Solucionario zill 4edicion capitulo 15

Chapter 15

Vector Integral Calculus 15.1

Line Integrals

Z 1.

Z

π/4

Z

π/4

2(5 cos t)(5 sin t)(−5 sin t) dt = −250

2xy dx = C

0

C

0

sin2 t cos t dt

0

√  π/4  125 2 1 =− sin3 t = −250 3 6 0  π/4  Z Z π/4 Z π/4 1 2xy dy = 2(5 cos t)(5 sin t)(5 cos t) dt = 250 cos2 t sin t dt = 250 − cos3 t 3 C 0 0 0 √ ! √ 250 2 125 = 1− = (4 − 2) 3 4 6 Z Z π/4 Z π/4 p 2 2 2xy ds = 2(5 cos t)(5 sin t) 25 sin t + 25 cos t dt = 250 sin t cos t dt 0

 = 250

Z 2.

 π/4 125 1 = sin2 t 2 2 0

(x3 + 2xy 2 + 2x) dx =

1

Z

C

[8t3 + 2(2t)(t4 ) + 2(2t)]2 dt = 2

0

1

Z

(8t3 + 4t5 + 4t) dt

0 1

Z

(8t3 + 4t5 + 4t) dt

=2 0

  1 2 6 28 4 2 = 2 2t + t + 2t = 3 3 0 Z Z 1 Z 1 (x3 + 2xy 2 + 2x) dy = [8t3 + 2(2t)(t4 ) + 2(2t)]2t dt = 2 (8t4 + 4t6 + 4t2 ) dt C

0

 =2

0

 1 8 5 4 7 4 3 736 t + t + t = 5 7 3 105 0 218

15.1. LINE INTEGRALS Z

219

(x3 + 2xy 2 + 2x) ds =

Z

1

Z p [8t3 + 2(2t)(t4 ) + 2(2t)] 4 + 4t2 dt = 8

0

C

1

t(1 + t2 )5/2 dt

0

 1 8 7/2 1 2 7/2 = (1 + t ) = 7 (2 − 1) 7 0 Z Z 0 Z 0 0 3. (3x2 + 6y 2 ) dx = [3x2 + 6(2x + 1)2 ] dx = (27x2 + 24x + 6) dx = (9x3 + 12x2 + 6x) −1 

−1

C

−1

= −(−9 + 12 − 6) = 3 Z Z 0 [3x2 + 6(2x + 1)2 ]2 dx = 6 (3x2 + 6y 2 ) dy = −1 C Z Z 0 √ √ 2 2 (3x + 6y ) ds = [3x2 + 6(2x + 1)2 ] 1 + 4 dx = 3 5 −1

C

Z 8 8 x2 x2 8 56 dx = dx = dx = 3 2 /8 y 27x 27 27 1 8 ZC 2 Z 18 Z 8 x 8 4 x2 4 2/3 −1/3 −1/3 dy = x dx = x dx = x = 3 2 y 27x /8 27 1 9 3 1 8 ZC 2 Z 18 Z 8 p x x2 p 8 −1/3 2/3 3/2 −2/3 2/3 ds = 1+x dx = x 1+x dx = (1 + x ) 3 2 27 C y 1 27x /8 1 1 8 3/2 = (5 − 23/2 ) 27 Z Z 2π Z 2π p 2 2 2 2 t 2 5. (x + y )ds = (25 cos −25 sin t) 25 sin t + 25 cos tdt = 125 (cos2 t − sin2 t)dt Z

Z

4.

C

0

0

Z = 125 0

Z 6.

Z

2π 125 sin 2t = 0 cos 2tdt = 2 0

π

(2x + 3y) d = C

2π

(6 sin 2t + 6 cos 2t)(−4 sin 2t) dt Z0 π

−24 sin2 2t − 24 sin 2t cos 2t dt   Z0 π 1 = −24 (1 − cos 2t) − 24 sin 2t cos 2t dt 2 0 π = −12t + 6 sin2 2t − 12 sin2 2t 0 =

= −12π Z 7.

Z z dx =

C

π/2

t(− sin t) dt

Integration by parts

0 π/2

Z C

= (t cos t − sin t)|0 = −1 Z π/2 z dy = t cos t dt Integration by parts 0 π/2

= (t sin t + cos t)|0

=

π −1 2

220

CHAPTER 15. VECTOR INTEGRAL CALCULUS π/2 π2 1 2 = z dz = t dt = t 2 0 8 ZC Z0 π/2 p √ Z z dx = t sin2 t + cos2 t + 1 dt = 2 Z

Z

C

8.

π/2

4xyz 4xyz ZC 4xyz C

Z 4xyz C

√ π2 2 8

1 8 9 8 1 3 8 1 8 2 2 t dt = t (t )(2t)t dt = t = dx = 4 3 3 27 27 0 01  Z 01  Z 1 1 3 2 2 16 dy = 4 t7 dt = t8 = t (t2 )(2t)2t dt = 3 3 3 3 1 0  Z0 Z0 1  16 16 1 3 16 1 6 2 t dt = 4 t (t )(2t)2 dt = dz = = 21 3 21 21 0 0    1 Z 1 Z 01  p 8 1 9 2 7 8 200 1 3 6 2 t (t + 2) dt = t (2t) t4 + 4t2 + 4 dt = t + t = ds = 4 3 3 3 9 7 189 0 0 0 Z

ZC

t dt =

0

0

Z

π/2

1





Z

9. Using x as the parameter, dy = dx and Z

Z

2

Z

2

= (2x + y) dx + xy dy = −1

C

 =

 2 1 3 x + 3x2 + 3x = 21. 3 −1

−1

 =

(x2 + 6x + 3) dx

−1

10. Using x as the parameter, dy Z Z 2= 2x dx and Z 2 (2x + y) dx + xy dy = (2x + x + 1) dx + C

2

(2x + x + 3 + x + 3x) dx =

2 2

Z

2

x(x + 1)2x dx =

−1

(2x4 + 3x2 + 2x + 1) dx

−1

 2 141 2 5 x + x3 + x2 + x = . 5 5 −1

11. From (−1, 2) to (2, 2) we use x as a parameter with y = 2 and dy = 0. From (2, 2) to (2, 5) we use y as a parameter with x = 2 and dx = 0.

Z

Z

2

Z

(2x + y) d + xy dy =

(2x + 2) dx + −1

C

2

5

2 2 2y dy = (x2 + 2x) −1 + y 2 −1 = 9 + 21 = 30

12. From (−1, 2) to (−1, 0) we use y as a parameter with x = −1 and dx = 0. From (−1, 0) to (2, 0) we use x as a parameter with y = dy = 0. From (2, 0) to (2, 5) we use y as a parameter with x = 2 and dx = 0.

Z

Z (2x + y) d + xy dy =

C

0

Z

2

(−1)y dy + 2

= 2 + 3 + 25 = 30

Z 2x dx +

−1

0

5

0 2 2 1 2y dy = − y 2 + x2 −1 + y 2 0 2 2

15.1. LINE INTEGRALS

221

13. Using x as a the parameter, dy = 2xdx. Z

1

Z

x2 dx +

y dx + x dy =

1

Z x(2x) dx = 0

0

0

C

1

Z

1 3x2 dx = x3 0 = 1

14. Using x as a the parameter, dydx. Z

Z

1

y dx + x dy =

1

Z x dx +

C

1

Z x dx =

0

0

0

1 2x dx = x2 0 = 1

15. From (0, 0) to (0, 1) we use y as a parameter with x = dx = 0. From (0, 1) to (1, 1) we use x as a parameter with y = 1 and dy = 0. Z

1

Z y dx + x dy = 0 +

1 dx = 1

C

0

16. From (0, 0) to (1, 0) we use x as a parameter with y = dy = 0. From (1, 0) to (1, 1) we use y as a parameter with x = 1 and dx = 0. Z

Z y dx + x dy = 0 +

C

Z

1

1 dy = 1 0

Z

9

17.

9

Z

1 (6x + 2y + 2) dx + 4xy dy = (6t + 2t ) t−1/2 dt + 2 C 4 9 = (2t3/2 + 2t3/2 ) = 460 2

2

Z 9 √ 4 tt dt + (3t1/2 + 5t3/2 ) dt

4

4

4

18.

R

(−y 2 ) dx + xy dy = C

Z 19.

R2

(−t6 )2 dt + 0

2x3 y d + (3x + y) dy =

Z 

= Z

Z

1

2(y 6 )y2y dy +

0

4t6 dt =

1

(3y 2 + y) dy =

Z

2 512 4 7 t = 7 0 7

1

(4y 8 + 3y 2 + y) dy

−1

4 9 1 26 y + y 3 + y 2 = 9 2 9 −1

4(y 3 + 1)3y 2 dy +

−1 6

Z

R2

−1  1

2

4x dx + 2y dy = C

(2t)(t3 )3t2 dt = 0

−1

C

20.

R2

Z

2

−1 3

= 2y + 4y +

2 y 2 −1



Z

2

2y dy =

(12y 5 + 12y 2 + 2y) dy

−1

= 165

21. From (−2, 0) to (2, 0) we use x as a parameter with y = dy = 0. From (2, 0) to (−2, 0) we parameterize the semicircle as x = 2 cos θ and y = 2 sin θ for 0 ≤ θ ≤ π.

