2π

θ=0

Z r=4

r=0

rp1 + r2dr

= 2π

Z r=4

r=0

rp1 + r2dr.

We set u = 1 + r2 so that du = 2rdr. Thus,

2π

Z r=4

r=0

rp1 + r2dr =

2π

2

Z u=17

u=1

u1/2du = π

Ã2

3u3/2

¯171

!=2π

3

³173/2 − 1

´.

9. S is the part of the graph of x = g (y, z) = 1 − y −z over the triangular region D in the


yz-plane that is bounded by the axes and the line z + y = 1. We haveZ ZS

xdS =

Z ZD

(1− y − z)√1 + 1 + 1dzdy

=√3

Z y=1

y=0

Z z=1−y

z=0

(1− y − z) dzdy

=√3

Z y=1

y=0

Ãz − yz − 1

2z2¯1−y0

dy

!

=√3

Z y=1

y=0

µ(1− y)− y (1− y)− 1

2(1− y)2

¶dy

=√3

Z y=1

y=0

µ1

2(y − 1)2

¶dy =

1

6

√3

10. The surface S is the graph of

z =p1− x2 − y2

over the region D that is the part of the unit circle x2 + y2 = 1 in the first quadrant. Thus,

Z ZS

z2dS =

Z ZD

¡1− x2 − y2¢

s1 +

µ∂z

∂x

¶2+

µ∂z

∂y

¶2dxdy

=

Z ZD

¡1− x2 − y2¢

vuut1 +Ã− xp1− x2 − y2

!2+

Ã− yp

1− x2 − y2

!2dxdy

=

Z ZD

¡1− x2 − y2¢s1 + x2 + y2

1− x2 + y2 dxdy

=

Z ZD

p1− x2 − y2dxdy.

In polar coordinates,Z ZD

p1− x2 − y2dxdy =

Z θ=π/2

θ=0

Z r=1

r=0

p1− r2rdrdθ

=π

2

µ1

3

¶=

π

6

(via the substitution u = 1− r2).11. Let D be the rectangle [0,π]× [−2, 2] in the uv-plane. We have

N (u, v) =

¯¯ i j k

− sin (u) 0 cos (u)0 1 0

¯¯

=

¯0 cos (u)1 0

¯i−

¯ − sin (u) cos (u)0 0

¯j+

¯ − sin (u) 00 1

¯k

= − cos (u) i− sin (u)k.

(Note that N is multiplied by −1 if we interchange the order of u and v).


Therefore,Z ZS

(xi+ zk) · dS =

Z ZD

(cos (u) i+ sin (u)k) ·N (u, v) dudv

=

Z ZD

(cos (u) i+ sin (u)k) · (− cos (u) i− sin (u)k) dudv

=

Z ZD

¡− cos2 (u)− sin2 (u)¢ dudv= −

Z ZD

dudv = −4π.

12. Let D be the rectangle [0, 3]× [0, 4] in the yz-plane. We have

N (y, z) =

¯¯ i j k

0 1 00 1 1

¯¯ = i.

(Note that N is multiplied by −1 if we interchange the order of y and z).Therefore, Z Z

S

(yi+ xj+ zk) · dS =

Z ZD

(yi+ 2j+ zk) ·N (y, z) dydz

=

Z ZD

(yi+ 2j+ zk) · idydz

=

Z ZD

ydydz

=

Z z=4

z=0

Z y=3

y=0

ydydz

= 4

Ã1

2y2¯30

!= 18.

13. As in the lecture notes, we have

N (φ, θ) = 9 sin (φ) (sin (φ) cos (θ) i+ sin (φ) sin (θ) j+ cos (φ)k) = 3 sin (φ)Φ (φ, θ) ,

Therefore, ZS

xi+ yj+ zk

(x2 + y2 + z2)· dS =

Z π

θ=0

Z π

φ=0

Φ (φ, θ)

9·N (φ, θ) dφdθ

=

Z π

θ=0

Z π

φ=0

Φ (φ, θ)

9· 3 sin (φ)Φ (φ, θ) dφdθ

=

Z π

θ=0

Z π

φ=0

1

3sin (φ)Φ (φ, θ) ·Φ (φ, θ) dφdθ

=

Z π

θ=0

Z π

φ=0

1

3sin (φ) ||Φ (φ, θ)||2 dφdθ

=

Z 2π

θ=0

Z π

φ=0

1

3sin (φ) (9) dφdθ

= 3 (2π)

Z π

φ=0

sin (φ) dφ

= 6π (2) = 12π


14. We have

1− 12x− 1

2y = 0⇔ x+ y = 2.