222

CHAPTER 15. VECTOR INTEGRAL CALCULUS

Z

2

Z

2

2

(x + y ) dx − 2xy dy =

x

2

π

Z 4(−2 sin θ dθ) −

dx +

−2

C

π

Z

8 cos θ sin θ(2 cos θ dθ) 0

0 π

2 Z 1 3 (sin θ + 2 cos2 θ sin θ) dθ x −8 3 −2 0   π 16 2 16 80 64 3 = − 8 − cos θ − cos θ = − =− 3 3 3 3 3 0 =

22. We start at (0, 0) and use x as a parameter. Z

2

Z

1

Z

1

Z

0

(x2 + x) dx xx (2x dx) + (x + x ) dx − 2 1 0 0   Z 0 √ 1 −1/2 x x −2 x dx 2 1 1 Z 1 Z 0 Z 1 3 3 = (x2 − 3x4 ) dx + x2 dx = (−3x4 ) dx = − x5 = − 5 5 0 1 0 0

2

(x + y ) dx − 2xy dy = C

2

4

2

23. From (1, 1) to (−1, 1) and (−1, −1) to (1. − 1) we use x as a parameter with y = 1 and y = −1, respectively, and dy = 0. From (−1, 1) to (−1, −1) and (1, −1) to (1, 1) we use y as a parameter with x = −1 and z = 1, respectively, and dx = 0. Z

x2 y 3 dx − xy 2 dy =

C

Z

−1

x2 (1) dx +

1

−1

Z

−(−1)y 2 dy +

Z

1

x2 (−1)3 dx +

Z

−1

1

1

−(1)y 2 dy

−1

−1 1 1 1 1 3 1 3 1 3 8 1 3 = x + y − x − y =− 3 1 3 −1 3 −1 3 −1 3 24. From (2, 4) to (0, 4) we use x as a parameter with y = 4 and dy = 0. From (0, 4) to (0, 0) we use y as a parameter with x = dx = 0. From (0, 0) to (2, 4) we use y = 2x and dy = 2dx.

Z

x2 y 3 dx − xy 2 dy =

C

Z

0

x2 (64) dx −

0

Z

Z 0 dy +

2

4

2

x2 (8x3 ) dx −

Z

0

2

x(4x2 )2 dx

0

0 2 2 64 3 4 6 512 256 352 = x + x − 2x4 0 = − + − 32 = − 3 2 3 0 3 3 3 Z

Z

π

Z

y dx − x dy =

25.

3 sin t(−2 sin t) dt − 0 Z π = −6 dt = −6π

C

Z

0

y dx − x dy = 6π.

Thus, −C

π

Z 2 cos t(3 cos t) dt = −6

0

0

π

(sin2 t + cos2 ) dt

15.1. LINE INTEGRALS Z 26.

x2 y 3 + x3 y 2 dy =

223 Z

1

x

= −1 Z 1

=

1

x3 (x8 )(4x3 ) dx

−1

−1 Z 1

C

Z

x2 (x12 ) dx + 14

1

Z

4x14 dx

dx + −1

1 5 15 x 15 −1

5x14 dx =

−1

5 5 2 = + = 15 15 3 27. We parameterize the line segment from (0, 0, 0) to (2, 3, 4) by x = 2t, y = 3y, z = 4t for 0 ≤ t ≤ 1. We parameterize the line segment from (2, 3, 4) to (6, 8, 5) by x = 2 + 2t, y = 3 + 5t, z = 4 + t, 0 ≤ t ≤ 1.

Z

1

Z

Z

0 Z 1

C

1

3t(2 dt) +

y dx + z dy + x dz =

Z

0

(3 + 5t)(4 dt) 0

(2 + 4t) dt 0

1

 (55t + 34) dt =

0

Z

y dx + z dy + x dz = C

2

Z

t3 (3 dt) +

2

 1 123 55 2 t + 34t = 2 2 0

   Z 2 5 2 5 t (3t2 dt) + (3t) t dt 4 2 0 0 0    2 Z 2 15 4 15 2 3 4 3 5 5 3 3 = 3t + t + t dt = t + t + t = 56 4 2 4 4 2 0 0

Z 28.

1

1

Z

0

=

Z 2t(4 dt) +

0

(4 + t)(5 dt) + Z

1

4t(3 dt) +



29. From (0, 0, 0) to (6, 0, 0) we use x as a parameter with y = dy = 0 and z = dz = 0. From (6, 0, 0) to (6, 0, 5) we use z as a parameter with x = 6 and dx = 0 and y = dy = 0. From (6, 0, 5) to (6, 8, 5) we use y as a parameter with x = 6 and dz = 0 and z = 5 and dz = 0. Z Z 6 Z 5 Z 8 y dx + z dy + z dz = 0 dx + 6 dz + 5 dy = 70 C

0

0

0

30. We parameterize the line segment from (0, 0, 0) to (6, 8, 0) by x = 6t, y = 8t, z = 0 for 0 ≤ t ≤ 1. From (6, 8, 0) to (6, 8, 5) we use z as a parameter with x = 6, dx = 0, and y = 8, dy = 0. Z

Z y dx + z dy + z dz =

C

Z 31. C

1

Z

0

10x dx − 2xy 2 dy + 6xz dz =

Z

1

0

Z 10(t) dt −

0

5

8t(6 dt) + 1

2(t)(t2 )2 (2t) dt +

0

1 1 1 = 5t2 0 − 4t6 0 + 18t6 0 = 5 − 4 + 18 = 19

1 6 dz = 24t2 0 + 30 = 54 Z 0

1

6(t)(t3 )(3t2 ) dt

224

CHAPTER 15. VECTOR INTEGRAL CALCULUS

32. Parametrize the line segments as follows: C1 : r1 (t) = ti + tj, 0 ≤ t ≤ 1 C2 : r2 (t) = i + j + tk, 0 ≤ t ≤ 1 C3 : r3 (t) = (1 − t)i + (1 − t)j + (1 − t)k, 0 ≤ t ≤ 1 We Z then have Z Z 1

3x dx − y 2 dy + z 2 dz =

C1

1

t2 dt

3t dt − 0

0

7 3 1 − = 2 3 6 Z 1 Z 1 Z 3(1)(0) dt − (1)2 (0) dt + 3x dx − y 2 dy + z 2 dz = =

Z

0

C2

0

1

t2 dt

0

1 = 3 Z 1 Z 1 Z 1 Z 2 2 2 (1 − t)2 (−1) dt (1 − t) (−1) dt + 3(1 − t)(−1) dt − 3x dx − y dy + z dz = 0 0 0 C3       3 1 1 3 = − − − + − =− 2 3 3 2 Z 33.

Z

2

y dx + xydy = C1

1 2

Z

(4t + 2) 2dt + 0

1

Z (2t + 1)(4t + 2)4dt =

0

1

(64t2 + 64t + 16)dt

0

 1 64 208 64 3 t + 32t2 + 16t = + 32 + 16 = 3 3 3 0 √3 Z Z √3 Z √3 Z √3 8 8 208 y 2 dx + xydy = 4y 4 (2t)dt + 2t4 (4t)dt = 16t5 dt = t6 = 72 − = 3 3 3 C2 1 1 1 1 e3 Z Z e3 Z e3 Z e3 1 8 8 2 y 2 dx + xydy = 4(ln t)2 dt + 2(ln t)2 dt = (ln t)2 dt = (ln t)3 t t t 3 C3 e e e e 8 208 = (27 − 1) = 3 3 √   2 √ R2 2 √ R R2 √ 16 5 1 3 34. C1 xyds = 0 t(2t) 1 + 4dt = 2 5 0 t dt = 2 5 t = 3 3 0 Z Z 2 Z 2 p p 1 xyds = t(t2 ) 1 + 4t2 dt = t3 1 + 4t2 dt u = 1 + 4t2 , du = 8tdt; t2 = (u − 1) 4 C2 0 0 Z 17 Z 17 1 1 1 = (u − 1)u1/2 du = (u3/2 − u1/2 )du 4 8 32 1 1   17 1 2 5/2 2 3/2 = u − u 32 5 3 1 √ 391 17 + 1 = √   3 Z Z 3 120 √ Z 3 √ 1 √ 16 5 xyds = (2t − 4)(4t − 8) 4 + 16dt = 16 5 (t − 2)2 dt 16 5 (t − 2)3 = 3 3 C3 2 2 2 C1 and C3 are different parameterization of the same curve, while C1 and C2 are different 

=

15.2. LINE INTEGRALS OF VECTOR FIELDS

225

curves.

35. We are given ρ = kx. Then Z m=

Z

π

C

Z kx ds = k

ρ dx =

π

Z π p (1 + cos t) sin2 t + cos2 t dt = k (1 + cos t) dt

0

0 π

0

= k (t + sin t)|0 = kπ.

36. From Problem 35,Zm = kπ and Zds = dt.   π Z π 1 (1 + cos t) sin tdtk − cos t + sin2 t = 2k Mx = yρds = kxyds = k 2 0Z C C 0 Z π π 2 2 My = intC xρds = intC kx ds = k (1 + cos t) dt = k (1 + 2 cos t + cos2 t)dt 0 0   π 1 3 1 = k t + 2 sin t + t + sin 2t = kπ 2 4 2 0 3 2 3kπ/2 2k = ; y = Mx /m = = . The center of mass is (3/2, 2/π). x = My /m = kπ 2 kπ π

15.2 1.

Line Integrals of Vector Fields 2.

y

x

y

x

226

3.

CHAPTER 15. VECTOR INTEGRAL CALCULUS

4.

y

y

x

5.

x

6.

y

y

x

7. Since each vector points in a northeasterly direction, the vector field must have positive i and j components. Therefore, the answer is (b). 8. Since each vector points in a northwesterly direction, the vector field must have negative i and positive j components. Therefore, the answer is (a).

x

15.2. LINE INTEGRALS OF VECTOR FIELDS

227

9. Since each vector points in a southwesterly direction, the vector field must have negative i and j components. Therefore, the answer is (d). 10. Since each vector points in a southeasterly direction, the vector field must have positive i and negative j components. Therefore, the answer is (c). 11. Note that the k component of each vector is always positive. Therefore, the answer is (d). 12. Note that the i component of each vector is always positive. Therefore, the answer is (c). 13. Note that each vector points directly away from the origin. Therefore, the answer is (a). 14. Note that the i and j components of each vector are zero. Therefore, the answer is (b). F = e3t i − (e−4t )et j = e3t i − e−3t j; dr = (−2e−2t i + et j)dt; F · dr = (−2et − e−2t )dt; ln 2 Z Z ln 2 15. 31 3 19 1 −2t t −2t t = − − (− ) = − (−2e − e )dt = (−2e + e ) F · dr = 2 8 2 8 0 c 0 16. F = 2(t)(t2 )i + t2 j = 2t3 i + t2 j; dr = (i + 2tj)dt; F · dr = 4t3 dt; 2 R R2 F · dr = 0 4t2 dt = t4 0 = 16 C 17. F = 2(2t − 1)i − 2(6t + 1)j = (4t − 2)i + (12t + 2)j; dr = (2i + 6j)dt; F · dr = −64t − 16; 1 R R1 F · dr = −1 (−64t − 16) dt = −32t2 − 16t −1 = −32 C 18. F = cos2 ti + sin tj; dr = (− sin ti + cos tj); F · dr = (− cos2 t sin t + sin t cos t)dt; Z Z π/6 F · dr = (− cos2 t sin t + sin t cos t) dt C

0

π/6 cos3 t sin2 t + = 3 2 0 √ !3  2   1 1 1 1 3 = + − +0 3 2 2 2 3 √ √ 5 3 1 1 3 = + − = − 8 8 3 8 24 19. F = −3 sin ti + 2 cos tj + 6tk; dr = (−2 sin ti + 3 cos tj + 3k)dt; 2 2 F Z · dr = (−6Zsin t + 6 cos t + 18t)dt; π

(−6 sin2 t + 6 cos2 t + 18t) dt     Z π π 1 1 = −6 (1 − cos 2t) + 6 (1 + cos 2t) + 18t dt = 3 sin 2t + 9t2 0 2 2 0

F · dr = C

0

= 9π 2

228

CHAPTER 15. VECTOR INTEGRAL CALCULUS 3

6

20. F = et i + tet j + t3 et k; dr = (i + 2tj + 3t2 k)dt; 1 Z Z 1 13 2 t3 1 t6 t 2 t3 5 t6 t (e − 1) F · dr = (e + 2t e + 3t e )dt = (e + e + e ) = 3 2 6 C 0 0 21. Using x as a parameter, r(x) = xi + ln xj. Then F = ln xi + xj, dr = (i + Z

Z F · dr =

W =

1

c

e

1 j)dx, and x

e

(ln x + 1)dx = (x ln x)|1 = e.

22. Let r1 = (−2+2t)i+(2−2t)j and r2 = 2ti+3tj for 0 ≤ t ≤ 1. Then dr1 = 2i − 2j, dr2 = 2i + 3j,

y

F1 = 2(−2 + 2t)(2 − 2t)i + 4(2 − 2t)2 j = (−8t2 + 16t − 8)i + (16t2 − 32t + 16)j,

c1

F2 = 2(2t)(3t)i + 4(3t)2 j = 12t2 i + 36t2 j, and Z W =

Z F1 · dr1 +

C1 Z 1

F2 · dr2

x

cC2 2

Z

2

(−16t + 32t − 16 − 32t + 64t − 32)dt +

= 0

Z

c2

1

(24t2 + 108t2 )dt

0 1

= 0

1 (84t2 + 96t − 48)dt = (28t3 + 48t2 − 48t) 0 = 28.

23. Let r1 = (1 + 2t)i + j, r2 = 3i + (1 + t)j, and r3 = (3 − 2t)i + (2 − t)j for 0 ≤ t ≤ 1. Then dr1 = 2i,

dr2 = j,

dr3 = −2i − j,

y

c3

F1 = (1 + 2t + 2)i + (6 − 2 − 4t)j = (3 + 2t)i + (4 − 4t)j, F2 = (3 + 2 + 2t)i + (6 + 6t − 6)j = (5 + 2t)i + 6tj, F3 = (3−2t+4−2t)i+(12−6t−6+4t)j = (7−4t)i+(6−2t)j, and Z

Z

Z

F1 · dr1 +

W = C1 Z 1

=

F2 · dr2 + Z

(6 + 4t)dt + 0

Z = 0

c3 1

Z

1

(−14 + 8t − 6 + 2t)dt

6tdt + 0

1

F3 · dr3

c2

0

1 (−14 + 20t)dt = (−14t + 10t2 ) 0 = −4.

24. F = t3 i + t4 j + t5 k; dr = 3t2 i + 2tj + k; 3 R R3 R3 W = C F · dr = 1 (3t5 + 2t5 = t5 )dt = 1 6t5 dt = t6 1 = 728

c2

c1 x

15.2. LINE INTEGRALS OF VECTOR FIELDS

229

25. r = 3 Rcos ti + 3 sinRtj, 0 ≤ t ≤ 2π; dr = −3 sin ti + 3 cos tj; F = ai + bj; 2π 2π W = C F · dr = 0 (−3a sin t + 3b cos t)dt = (3a cos t + 3b sin t)|0 = 0 26. Let r = ti + tj + tk for 1 ≤ t ≤ 3. Then dr = i + j + k, and ct c (ti + tj + tk) = √ (i + j + k) = 3 |r| ( 3t2 )3 Z 3 Z Z 3 c 1 c √ (1 + 1 + 1)dt = √ dt = W = F · dr = 2 t 3 1 2 C 1 3 3t c 1 2c = √ (− + 1) = √ . 3 3 3 3 F=

c √ (i + j + k), 3 3t2   3 1 c √ − t 1 3

27. F = 10 cos ti − 10 sin tj; dr = (−5 sin ti + 5 cos tj)dt; F · dr = (−100 cos t sin t)dt; 2π R R 2π F · dr = 0 (−100 cos t sin t) dt = 50 cos2 t 0 = 0 C 28. F = 10 cos ti + 2 sin tj; dr = (2 cos ti − 10 sin tj)dt; F · dr = (10 cos2 t − 20 sin2 t)dt; Z Z 2π F · dr = (10 cos2 t − 20 sin2 t) dt C 0 π    Z 2π  1 1 15 = 10 (1 + cos 2t) − 20 (1 − cos 2t) dt = −5t + sin 2t 2 2 2 0 0 = −10π 29. On C1 , T = i and F · TcompT F ≈ 1. On C2 , T = −j and F · T = compT F ≈ 2. OnC3 , T = −i and F · T = compT F ≈ 1.5. Using the fact that the lengths of C1 , C2 , and C3 are 4, 5, and 5, respectively, we have R R R R W = C F · Tds = C1 F · Tds + C2 F · Tds + C3 F · Tds ≈ 1(4)+2(5)+1.5(5)=21.5 ft-lb. Z

Z F · dr =

30. W = C b

b

(ma · r0 (t)) dt

a b



 dv m(a · v) dt = m = · v dt dt a a   Z b Z b m dv dv m d = ·v+v· dt = (v · v) dt dt dt a 2 a 2 dt Z b m d 2 m 2 b = (v ) dt = v 2 a a 2 dt 1 1 = m[v(b)]2 − m[v(a)]2 2 2 = K(B) − K(A) Z

Z

1 1 (3x − 6y)3i + (3x − 6y)(−6)j 3 3 = (3x − 6y)i + (−6x + 12y)j

31. ∇f (x, y) =

230

CHAPTER 15. VECTOR INTEGRAL CALCULUS

32. ∇f (x, y) = (1 + 2 cos 5xy − 10xy sin 5xy)i + (−1 − 10x2 sin 5xy)j

33. ∇f (x, y, z) = tan−1 zyi +

xz xy j+ 2 2 k +1 y z +1

y2 z2

34. ∇f (x, y, z) = (1 − 2xyz 4 )i − x2 z 4 j − 4x2 yz 3 k

  2 2 35. ∇f (x, y, z) = e−y i + 1 + 2xye−y j + k

36. ∇f (x, y, z) =

8y 3 18z 5 2x i + j + k x2 + 2y 4 + 3z 6 x2 + 2y 4 + 3z 6 x2 + 2y 4 + 3z 6


37. ∇ x2 + 21 y 2 = 2xi + yj = F(x, y). Therefore, the answer is (b).

38. ∇

1 2 2x


+ y 2 − 4 = xi + 2yj = F(x, y). Therefore, the answer is (c).


39. ∇ 2x + 12 y 2 + 1 = 2i + yj = F(x, y). Therefore, the answer is (d).

40. ∇

1 2 2x


+ 13 y 3 − 5 = xi + y 2 j = F(x, y). Therefore, the answer is (a).

41. φ(x, y) = sin x + y + cos y

42. φ(x, y) = xe−y

43. φ(x, y) = x + y 2 − 4z 3

44. φ(x, y) = xy 2 z 3

15.2. LINE INTEGRALS OF VECTOR FIELDS

45.

231

46.

y

y

x

47.

x

48.

y

x

y

x

232

CHAPTER 15. VECTOR INTEGRAL CALCULUS

49.

50.

y

y

x

x

51. Let φ(x, y, z) = −c(x2 + y 2 + z 2 )−1/2 . Then cx cy cz i+ 2 j+ 2 k (x2 + y 2 + z 2 )3/2 (x + y 2 + z 2 )3/2 (x + y 2 + z 2 )3/2 c(xi + yj + zk) = 2 (x + y 2 + z 2 )3/2 cr = 3 =F |r|

∇φ(x, y, z) =

52. Yes; if f and g differ by a constant, they will have the same gradient field.

15.3

Independence of the Path 1 3 x + g(y), φy = 3 (2,2) 1 3 1 3 1 3 R (2,2) 2 16 1 3 0 2 2 3 g (y) = y , g(y) = y , φ = x + y , (0,0) x dx + y dy = (x + y ) = 3 3 3 3 3

1. (a) Py = 0 = Qx and the integral is independent of path. φx = x2 , φ =

(0,0)

2 R (2,2) 2 R2 2 2 3 16 2 2 (b) Use y = x for 0 ≤ x ≤ 2. (0,0) x dx + y dy = 0 (x + x )dx = x = 3 0 3 2. (a) Py = 2x = Qx and the integral is independent of path. φx = 2xy, φ = x2 y +g(y), φy = (2,4) R (2,4) x2 + g 0 (y) = x2 , g(y) = 0, φ = x2 y, (1,1) 2xydx + x2 dy = x2 y (1,1) = 16 − 1 = 15 (b) Use y = 3x − 2 for 1 ≤ x ≤ 2. 2 R (2,4) R2 R2 2xydx + x2 dy = 1 [2x(3x − 2) + x2 (3)]dx = 1 (9x2 − 4x)dx = (3x3 − 2x2 ) 1 = 15 (1,1)