Therefore, the projection of S onto the xy-plane is the triangular region D that is bounded by

the coordinate axes and the line x+ y = 2. Thus,Z ZS

(xi− zk) ·dS =

Z ZD

µxi−

µ1− 1

2x− 1

2y

¶k

¶·µ−∂z∂xi− ∂z

∂yj+ k

¶dxdy

=

Z ZD

µxi−

µ1− 1

2x− 1

2y

¶k

¶·µ1

2i+

1

2j+ k

¶dxdy

=

Z ZD

µ1

2x−

µ1− 1

2x− 1

2y

¶¶dxdy

=

Z ZD

µx+

1

2y − 1

¶dxdy

=

Z x=2

x=0

Z y=2−x

y=0

µx+

1

2y − 1

¶dydx

=

Z x=2

x=0

Ãxy +

1

4y2 − y

¯y=2−xy=0

!dx

=

Z x=2

x=0

µx (2− x) + 1

4(2− x)2 − (2− x)

¶dx

=

Z x=2

x=0

µ−34x2 + 2x− 1

¶dx = 0

15. The projection of S onto the xy-plane is the disk D of radius 2 centered at the origin. We

have

N (x, y) = −∂z∂xi− ∂z

∂yj+ k

=xp

4− x2 − y2 i+yp

4− x2 − y2 j+ k.

Thus, Z ZS

(xi+yj+ zk) · dS

=

Z ZD

³xi+yj+

p4− x2 − y2k

´·N (x, y) dxdy

=

Z ZD

³xi+yj+

p4− x2 − y2k

´·Ã

xp4− x2 − y2 i+

yp4− x2 − y2 j+ k

!dxdy

=

Z ZD

Ãx2p

4− x2 − y2 +y2p

4− x2 − y2 +p4− x2 − y2

!dxdy

=

Z ZD

Ãx2 + y2 + 4− x2 − y2p

4− x2 − y2

!dxdy

= 4

Z ZD

1p4− x2 − y2 dxdy.



4

Z ZD

1p4− x2 − y2 dxdy = 4

Z 2π

θ=0

Z 2

r=0

r√4− r2 drdθ

= 8π

Z 2

r=0

r√4− r2 dr

= 16π

(via the substitution u = 4− r2).16. The projection of S onto the xy-plane is the unit disk D. We haveZ Z

S

(yj+ k) · dS =

Z ZD

(yj+ k) ·N (x, y) dxdy

=

Z ZD

(yj+ k) · (2xi+ 2yj+ k) dxdy

=

Z ZD

¡2y2 + 1

¢dxdy

=

Z θ=2π

θ=0

Z r=1

r=0

¡2r2 sin2 (θ) + 1

¢rdrdθ

=

Z θ=2π

θ=0

Z r=1

r=0

2r3 sin2 (θ) drdθ +

Z θ=2π

θ=0

Z r=1

r=0

rdrdθ

=

µZ r=1

r=0

2r3dr

¶ÃZ θ=2π

θ=0

sin2 (θ) dθ

!+ 2π

µ1

2

¶=

µ1

2

¶(π) + π =

1

2π

(you can make use of the identity sin2 (θ) = (1− cos (θ)) /2).17. We haveZ Z

S

(yzi+ xzj+ xyk) · dS =Z Z

S1

(yzi+ xzj+ xyk) · dS+Z Z

S2

(yzi+ xzj+ xyk) · dS.

We will evaluate the integrals separately.

As in Problem 10, on S1,

N (x, y) = −∂z∂xi− ∂z

∂yj+ k =

xp4− x2 − y2 i+

yp4− x2 − y2 j+ k,

and D is the disk of radius 2 centered at the origin. We haveZ ZS1

(yzi+ xzj+ xyk) · dS

=

Z ZD

³yp4− x2 − y2i+ x

p4− x2 − y2j+ xyk

´·Ã

xp4− x2 − y2 i+

yp4− x2 − y2 j+ k

!

=

Z ZD

3xydxdy

=

Z θ=2π

θ=0

Z 2

r=0

3 (r cos (θ)) (r sin (θ)) rdrdθ

= 3

µZ r=2

r=0

r3dr

¶µZ 2π

0

cos (θ) sin (θ) dθ

¶= 3 (4) (0) = 0.