15.3. INDEPENDENCE OF THE PATH

233

1 3. (a) Py = 2 = Qx and the integral is independent of path. φx = x + 2y, φ = x2 + 2xy + 2 1 2 1 2 1 2 0 g(y), φy = 2x + g (y) = 2x − y, g(y) = − y , φ = x + 2xy − y , 2 2   2(3,2) R (3,2) 1 2 1 2 (x + 2y)dx + (2x − y)dy = = 14 x + 2xy − y (1,0) 2 2 (1,0) (b) Use y = x − 1 for 1 ≤ x ≤ 3. Z

(3,2)

3

Z (x + 2y)dx + (2x − y)dy =

[x + 2(x − 1) + 2x − (x − 1)]dx

(1,0)

1 3

Z

3 (4x − 1)dx = (2x2 − x) 1 = 14

= 1

4. (a) Py = − cos x sin y = Qx and the integral is independent of path. φx = cos x cos y, φ = sin x cos y + g(y), φy = − sin x sin y + g 0 (y) = 1 − sin x sin y, g(y) = y, φ = R (π/2,0) (π/2,0) sin x cos y + y, (0,0) cos x cos ydx + (1 − sin x sin y)dy = (sin x cos y + y)|(0,0) = 1 (b) Use y = 0 for 0 ≤ x ≤ π/2. Z

(π/2,0)

Z cos x cos ydx + (1 − sin x sin y)dy =

(0,0)

0

π/2

π/2

cos xdx = sin x|0

=1

x 1 5. (a) Py = 1/y 2 = Qx and the integral is independent of path. φx = − , φ = − + y y (4,4) R x x x x x 1 (4,4) =3 g(y), φy = 2 + g 0 (x) = 2 , g(y) = 0, φ = − , (4,1) − dx + 2 dy = (− ) y y y y y y (4,1) (b) Use x = 4 for 1 ≤ y ≤ 4. Z

(4,4)

(4,1)

1 x − dx + 2 dy = y y

Z 1

4

4 4 4 dy = − = 3 y2 y 1

x 6. (a) Py = −xy(x2 +y 2 )−3/2 = Qx and the integral is independent of path. φx = p , φ= x2 + y 2 p p y y x2 + y 2 + g(y), φy = p + g 0 (y) = p , g(y) = 0, φ = x2 + y 2 , 2 2 2 2 x +y x +y (3,4) Z (3,4) xdx + ydy p 2 p = x + y2 =4 x2 + y 2 (1,0) (1,0)

(b) Use y = 2x − 2 for 1 ≤ x ≤ 3. Z 3 Z 3 Z (3,4) x + (2x − 2)2 5x − 4 xdx + ydy p p √ = dx = 5x2 − 8x + 4 x2 + (2x − 2)2 x2 + y 2 1 1 (1,0) p 3 = 5x2 − 8x + 4 = 4 1

234

CHAPTER 15. VECTOR INTEGRAL CALCULUS

7. (a) Py = 4xy = Qx and the integral is independent of path. φx = 2y 2 x − 3, φ = x2 y 2 − 3x + g(y), φy = 2x2 y + g 0 (y) = 2x2 y + 4, g(y) = 4y, φ = x2 y 2 − 3x + (3,6) R (3,6) 4y, (1,2) (2y 2 x − 3)dx + (2yx2 + 4)dy = x2 y 2 − 3x + 4y) (1,2) = 330 (b) Use y = 2x for 1 ≤ x ≤ 3. Z

(3,6)

(2y 2 x − 3)dx + (2yx2 + 4)dy =

(1,2)

3

Z

([2(2x)2 x − 3] + [2(2x)x2 + 4]2)dx

1 3

Z = 1

3 (16x3 + 5)dx = (4x4 + 5x) 1 = 330

5 8. (a) Py = 4 = Qx and the integral is independent of path. φx = 5x + 4y, φ = x2 + 4xy + 2 5 g(y), φy = 4x + g 0 (y) = 4x − 8y 3 , g(y) = −2y 4 , φ = x2 + 4xy − 2y 4 , 2  (0,0) R (0,0) 7 5 = x2 + 4xy − 2y 4 (5x + 4y)dx + (4x − 8y 3 )dy = (−1,1) 2 2 (−1,1) (b) Use y = −x for −1 ≤ x ≤ 0. Z (−1, 1)

(0,0)

Z

3

0

[(5x − 4x) + (4x + 8x3 )(−1)]dx

(5x + 4y)dx + (4x − 8y )dy = −1 Z 0

0 7 3 (−3x − 8x3 )dx = (− x2 − 2x4 ) = 2 2 −1 −1

=

9. (a) Py = 3y 2 + 3x2 = Qx and the integral is independent of path. φx = y 3 + 3x2 y, ; φ = xy 3 + x3 y + g(y), φy = 3xy 2 + x3 + g 0 (y) = x3 + 3y 2 x + 1, g(y) = y, φ = (2,8) R (2,8) xy 3 + x3 y + y, (0,0) (y 3 + 3x2 y)dx + (x3 + 3y 2 x + 1)dy = .(xy 3 + x3 y + y) = 1096 (0,0)

(b) Use y = 4x for 0 ≤ x ≤ 2. Z (2,8) Z 2 3 2 3 2 (y + 3x y)dx + (x + 3y x + 1)dy = [(64x3 + 12x3 ) + (x3 + 48x3 + 1)(4)]dx (0,0)

0

Z = 0

2

2 (272x3 + 4)dx = (68x4 + 4x) 0 = 1096

10. 11. Py = 12x3 y 2 = Qx throughout the plane and the vector field is a conservative field. φx = 4x3 y 3 +3, φ = x4 y 3 +3x+g(y), φy = 3x4 y 2 +g 0 (y) = 3x4 y 2 +1, g(y) = y, φ = x4 y 3 +3x+y 12. Py = 6xy 2 = Qx throughout the plane and the vector field is a conservative field. φx = 2xy 3 , φ = x2 y 3 + g(y), φy = 3x2 y 2 + g 0 (y) = 3x2 y 2 + 3y 2 , g(y) = y 3 , φ = x2 y 3 + y 3 13. Py = −2xy 3 sin xy 2 + 2y cos xy 2 , Qx = −2xy 3 cos xy 2 − 2y sin xy 2 throughout the plane and the vector is not a conservative field.

15.3. INDEPENDENCE OF THE PATH

235

14. Py = −4xy(x2 + y 2 + 1)−3 = Qx throughout the plane and the vector field is a conservative 1 field. φx = x(x2 + y 2 + 1)−2 , φ = − (x2 + y 2 + 1)−1 + g(y), φy = y(x2 + y 2 + 1)−2 + g 0 (y) = 2 1 y(x2 + y 2 + 1)−2 , g(y) = 0, φ = − (x2 + y 2 + 1)−1 2 15. Py = 1 = Qx throughout the plane and the vector field is a conservative field. φx = x3 + 1 1 1 1 y, φ = x4 + xy + g(y), φy = x + g 0 (y) = x + y 3 , g(y) = − y 4 , φ = x4 + xy + y 4 4 4 4 4 16. Py = 4e2y , Qx = e2y throughout the plane and the vector field is not a conservative field. 17. Py = 0 = Qx , Px = 0 = Rx , Qz = −1 = Ry throughout 3-space and the vector field is a conservative field. ∂g φx = 2x, φ = x2 + g(y, z)φy = = 3y 2 − x, ∂y g(y, z) = y 3 − yz + h(z), φ = x2 + y 3 − yz + h(z), φz = −y + h0 (z) = −y, h(z) = 0, φ = x2 + y 3 − yz 18. Py = 2x = Qx , Pz = 0 = Rx , Qz = −e−y = Ry throughout 3-space and the vector field is a conservative field. ∂g φx = 2xy, φ = x2 y + g(y, z), φy = x2 + = x2 − ze−y , ∂y g = ze−y + h(z), φ = x2 y + ze−y + h(z), φz = e−y + h0 (z) = ey − 1, h(z) = −z, φ = x2 y + ze−y − z R 19. Since Py = −e−y = Qx , F is conservative and C F · dr is independent of the path. Thus, instead of the given curve we may use the simpler curve C1 : y = x, 0 ≤ x ≤ 1. Then Z (2x + e−y )dx + (4y − xe−y )dy W = C1 1

Z =

(2x + e−x )dx +

0

Z

1

(4x − e−x )dx

Integration by parts

0

1 1 = (x2 − e−x ) 0 + (2x2 + xe−x ) 0 = [(1 − e−1 ) − (−1)] + [(2 + e−1 + e−1 ) − (1)] = 3 + e−1 . R 20. Since Py = −e−y = Qx , F is conservative and C F · dr is independent of the path. Thus, instead of the given curve we may use the simpler curve C1 : y = 0 − 2 ≤ −x ≤ 2. Then dy = 0 and −2 R R −2 W = C1 (2x+e−y )dx+(4y −xe−y )dy = 2 (2x+1)dx = (x2 + x) 2 = (4−2)−(4+2) = −4. 21. Py = z = Qx , Qz = x = Ry , Rx = y = Pz , and the integral is independent of path. Parameterize the line segment between the points by x = 1 + t, y = 1 + 3t, z = 1 + 7t, for 0 ≤ t ≤ 1. Then dx = dt, dy = 3dt, dz = 7dt and Z (2,4,8) Z 1 yzdx + xzdy + xydz = [(1 + 3t)(1 + 7t) + (1 + t)(1 + 7t)(3) + (1 + t)(1 + 3t)(7)]dt (1,1,1)

0

Z = 0

1

1 (11 + 62t + 63t2 )dt = (11t + 31t2 + 21t3 ) 0 = 63.