The outward unit normal to S2 is -k. Therefore,Z ZS2

(yzi+ xzj+ xyk) · dS =

Z ZD

(yzi+ xzj+ xyk) · kdxdy

=

Z ZD

xydxdy = 0

Thus, Z ZS

(yzi+ xzj+ xyk) · dS = 0.

14.6 Green’s Theorem

1.

a) Let D be the disk that is bounded by the circle x2 + y2 = 4. ThenZC

y3dx− x3dy =

Z ZD

µ∂

∂x

¡−x3¢− ∂

∂y

¡y3¢¶dxdy

=

Z ZD

¡−3x2 − 3y2¢ dxdy= −3

Z ZD

¡x2 + y2

¢dxdy.

b) We transform the double integral to polar coordinates:

−3Z Z

D

¡x2 + y2

¢dxdy = −3

Z 2π

θ=0

Z 2

r=0

r2rdrdθ

= −6πÃ1

4r4¯20

!= −24π

2.

a) Let D be the triangular region with vertices (0, 0) , (2, 0) and (2, 4). ThenZC

xy2dx+ 2x2ydy =

Z ZD

µ∂

∂x

¡2x2y

¢− ∂

∂y

¡xy2

¢¶=

Z ZD

(4xy − 2xy) dxdy

=

Z ZD

2xydxdy

b) Since the equation of the line that joins (0, 0) to (2, 4) is y = 2x,Z ZD

2xydxdy =

Z x=2

x=0

µZ 2x

y=0

2xydy

¶dx =

Z x=2

x=0

³xy2

¯y=2xy=0

´dx

=

Z x=2

x=0

4x3dx = x4¯20= 16.

3.

14.6. GREEN’S THEOREM 141

a) Let D be the rectangular region with vertices (0, 0) , (5, 0) , (5,π) and (0,π). ThenZC

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

Z ZD

µ∂

∂x

¡x2 sin (y)

¢− ∂

∂ycos (y)

¶dxdy

=

Z ZD

(2x sin (y) + sin (y)) dxdy

=

Z ZD

(2x+ 1) sin (y) dxdy.

b) Z ZD

(2x+ 1) sin (y) dxdy =

Z x=5

x=0

Z y=π

y=0

(2x+ 1) sin (y) dydx

=

µZ x=5

x=0

(2x+ 1) dx

¶µZ y=π

y=0

sin (y) dy

¶=

³x2 + x

¯50

´(− cos (y)|π0 )

= 30 (2) = 60.

4.

a) Let D be annular region bounded by the circles x2 + y2 = 1 an x2 + y2 = 4. ThenZC

xe−2xdx+¡x4 + 2x2y2

¢dy =

Z ZD

µ∂

∂x

¡x4 + 2x2y2

¢− ∂

∂y

µZC

xe−2x¶¶

dxdy

=

Z ZD

¡4x3 + 4xy2

¢dxdy = 4

Z ZD

x¡x2 + y2

¢dxdy

b) Let’s transform the double integral to polar coordinates:

4

Z ZD

x¡x2 + y2

¢dxdy = 4

Z θ=2π

θ=0

Z r=2

r=1

r4 cos (θ) drdθ

= 4

ÃZ θ=2π

θ=0

cos (θ) dθ

!µZ r=2

r=1

r4dr

¶

= 4³sin (θ)|2π0

´Ã 15r5¯21

!

= 4 (0)

µ25 − 15

¶= 0.

5.

a) Let D be the triangular region with vertices (0, 0) , (2, 6) and (2, 0). Note that C is the

negatively oriented boundary of D. Thus,ZC

F · dσ =

ZC

y2 cos (x) dx+¡x2 + 2y sin (x)

¢dy

= −Z Z

D

µ∂

∂x

¡x2 + 2y sin (x)

¢− ∂

∂y

¡y2 cos (x)

¢dxdy

¶= −

Z ZD

(2x+ 2y cos (x)− 2y cos (x)) dxdy

= −Z Z

D

2xdxdy


b) Since the line that joins (0, 0) to (2, 6) is y = 3x,

−Z Z

D

2xdxdy = −Z x=2

x=0

Z 3x

y=0

2xdydx

= −Z x=2

x=0

2x

µZ y=3x

y=0

dy

¶dx

= −Z x=2

x=0

2x (3x) dx

= −6Z 2

0

x2dx = −6Ãx3

3

¯20

!= −16.

6.

a) Let D be the disk bounded by the circle x2 + y2 = 25. ThenZC

F · dσ =

Z ¡ex + x2y

¢dx+

¡ey − xy2¢ dy

=

Z ZD

µ∂

∂x

¡ey − xy2¢− ∂

∂y

¡ex + x2y

¢¶dxdy

=

Z ZD

¡−y2 − x2¢ dxdy = −Z ZD

¡x2 + y2

¢dxdy

b) We transform the double integral to polar coordinates:

−Z Z

D

¡x2 + y2

¢dxdy = −

Z 2π

θ=0

Z 5

r=0

r3drdθ

= −µZ 2π

θ=0

dθ

¶µZ 5

r=0

r3dr

¶= −2π

µ54

4

¶= −625

2π

7. Let C2 be the positively oriented circle of radius a centered at the origin, and let D be the

region between C1 and C2. By Greens theorem,ZC1

F · dσ −ZC2

F · dσ =

Z ZD

µ∂

∂x

µy

x2 + y2

¶− ∂

∂y

µx

x2 + y2

¶¶dxdy

=

Z ZD

Ã− 2xy

(x2 + y2)2 +

2xy

(x2 + y2)2

!dxdy = 0.

Thus, ZC1

F · dσ =ZC2

F · dσ.

We can parametrize the circle C2 by

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

dσ

dθ= − sin (θ) i+ cos (θ) j

and

F (σ (θ)) = F (cos (θ) , sin (θ)) = cos (θ) i+ sin (θ) j

14.7. STOKES’ THEOREM 143

Thus,

F · dσdθ

= (cos (θ) i+ sin (θ) j) · (− sin (θ) i+ cos (θ) j)= − cos (θ) sin (θ) + sin (θ) cos (θ) = 0

Therefore, ZC1

F · dσ =ZC2

F · dσ = 0.

8. Let D be the triangular region with vertices (0, 0) , (1, 0) and (1, 2). We have

∇ · F (x, y) = ∂

∂x

¡x2y3

¢− ∂

∂y(xy) = 2xy3 − x.

Therefore, ZC

F · nds =

Z ZD

∇ · F (x, y) dxdy

=

Z ZD

¡2xy3 − x¢ dxdy

=

Z x=1

x=0

Z y=2x

y=0

¡2xy3 − x¢ dydx

=

Z x=1

x=0

Ã1

2xy4 − xy

¯2x0

!dx

=

Z x=1

x=0

¡−2x2 + 8x5¢ dx= −2

3x3 +

4

3x6¯10

=2

3

9. Let D be the disk bounded by the circle x2 + y2 = 4. We have

∇ · F (x, y) = ∂

∂x

¡x3¢+

∂

∂y

¡y3¢= 3x2 + 3y2

Therefore, ZC

F · nds =Z Z

D

∇ · F (x, y) dxdy =Z Z

D

3¡x2 + y2

¢dxdy.

We transform the double integral to polar coordinates:Z ZD

3¡x2 + y2

¢dxdy = 3

Z θ=2π

θ=0

Z r=2

r=0

r3drdθ

= 3 (2π)

Ã1

4r4¯20

!= 24π

14.7 Stokes’ Theorem

1. Z ZM

(∇×F) · ndS =ZC

F · dσ.


We can parametrize C by

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

Thus,dσ

dθ= −4 sin (θ) i+ 4cos (θ) j,

and

F (σ (θ)) = 4 sin (θ) i+ 8 cos (θ) j

Therefore

F (σ (θ)) · dσdθ

= (4 sin (θ) i+ 8cos (θ) j) · (−4 sin (θ) i+ 4 cos (θ) j)= −16 sin2 (θ) + 32 cos2 (θ)

Thus ZC

F · dσ = −16Z 2π

0

sin2 (θ) dθ + 32

Z 2π

0

cos2 (θ) dθ = −16π + 32π = 16π.

2. Z ZM

(∇×F) · ndS =ZC

F · dσ.

We can parametrize C by

σ (θ) = (cos (θ) , sin (θ) , 1) , 0 ≤ θ ≤ 2π.