236

CHAPTER 15. VECTOR INTEGRAL CALCULUS

22. Py = 0 = Qx , Qz = 0 = Ry , Rx = 0 = Pz and the integral is independent of path. Parameterize the line segment between the points by x = t, y = t, z = t, for 0 ≤ t ≤ 1. Then dx = dy = dz = dt and 1 R (1,1,1) R1 2xdx + 3y 2 dy + 4z 3 dz = 0 (2t + 3t2 + 4t3 )dt = (t2 + t3 + t4 ) 0 = 3. (0,0,0) 23. Py = 2x cos y = Qx , Qz = 0 = Ry , Rx = 3e3z = Pz , and the integral is independent of path. Integrating φx = 2x sin y + e3z we find φ = x2 sin y + xe3z + g(y, z). Then φy = x2 cos y + gy = Q = x2 cos y, so gy = 0, g(y, z) = h(z), and φ = x2 sin y + xe3z + h(z). Now φz = 3xe3z + h0 (z) = R = 3xe3z + 5, so h0 (z) = 5 and h(z) = 5z. Thus φ = x2 sin y + xe3z + 5z and Z (2,π/2,1) (2x sin y + e3z )dx + x2 cos ydy( 3xe3z + 5)dz (1,0,0)

(2,π/2,1) = (x2 sin y + xe3z + 5z) (1,0,0) = [4(1) + 2e3 + 5] − [0 + 1 + 0] = 8 + 2e3 . 24. Py = 0 = Qx , Qz = 0 = Ry , Rx = 0 = Pz , and the integral is independent of path. Parameterize the line segment between the points by x = 1 + 2t, y = 2 + 2t, z = 1, for 0 ≤ t ≤ 1. Then dx = 2dt, dz = 0 and (3,4,1)

Z

(1,2,1)

Z 1 1 [(2 + 4t + 1)2 + 3(2 + 2t)2 2]dt (2x + 1)dx + 3y 2 dy + dz = z 0 Z 1 1 = (24t2 + 56t + 30)dt = (8t3 + 28t2 + 30t) 0 = 66. 0

25. Py = 0 = Qx ; Qz = 0 = Ry , Rx = 2e2z = Pz and the integral is independent of path. Parameterize the line segment between the points by x = 1 + t, y = 1 + t, z = ln 3, for 0 ≤ t ≤ 1. Then dx = dy = dt, dz = 0 and Z

(2,2,ln 3) 2z

2

Z

2z

e dx + 3y dy + 2xe dz = (1,1,ln 3)

0

1

1 [e2 ln 3 + 3(1 + t)2 ]dt = [9t + (1 + t)3 ] 0 = 16

26. Py = 0 = Qx , Qz = 2y = Ry , Rx = 2x = Pz and the integral is independent of path. Parameterize the line segment between the points by x = −2(1 − t), y = 3(1 − t), z = 1 − t, for 0 ≤ t ≤ 1. Then dx = 2dt, dy = −3dt, dz = −dt, and Z

(0,0,0) 2

2

1

Z

[−4(1 − t)2 (2) + 6(1 − t)2 (−3)

2xzdx + 2yzdy + (x + y )dz = (−2,3,1)

0

+ 4(1 − t)2 (−1) + 9(1 − t)2 (−1)]dt 1 Z 1 2 3 = −39(1 − t) dt = 13(1 − t) = −13. 0

0

27. Py = 1 − z sin x = Qx , Qz = cos x = Ry , Rx = −y sin x = Pz and the integral is independent of path. Integrating θx = y − yz sin x we find θ = xy + yz cos x + g(y, z). Then θy = x + z cos x + gy (y, z) = Q = x + z cos x, so gy = 0, g(y, z) = h(z), and

15.3. INDEPENDENCE OF THE PATH

237

θ = xy + yz cos x + h(z). Now θz = y cos x + h(z) = R = y cos x, so h(z) = 0 and θ = xy + yz cos x. Since r(0) = 4j and r(π/2) = πi + j + 4k, Z π,1,4) F · dr = (xy + yz cos x)|(0,4,0) = (π − 4) − (0 + 0) = π − 4. C

28. P − y = 0 = Qx , Qz = 0 = Ry , Rz = −ez = Pz and the integral is independent of path. Integrating φx = 2 − ez we find φ = 2x − xez + g(y, z). Then φ − Y = gy = 2y − 1, so g(y, z) = y 2 − y + h(z) and φ = 2x − xez + y 2 − y + h(z). Now φz = −xez + hZ0 (z) = R = 2 − xez , so h0 (z) = 2, h(z) = 2z, and φ = 2x − xez + y 2 − y + 2z. Thus (2,4,8) F · dr = (2x − xez + y 2 − y + 2z) (−1,1,−1)

C

= (4 − 2e + 16 − 4 + 16) − (−2 + e−1 + 1 − 1 − 2) = 36 − 2e8 − e−1 8

29. Since Py = Gm1 m2 (2xy/|r|5 ) = Qx , Qz = Gm1 m2 (2yz/|r|5 ) = Ry , and Rx = Gm1 m2 (2xz/|r|5 ) = Pz , the force field is conservative. x , θ = Gm1 m2 (x2 + y 2 + z 2 )−1/2 + g(y, z), θx = −Gm1 m2 2 2 (x + y + z 2 )3/2 y y θy = −Gm1 m2 2 + gy (y, z) = −Gm1 m2 2 , g(y, z) = h(z), (x + y 2 + z 2 )3/2 (x + y 2 + z 2 )3/2 θ = Gm1 m2 (x2 + y 2 + z 2 )−1/2 + h(z), z z θz = −Gm1 m2 2 + h0 (z) = −Gm1 m2 2 , 2 2 3/2 2 (x + y + z ) (x + y + z 2 )3/2 Gm1 m2 Gm1 m2 h(z) = 0, θ = p = 2 2 2 |r| x +y +z 30. Since Py = 24xy 2 z = Qx , Qz = 12x2 y 2 = Ry , and Rx = 8xy 3 = Pz , F is conservative. Thus, the work done between two points is independent of the path. From θx = 8xy 3 z we obtain θ = 4x2 y 3 z which is a potential function for F. Then Z

√ (1, 3,π/3)

W = (2,0,0)

√ (1,√3,π/3) F · dr = 4x2 y 3 z (2,0,0) = 4 3π

R (0,2,π/2)

F · dr = 0. R R 31. Since F is conservative, C1 F · dr = −C2 F · dr. Then, since the simply closed curve C is composed of C1 and C2 , Z Z Z Z Z F · dr = F · dr + F · dr = F · dr − F · dr = 0. and W =

(2,0,0)

C

C1

C2

C1

c2

c1

−C2

32. From F = (x2 + y 2 )n/2 (xi + yj) we obtain Py = nxy(x2 + y 2 )n/2−1 = Qx , so that F is conservative. From θx = x(x2 + y 2 )n/2 we obtain the potential function θ + (x2 + y 2 )(n+2)/2 /(n + 2). Then Z

(x2 ,y2 )

W = (x1 ,y1 )

(x2 ,y2 ) i 1 h 2 (x2 + y 2 )(n+2)/2 F·dr = ( ) = (x2 + y22 )(n+2)/2 − (x21 + y12 )(n+2)/2 . n+2 n+2 (x1 ,y1 )

238

CHAPTER 15. VECTOR INTEGRAL CALCULUS

33. P − y = −2x sin y = Qx throughout the plane and the vector field F is a conservative field. The path starts at point (1, 0) and ends at point (2, 1). Since F is conservative, the integral is path independent so we can use any path C starting at (1, 0) and ending at (2, 1). Use the path y = x − 1, 1 ≤ x ≤ 2. Then Z

Z F · dr =

C

2

(2x cos(x − 1) − x2 sin(x − 1))dx

1

2 = x2 cos(x − 1) 1 = 4 cos 1 − 1 34. Py = cos y + Qx , P − z = 0 = Rx , Qz = 0 = Ry throughout the plane and the vector field F is a conservative field. The path starts at the point (0, 0, 0) and ends at the point (1, 1, 1). Since F is conservative, the integral is path independent so we can use any path C starting at (0, 0, 0) and ending at (1, 1, 1). Use the path y = z = x, 0 ≤ x ≤ 1. Then Z

Z F · dr =

1

(sin x + x cos x + x2 )dx

0

C

1 1 x3 = x sin x + = sin 1 + 3 0 3 35. F cannot be a conservative field in the region. 36. (a) Py =

y 2 − x2 = Qx . Using the hint, we have (y 2 + x2 )2 Z Z 2π Z F · dr = (sin2 t + cos2 t)dt = C

0

2π

dt = 2π.

0

Since C is a closed path, the integral would be zero if F were conservative. (b) Any simply connected region containing the path C would have to contain the origin. But F and the partials Py and Qx are not defined at the origin. Therefore, the theorem does not apply. dv dr dv 1 d 2 dp ∂p dx 37. From Problem 45 in Exercises 15.2, · = ·v = v . Then, using = + dt dt dt 2 dt dt ∂x dt dr ∂p dy = ∇p · , we have ∂y dt dt Z Z Z dv dr m · drdtdt + ∇p · = 0dt dt dt Z Z 1 d 2 dp m v dt + dt = constant 2 dt dt 1 mv 2 + p = constant. 2 38. By Problem 37, the sum of kinetic and potential energies in a conservative force field is constant. That is, it is independent of points A and B, so p(B) + K(B) = p(A) + K(A).

15.4. GREEN’S THEOREM

15.4

239

Green’s Theorem

1. The sides of the triangle are C1 : y = 0, 0 ≤ x ≤ 1; C2 : x = 1, 0 ≤ y ≤ 3; C Z3 : y = 3x, 0 ≤ −x ≤Z 1.1 Z 3 Z 0 Z (x − y)dx + xydy = xdx + ydy + (x − 3x)dx + C

0

0

1

y

0

x(3x)dx

x=1

1

 1   3 1 0 1 2 1 2 x + y + (−x2 ) 0 + (3x2 ) 1 = 2 2 0 0 1 9 1 = + +1−3=3 2 2  3x  Z 1 Z Z Z 1 Z 1 Z 3x 9 2 1 2 (y + 1)dydx = (y + 1)dA = y + y dx = x + 3x dx 2 2 0 0 0 R 0 0  1  3 3 3 2 x + x =3 = 2 2 0 

x

2. The sides of the rectangle are C1 : y = 0, −1 ≤ x ≤ 1; C2 : x = 1, 0 ≤ y ≤ 1; C3 : yZ = 1, 1 ≥ x ≥ −1; C4 : Z x = −1, 1Z≥ y ≥ 0. Z 1 Z 1 1 1 2 2 3x ydx+(x − 5y)dy = 0dx + (1 − 5y)dy = 0dx + (1 − 5y)dy −1

C

−1

0

0

   1 −1 5 = 3x2 dx + (1 − 5y)dy = y − y 2 + x3 1 + y − 2 1 1 01 RR R1R1 R1 2 R1 2 2 3 (2x − 3x )dA = (2x − 3x )dxdy = (x − x ) dy = 0 (−2)dy R 0 −1 0 Z

−1

Z

0

−1

Z

2

2

Z

2π

−y dx + x dy =

3. C

Z

2

= −2

2π

(−9 sin t)(−3 sin t)dt + 0

 1 5 2 y = −2 2 0

9 cos2 t(3 cos t)dt

0 2π

Z

[(1 − cos2 t) sin t + (1 − sin2 t) cos t]dt

= 27 0

 2π  1 1 = 27 − cos t + cos3 t + sin t − sin3 t = 27(0) = 0 3 3 Z Z Z 2π Z 3 Z 2π0 Z 3 (2x + 2y)dA = 2 (r cos θ + r sin θ)rdrdθ = 2 r2 (cos θ + sin θ)drdθ R

0

Z =2 0

0 2π



0

0

 3 Z 2π 1 3 r (cos θ + sin θ) dθ = 18 (cos θ + sin θ)dθ 3 0 0 2π

= 18(sin θ − cos θ)|0 = 18(0) = 0

240

CHAPTER 15. VECTOR INTEGRAL CALCULUS

4. The sides of the region are C1 : y √ = 0, 0 ≤ x ≤ 2; C2 : y = −x + 2, 2 ≥ x ≥ 1; C3 : y = x, 1 ≥ x ≥ 0. Z 1 Z Z 2 −2(−x + 2)2 dx 0dx + −2y 2 dx + 4xydy = C

y 2

2

0 1

Z

4x(−x + 2)(−dx)

+

R

2 0

Z

Z −2xdx +

+

0

√



4x x

1

1

1 √ 2 x

 dx

2

x

2 8 10 + +1−1= 3 3 3   1 RR R 1 R 2−y R1 10 8 3 2 2 4 8ydA = 8ydxdy = 8y(2 − y − y )dy = 8y − y − 2y = 3 R 0 y2 0 3 0 =0+

5. PZ = 2y, Py = 2, Q Z =Z 5x, Qx = 5 2ydx + 5xdy = (5 − 2)dA C Z RZ =3 dA = 3(25π) = 75π

y

x

R

R

6. PZ = x + y 2 , Py = 2y, Q = 2x2Z−Zy, Qx = 4x (x + y 2 )dx + (2x2 − y)dy = (4x − 2y)dA C

y

4

R

Z

2

4

Z

R

(4x − 2y)dydx

= x2

−2 2

Z

4 (4xy − y ) dx 2

= −2 Z 2

2

x

x2

(16x − 16 − 4x3 + x4 )dx

= −2

 =

 2 96 1 5 8x − 16x − x + x = − 5 5 −2 2

4

y

7. P = x4 − 2y 3 , Py = −6y 2 , Q = 2x3 − y 4 , Qx = 6x2 . Using polar coordinates, Z Z Z 4 3 3 4 (x − 2y )dx + (2x − y )dy = (6x2 + 6y 2 )dA C

R 2π Z 2

Z =

2

6r rdrdθ 0

Z = 0

0 2π



 2 Z 2π 3 4 r dθ = 24dθ = 48π. 2 0 0

2

R

r=2

2

x

15.4. GREEN’S THEOREM

241 y

8. RP = x − 3y, Py = −3, Q = 4x R R+ y, Qx = 4 (x − 3y)dx + 4(x + y)dy = (4 + 3)dA = 7(10) = 70 C R

3

R

3

y

9. P = 2xy, Py = 2x, Q = 3xy 2 , Qx = 3y 2 Z Z Z Z 2xydx + 3xy 2 dy = (3y 2 − 2x)dA = C

R

Z = 1

=

2

Z

2x

(3y 2 − 2x)dydx

4

2

1

y=2x

2x Z 2 2 3 (8x3 − 4x2 − 8 + 4x)dx (y − 2xy) dx =

R

1

2

    2 40 4 16 56 − − = 2x4 − x3 − 8x + 2x2 = 3 3 3 3 1

2x 2x 10. RP = e2x sin 2y, Py = 2e2x cos 2y, Q R R= e cos 2y, Qx = 2e cos 2y 2x 2x = e sin 2ydx + e cos 2ydy = 0dA = 0 C R

11. P = xy, Py = x, Q = x2 , Qx = 2x. Using polar coordinates, Z Z Z Z π/2 Z 1 2 xydx + x dy = (2x − x)dA = r cos θrdrdθ C

−π/2

R

Z

x

π/2

= −π/2



1 3 r cos θ 3

2

x

y

1

r=1

0

 1 Z π/2 1 cos θdθ dθ = −π/2 3

R

0

1

x

π/2 2 1 = = sin θ 3 3 −π/2 2 2 y 12. P = ex , Py = 0, Q = 2 tan−1 x, Qx = 2 1 + Z Z Z Zx 0 Z 1 2 2 2 1 ex dx + 2 tan−1 xdy = dA = dydx 2 2 1 + x 1 + x R C R −1 −x y=-x  1 Z 0 2y dx = 1 + x2 −x −1 -1  Z 0 2x 2 + dx = 1 + x2 1 + x2 −1  π  π 0 = [2 tan−1 x + ln(1 + x2 )] −1 = 0 − − + ln 2 = − ln 2 2 2

x

242

CHAPTER 15. VECTOR INTEGRAL CALCULUS

1 13. P = y 3 , Py = y 2 , Q = xy + xy 2 , Qx = y + y 2 3 Z Z Z Z 1/√2 Z 1−y2 1 3 2 ydxdy y dx + (xy + xy )dy = ydA = C 3 0 y2 R 1−y2 Z 1/√2 = (xy) dy 2 0

0

=

Z =− 0

 =

1

R x2

(xy)

x=1-y2

1

x

 1/√2 1 1 1 2 1 4 1 = − = y − y 2 2 4 8 8 0

14. P = xy 2 , Py = 2xy, Q = 3 cos y, Qx = 0 Z Z Z Z xy 2 dx + 3 cos ydy = (−2xy)dA = − C

R

(y − y 2 − y 3 )dy

= 

x=y2

1

y

√ 1/ 2

Z

y

0

Z

y 1

Z

x2

2xydydx x3

y=x2

1

dx = −

(x3 − x4 )dx

R

0

x3

 1 1 1 4 1 5 x − x =− 4 5 20

y=x3

x

0

15. P R = ay, Py = a, RQR= bx, Qx = b. aydx + bxdy = (b − a)dA = (b − a) × (area bounded by C) C R 16. P = P (x), Py = 0, Q = Q(y), Qx = 0.

R C

P (x)dx + Q(y)dy =

RR R

0dA = 0

R RR 17. For the first integral: P = 0, Py = 0, Q = x, Qx = 1; C Rxdy = − RR R1dA =area of R. For the second integral: R R P = y, Py = 1, Q = 0, Q = 0; − C ydx = − R −dA =area of R. Thus, C xdy = − C ydx. 18. P = −y, Py = −1, Q + x, Qx = 1. Z Z

Z

19. A =

dA = R



= 3a2

= ab

a cos3 t(3a sin2 t cos tdt) = 3a2

0

Z

2π

0

 2π 1 1 3 1 t− sin 4t + sin3 2t = πa2 16 64 48 8 0 Z dA =

R

2π

xdy = C

Z Z 20. A = 

Z

RR 1RR 1R −ydx + xdy = 2dA = dA =area of R C R R 2 2

Z xdy =

C

2π

Z a cos t(b cos tdt) = ab

0

 2π 1 1 t + sin 2t = πab 2 4 0

0

2π

cos2 tdt

sin2 t cos4 tdt

15.4. GREEN’S THEOREM

243

21. (a) Parameterize C by x = x1 + (x2 − x1 )t and y = y1 + (y2 − y1 )t for 0 ≤ t ≤ 1. Then Z Z 1 Z 1 −ydx + xdy = −[y1 + (y2 − y1 )t](x2 − x1 )dt + [x1 + (x2 − x1 )t](y2 − y1 )dt C

0

0

1 1 1 1 2 2 = −(x2 − x1 )[y1 t + (y2 − y1 )t ] + (y2 − y1 )[x1 t + (x2 − x1 )t ] 2 2 0   0   1 1 = −(x2 − x1 ) y1 + (y2 − y1 ) + (y2 − y1 ) x1 + (x2 − x1 ) 2 2 = x1 y2 − x2 y1 . (b) Let Ci be the line segment from (xi , yi ) to (xi+1 , yi+1 ) for i = 1, 2, · · · , n − 1, and C2 the lineZsegment from (xn , yn )to (x1 , y1 ). Then 1 A= −ydx + xdy Problem 18 2 C "Z # Z Z Z 1 = −ydx + xdy + −ydx + xdy + · · · + −ydx + xdy + −ydx + xdy 2 C1 C2 Cn−1 Cn =

1 1 1 1 (x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + (xn−1 yn − xn yn−1 ) + (xn y1 − x1 yn ). 2 2 2 2

22. From part  (b) of Problem 21    1 1 1 1 A= (−1)(1) − (1)(3)] + [(1)(2) − (4)(1) + (4)(5) − (3)(2)] + [(3)(3) − (−1)(5) 2 2 2 2 1 = (−4 − 2 + 14 + 14) = 11. 2

23. P = 4x2 − y 3 , Py = −3y 2 ; Q = x3 + y 2 , Qx = 3x2 . Z Z Z Z (4x2 − y 3 )dx + (x3 + y 2 )dy = (3x2 + 3y 2 )dA = C

Z = 0

R 2π

2π

Z

0

2 2

Z

3r (rdrdθ) = 1

0

2π



 2 3 4 r dθ 4 1

45 45π dθ = 4 2

p 24. PI = cos2 x − y, Py = −1; Q = Zy 2Z+ 1, Qx = 0 Z Z p (cos2 x − y)dx + y 2 + 1dy = (0 + 1)dA = dA C R R √ = (6 2)2 − π(2)(4) = 72 − 8π

1 25. We first observe that Py + (y 4 − 3x2 y 2 )/(x2 + y 2 )3 = Qx . Letting C 0 be the circle x2 + y 2 = 4 we have