Thus,dσ

dθ= − sin (θ) i+ cos (θ) j,

and

F (σ (θ)) = sin (θ) i− cos (θ) j+ kTherefore

F (σ (θ)) · dσdθ

= (sin (θ) i− cos (θ) j+ k) · (− sin (θ) i+ cos (θ) j)= − sin2 (θ)− cos2 (θ) = −1.

Thus ZC

F · dσ = −Z 2π

0

dθ = −2π.

3. ZC

F · dσ =Z Z

M

(∇×F) · ndS

We have

∇×F (x, y, z) =

¯¯ i j k

∂x ∂y ∂zz x3 y2

¯¯

=

¯∂y ∂zx3 y2

¯i−

¯∂x ∂zz y2

¯j+

¯∂x ∂yz x3

¯k

= 2yi− j+ 3x2k

14.7. STOKES’ THEOREM 145

We also have n = k. ThusZM

(∇×F) · ndS =

Z ZM

¡2yi− j+ 3x2k¢ · kdS

=

Z ZM

3x2dS.

M can be parametrized by

Φ (x, y) = (x, y, 4) , (x, y) ∈ D,where D is the circle of radius 3 centered at the origin in the xy-plane. ThereforeZ Z

M

3x2dS =

Z ZD

3x2dxdy =

Z θ=2π

θ=0

Z r=3

r=0

3r2 cos2 (θ) rdrdθ

=

ÃZ θ=2π

θ=0

cos2 (θ) dθ

!µZ r=3

r=0

3r3dr

¶= (π)

µ243

4

¶=243

4π

4. ZC

F · dσ =Z Z

M

(∇×F) · ndS

We have

∇×F (x, y, z) =

¯¯ i j k

∂x ∂y ∂zz −x 2y

¯¯

=

¯∂y ∂z−x 2y

¯i−

¯∂x ∂zz 2y

¯j+

¯∂x ∂yz −x

¯k

= 2i+ j− k

M can be parametrized by

Φ (x, y) = (x, y, y + 2) , (x, y) ∈ D,

where D is the circle of radius 2 centered at the origin in the xy-plane. Therefore

N = −j+ k

Thus, ZM

(∇×F) · ndS =

ZD

(∇×F) ·Ndxdy

=

Z ZD

(2i+ j− k) · (−j+ k) dxy

=

Z ZD

−2dxdy= −2 (area of the disk of radius 2 )= −2 (4π) = −8π.


14.8 Gauss’ Theorem

1. We have

∇ · F (x, y, z) =∇ · (xi+ yj+ zk) = ∂

∂x(x) +

∂

∂y(y) +

∂

∂z(z) = 3.

Therefore Z ZM

F · ndS =

Z Z ZW

∇ · FdV

=

Z Z ZW

3dV

= 3 (volume of W ) = 3

µ4

3π¡43¢¶= 44π = 256π

2. We have

∇ · F (x, y, z) =∇ · ¡z3k¢ = ∂

∂z

¡z3¢= 3z2

Therefore Z ZM

F · ndS =Z Z Z

W

∇ · FdV =Z Z Z

W

3z2dV

We transform to spherical coordinates:Z Z ZW

3z2dV = 3

Z θ=2π

θ=0

Z φ=π

φ=0

Z ρ=2

ρ=0

(ρ cos (φ))2ρ2 sin (φ) dρdφdθ

= 2

ÃZ θ=2π

θ=0

dθ

!ÃZ φ=π

φ=0

cos2 (φ) sin (φ) dφ

!µZ ρ=2

ρ=0

ρ3dρ

¶= 2 (2π)

µ2

3

¶(4) =

32

3π

3. We have

∇ · F (x, y, z) =∇ · (xi+ yj+ zk) = ∂

∂x(x) +

∂

∂y(y) +

∂

∂z(z) = 3.

Therefore Z ZM

F · ndS =

Z Z ZW

∇ · FdV

=

Z Z ZW

3dV = 3 (volume of W ) = 3¡43¢= 192

4. We have

∇ · F (x, y, z) =∇ · ¡x2i¢ = ∂

∂x

¡x2¢= 2x

Therefore Z ZM

F · ndS =Z Z Z

W

∇ · FdV =

Z Z ZW

2xdV

=

Z x=1

x=0

Z y=1

y=0

Z z=1

z=0

2xdzdydx

=

Z x=1

x=0

2xdx = 1.

Calculus 3 Solution Manual

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