244

CHAPTER 15. VECTOR INTEGRAL CALCULUS Z C

−y 3 dx + xy 2 dy = (x2 + y 2 )2

−y 3 dx + xy 2 dy (x2 + y 2 )2

Z C0

1 1 1 1 cos t, dx = − sin tdt, y = sin t, dy = cos tdt 4 4 4 4      1 1 1 1 1 3 2 Z 2π − sin t − sin tdt + cos t sin t cos tdt 64 4 4 16 4 = 1/256 0 Z 2π Z 2π = (sin4 t + sin2 t cos2 t)dt = (sin4 t + (sin2 t − sin4 t)dt x=

0

0 2π

Z

sin2 tdt =

=



0

 2π 1 1 t − sin 2t = π 2 4 0

26. We first observe that Py = [4y 2 − (x + 1)2 ]/[(x + 1)2 + 4y 2 ]2 = Qx . Letting C 0 be the ellipse 2 2 (x Z + 1) + 4y = 4 we have Z −y x+1 −y x+1 dx+ dy = dx + dy 2 2 2 2 2 2 (x + 1) + 4y (x + 1)2 + 4y 2 C (x + 1) + 4y C 0 (x + 1) + 4y x + 1 = 2 cos t, dx = −2 sin tdt, y = sin t, dy = cos tdt  Z 2π  Z 2 cos t 1 2π − sin t = (−2 sin t) + cos t dt = (sin2 t + cos2 t)dt = π. 4 4 2 0 0 RR 2 RR 27. Writing x dA = (Qx − Py )dA we identify Q = 0 and P = −x2 y. Then, with R R C : x = 3 cos t, y = 2 sin t, 0 ≤ t ≤ 2π, we have Z Z

x2 dA = R

Z

Z P dx + Qdy =

C

=

54 4

27 = 4

−x2 ydx = −

C

2π

9 cos2 t(2 sin t)(−3 sin t)dt

0

2π

Z

Z

4 sin2 t cos2 tdt =

0

27 2

Z

2π

sin2 2tdt =

0

27 4

Z

2π

(1 − cos 4t)dt 0

  2π 27π 1 t − sin 4t = . 4 2 0

RR RR 28. Writing [1 − 2(y − 1)]dA = (Qx − Py )dA we identify Q = x and P = (y − 1)2 . Then, R R with Z Z C1 : x = cos t, y −Z1 = sin t, −π/2 ≤ Z t ≤ π/2, and C2Z: x = 0, 2 ≥ y ≥ 0, Z [1 − 2(y − 1)]dA = R

P dx + Qdy + C1 Z π/2

=

[sin2 t(− sin t) + cos t cos t]dt =

−π/2 π/2

P dx + Qdy = C2

(y − 1)2 dx + xdy +

C1 Z π/2

[cos2 t − (1 − cos2 t) sin t]dt

−π/2

 1 (1 + cos 2t) − sin t + cos2 t sin t dt = −π/2 2   π/2 1 1 1 π  π π = t + sin 2t + cos t − cos3 t = − − = . 2 4 3 4 4 2 −π/2 Z



0dy C2

15.5. PARAMETRIC SURFACES AND AREA

245

29. P = x − y, Py = −1, Q = x + y, Qx = 1; R RR 3π 3 W = C F · dr = 2dA = 2 × area = 2( ) = π R 4 2 30. P = −xy 2 , Py = −2xy, Q = x2 y, Qx = 2xy. Using polar coordinates, 2 Z Z Z Z π/2 Z π/2 Z 2 W = F · dr = 4(r cos θ)(r sin θ)rdrdθ = (r4 cos θ sin θ) dθ 4xydA = C 1 0 R 0 1 π/2 Z π/2 15 15 = sin2 θ . = 15 sin θ cos θdθ = 2 2 0 0 31. Let P = 0 and W = x2 . Then Qx − Py = 2x and RR I Z Z xdA 1 1 2 R x dy = 2xdA = = x. 2A C 2A A R Let P = y 2 and Q = 0. Then Qx − Py = −2y and RR I I Z Z ydA 1 1 2 R y dx = − −2ydA = = y. − 2A C 2A C A R 32. Using Green’s Theorem, Z Z Z Z Z W = F · dr = −ydx + xdy = 2dA = 2 C

C

R

2π

0

Z

1+cos θ

rdrdθ 0

 1+cos θ Z 2π 1 2 dθ = (1 + 2 cos θ + cos2 θ)dθ r 2 0 0 0   2π 1 1 = θ + 2 sin θ + θ + sin 2θ = 3π. 2 4 0 Z

2π



=2

RB 33. Since A P dx + Qdy is independent of path, Py = Qx by Theorem 17.3. Then, by Green’s Theorem Z Z Z Z Z P dx + Qdy = (Qx − Py )dA = 0dA = 0. C

R

R

34.

15.5

Parametric Surfaces and Area

1. x = u, y = v, z = 4u + 3v − 2, −∞ < u < ∞, −∞ < v < ∞ 2. x = u, y = 1 − 2u, z = v, −∞ < u < ∞, −∞ < v < ∞ √ 3. x = u, y = − 1 + u2 + v 2 , z = v, −∞ < u < ∞, −∞ < v < ∞ 4. x = u, y = v, z = 5 − u2 − v 2 , −∞ < u < ∞, −∞ < v < ∞ 5. r(u, v) = ui + vj + (1 − v 2 )k, −2 ≤ u ≤ 2, −3 ≤ v ≤ 3

246

CHAPTER 15. VECTOR INTEGRAL CALCULUS

6. r(u, v) = 2 cos ui + 3 sin uj + vk, 0 ≤ u ≤ 2π, −∞ < v < ∞ 7. x2 + y 2 = cos2 u + sin2 u = 1, circular cylinder 8. z = x2 + y 2 , paraboloid 9. x = sin u, y = sin u cos v, z = sin u sin v y 2 + z 2 = sin2 u cos2 v + sin2 u sin2 v = sin2 (cos2 v + sin2 v) = sin2 u = z 2 , so x2 = y 2 + z 2 , portion of a circular cone 10. x = 2 sin φ cos θ, y = 3 sin φ sin θ, z = 4 cos φ, y2 z2 x2 + + = sin2 φ cos2 θ + sin2 φ sin2 θ + cos2 φ 4 9 16 = sin2 (cos2 θ + sin2 θ) + cos2 φ = sin2 φ + cos2 φ = 1, ellipsoid 11. Surface is parameterzied by x = u, y = sin v, z = cos v so R is defined by 0 ≤ u ≤ 4, 0 ≤ v ≤ π2 12. Surface is parameterzied by x = u, 2, − π2 ≤ v ≤ π2

y = sin v,

z = cos v so R is defined by −2 ≤ u ≤

13. Surface is parameterzied by x = sin φ cos θ, y = sin φ sin θ, z = cos φ so R is defined by 0 ≤ θ ≤ 2π, π2 ≤ φ ≤ π 14. Surface is parameterzied by x = sin φ cos θ, y = sin φ sin θ, z = cos φ so R is defined by 0 ≤ θ ≤ π, 0 ≤ φ ≤ π2 √ 15. At u = π/6, v = 2, we have x = 5, y = 5 3, z = 2. √ ∂x π
∂z π
∂y π
, 2 = 5 3, , 2 = −5, ,2 = 0 ∂u 6 ∂u 6 ∂u 6 ∂y π
∂z π
∂x π
, 2 = 0, , 2 = 0, , 2 = 1. ∂v 6 ∂v 6 ∂v 6 i j k √ √ A normal vector is given by n = 5 3 −5 0 = −5i − 5 3j. 0 0 1 √ √ √ The tangent plane is −5(x − 5) − 5 3(y − 5 3) = 0 or x + 3y = 20. 16. At u = 1, v = 0, we have x = 1, y = 0, z = 1. ∂y ∂z ∂x (1, 0) = 1, (1, 0) = 0, (1, 0) = 2 ∂u ∂u ∂u ∂x ∂y ∂z (1, 0) = 0, (1, 0) = 1, (1, 0) = 0. ∂v ∂v ∂v i j k A normal vector is given by n = 1 0 2 = −2i + k. 0 1 0 The tangent plane is −2(x − 1) + (z − 1) = 0 or −2x + z = −1.

15.5. PARAMETRIC SURFACES AND AREA

247

17. At u = 1, v = 2, we have x = 3, y = 3, z = −3. ∂x ∂y ∂z (1, 2) = 2, (1, 2) = 1, (1, 2) = 2 ∂u ∂u ∂u ∂x ∂y ∂z (1, 2) = 1, (1, 2) = 1, (1, 2) = −4. ∂v ∂v ∂v i j k A normal vector is given by n = 2 1 2 = −6i + 10j + k. 1 1 −4 The tangent plane is −6(x − 3) + 10(y − 3) + (z + 3) = 0 or −6x + 10y + z = 9. √

we have x = −4, y = 32 , z = 3 2 3 . √

∂y ∂z −1, π3 = −3, −1, π3 = −3 3 ∂u ∂u √
−3 3 ∂z
3 ∂y π , −1, 3 = −1, π3 = . ∂v 2 ∂v 2 i j k √ 4 √ −3 −3 3 √ = −18i − 6j − 6 3k. A normal vector is given by n = 3 0 −3 3 2 2  √  √ √ The tangent plane is −18(x + 4) − 6(y − 32 ) − 6 3 z − 3 2 3 = 0 or 3x + y + 3z = −6.

18. At u = −1, v =
∂x −1, π3 = 4, ∂u
∂x −1, π3 = 0, ∂v

π 3,

19. At u = 3, v = 3, we have x = 3, y = 3, z = 9. ∂x ∂y ∂z (3, 3) = 1, (3, 3) = 0, (3, 3) = 3 ∂u ∂u ∂u ∂x ∂y ∂z (3, 3) = 0, (3, 3) = 1, (3, 3) = 3. ∂v ∂v ∂v i j k A normal vector is given by n = 1 0 3 = −3i − 3j + k. 0 1 3 The tangent plane is −3(x − 3) − 3(y − 3) + (z − 9) = 0 or 3x + 3y − z = 9. √

√

20. At u = 1, v = π/4, we have x = 22 , y = 22 , z = 1. √ √


2 ∂y 2 ∂z ∂x 1, π4 = , 1, π4 = , 1, π4 = 1 ∂u ∂u 2√ ∂u √2


2 ∂y 2 ∂z ∂x π π 1, 4 = , 1, 4 = − , 1, π4 = 0. ∂v 2 ∂v ∂v 2 i √ √ j k √ √ 2 2 2 2 1 = A normal vector is given by n = √2 i+ j − k. 2 √ 2 2 2 2 − 22 0 √ √ ! √ √ ! √ 2 2 2 2 The tangent plane is x− + y− − (z − 1) = 0 or x + y − 2z = 0. 2 2 2 2 21. At u = −2, v = 1, we have x = −1, y = 3, z = −2. ∂x ∂y ∂z (−2, 1) = 1, (−2, 1) = −1, (−2, 1) = 1 ∂u ∂u ∂u ∂x ∂y ∂z (−2, 1) = 1, (−2, 1) = 1, (−2, 1) = −2. ∂v ∂v ∂v

248

CHAPTER 15. VECTOR INTEGRAL CALCULUS i j k A normal vector is given by n = 1 −1 1 = i + 3j + 2k. 1 1 −2 The tangent plane is (x + 1) + 3(y − 3) + 2(z + 2) = 0 or x + 3y + 2z = 4.

22. At u = 0, v = ln 3, we have x = 0, y = ln 3 + 1, z = 3. ∂x ∂y ∂z (0, ln 3) = ln 3, (0, ln 3) = 1, (0, ln 3) = 1 ∂u ∂u ∂u ∂x ∂y ∂z (0, ln 3) = 0, (0, ln 3) = 1, (0, ln 3) = 3. ∂v ∂v ∂v i j k A normal vector is given by n = ln 3 1 1 = 2i − 3 ln 3j + ln 3k. 0 1 3 The tangent plane is 2(x − 0) − 3 ln 3(y − ln 3 − 1) + ln 3(z − 3) = 0 or 2x − 3(ln 3)y + (ln 3)z = −3(ln 3)2 . 23. At (1, 7, 5), we have u = 2, v = 1. ∂x ∂y ∂z (2, 1) = 1, (2, 1) = 2, (2, 1) = 4 ∂u ∂u ∂u ∂x ∂y ∂z (2, 1) = −1, (2, 1) = 3, (2, 1) = 2. ∂v ∂v ∂v i j k A normal vector is given by n = 1 2 4 = −8i − 6 ln 3j + 5k. −1 3 2 The tangent plane is −8(x − 1) − 6(y − 7) + 5(z − 35) = 0 or 8x + 6y − 5z = 25. 24. At (1, 3, 16), we have u = 4, v = 1. ∂y ∂z ∂x (4, 1) = 0, (4, 1) = 1, (4, 1) = 8 ∂u ∂u ∂u ∂x ∂y ∂z (4, 1) = 2, (4, 1) = −1, (4, 1) = 0. ∂v ∂v ∂v i j k A normal vector is given by n = 0 1 8 = 8i − 16j − 2k. 2 −1 0 The tangent plane is 8(x − 1) − 16(y − 3) − 2(z − 16) = 0 or 4x − 8y − z = −36. 25.

∂r ∂r = h2, 1, i, = h−1, 1, 0i ∂u ∂v i j k ∂r ∂r × = 2 1 1 = h−1, −1, 3i ∂u ∂v −1 1 0 √ ∂r √ ∂r ∂u × ∂v = 1 + 1 + 9 = 11 √ R2R1 √ A = 0 −1 11 dv du = 4 11

26. Let x = u, y = √ v, z = 1 − u − v. √ R 1 R 1−u2 p Then A = −1 −√1−u2 1 + (−1)2 + (−1)2 dv du = 3π 27.

∂r ∂r = h1, 0, 2ui, = h0, 1, 2vi ∂u ∂v

15.5. PARAMETRIC SURFACES AND AREA i j k ∂r ∂r × = 1 0 2u = h−2u, −2v, 1i ∂u ∂v 0 1 2v ∂r √ ∂r 2 2 ∂u × ∂v = 4u + 4v + 1 Since 0 ≤ z √≤ 4, we have 0 ≤ u2 + v 2 ≤ 4. So Z 2 Z 4−u2 p A= 4u2 + 4v 2 + 1 dv du √ −2

− 4−u2

√4−u2 p vp 2 (4u2 + 1) 2 2 2 4u + 4v + 1 + = ln |2v + 4u + 4v + 1| √ 4 −2 2 − 4−u2 i h p p √ 1 = 2(4u2 + 1) ln |2 4 − u2 + 17| − (4u2 + 1) ln(4u2 + 1) + 4 −17(u2 − 4) 4 Z 2π Z 2 p = 4r2 + 1r drdθ polar transformation Z

2

0

0

2 Z 2π √ (4r2 + 1)3/2 17 17 − 1 = dθ dθ = 12 12 0 0 0 2π √ √ 17 17 − 1 (17 17 − 1)π = θ = 12 6 Z

2π

0

28.

∂r ∂r = hcos θ, sin θ, 1i, = h−r sin θ, r cos θ, 0i ∂r ∂θ i j k ∂r ∂r sin θ 1 = h−r cos θ, −r sin θ, ri × = cos θ ∂r ∂θ r sin θ r cos θ 0 ∂r ∂r p √ √ 2 2 2 2 2 2 2 ∂r × ∂θ = r cos θ + r sin θ + r = r + r = r 2. √ R 2 R 2π √ A = 0 0 r 2 dθ dr = 4π 2

29. r = (r cos θ)i + (r sin θ)j + θk ∂r ∂r = hcos θ, sin θ, 0i, = h−r sin θ, r cos θ, f i ∂r ∂θ i j k ∂r ∂r cos θ sin θ 0 = hsin θ, − cos θ, ri × = ∂r ∂θ −r sin θ r cos θ 1 ∂r ∂r p 2 √ 2 2 2 ∂r × ∂θ = sin θ + cos θ + r = 1 + r . √ √ R 2π R 2 √ A = 0 0 1 + r2 dr dθ = 2 5π + π ln(2 + 5) 30. r = (a sin φ cos θ)i + (a sin φ sin θ)j + (a cos φ)k ∂r ∂r = h−a sin φ sin θ, a sin φ cos θ, 0i, = ha cos φ cos θ, a cos φ sin θ, −a sin φi ∂θ ∂φ

249

250

CHAPTER 15. VECTOR INTEGRAL CALCULUS i ∂r ∂r × = − sin φ sin θ ∂θ ∂φ a cos φ cos θ

j a sin φ cos θ a cos φ sin θ

k 0 −a sin φ



= h−a2 sin2 φ cos θ, −a sin2 φ sin θ, −a2 sin φ cos φi q ∂r ∂r 4 4 2 2 4 2 4 4 2 ∂θ × ∂φ = a sin φ cos θ + a sin φ sin θ + a sin φ cos θ q q 4 2 4 4 2 = a sin φ + a sin φ cos φ = a4 sin2 φ = a2 sin φ R π R 2π 2 A = 0 0 a sin φdθdφ = 4a2 π 31. We have a = 2, so x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ, R π R 2π A = π/3 0 4 sin φ dθ dφ = 12π 32. x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ, R π R 2π A = π/3 0 4 sin φ dθ dφ = 4π

π 3

π 2,

≤φ≤

33. x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ, 0 ≤ φ ≤ √ R π/4 R 2π A= 0 4 sin φ dθ dφ = 4π(2 − 2) 0

π 4,

π 3

≤ φ ≤ π, 0 ≤ θ ≤ 2π,

0 ≤ θ ≤ 2π,

0 ≤ θ ≤ 2π,

34. x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ; the spehre intersects the cylinder when z 2 = 2, so the region outside the cylinder is described by π4 ≤ φ 3π 4 , 0 ≤ θ ≤ 2π √ R 3π/4 R 2π A = π/4 0 4 sin φdθdφ = 8π 2 35.

(b)

(a)

(c)

z

z

1

1

1

1

y

1

y

y

x

(b)

(c)

z

z

y x

1

1 x

(a)

37. (f )

1

1

x

36.

z

z

y x

y x

15.5. PARAMETRIC SURFACES AND AREA

251

38. (e) 39. (d) 40. (a) 41. (c) 42. (b) 43. z

y

x

44.

∂r = h− cos φ cos θ, − cos φ sin θ, − sin φi, ∂φ ∂r = h(sin φ − R) sin θ, (R − s ∈ φ) cos θ, 0i ∂θ i j ∂r ∂r − cos φ sin θ × = − cos φ cos θ ∂φ ∂θ (sin φ − R) sin θ (R − sin φ) cos θ

k − sin φ 0



= h(r − sin φ) sin φ cos θ, (R − sin φ) sin φ sin θ, −(R − sin φ) cos φi q ∂r ∂r 2 2 2 2 2 2 2 2 ∂φ × ∂θ = (R − sin φ) sin φ cos θ + (R − sin φ) sin φ sin θ + (R − sin φ) cos φ q = (R − sin φ)2 sin2 φ + (R − sin φ)2 cos2 φ p = (R − sin φ)2 = R − sin φ R 2π R 2π A = 0 0 R − sin φdθdφ = 4π 2 R 45. x = 2u, y = 2v, z = 8u + 6v − 2, −∞ < u < ∞, −∞ < v < ∞ 46. The surface area of a circular cylinder with height h and radius r is A = 2πrh. The surface pictures is one quarter of a circular cylinder with height 4 and radius 1. Therefore, A = 2π. 47. We have r = ui + f (u) cos vj + f (u) sin vk. ∂r ∂r = h1, f 0 (u) cos v, f 0 (u) sin vi, = h0, −f (u) sin v, f (u) cos vi, ∂u ∂v i j k ∂r ∂r × = 1 f 0 (u) cos v f 0 (u) sin v ∂u ∂v 0 −f (u) sin v f (u) cos v = hf (u)f 0 (u), −f (u) cos v

−f (u) sin vi

252

CHAPTER 15. VECTOR INTEGRAL CALCULUS q ∂r ∂r 2 0 2 2 2 2 ∂u × ∂v = [f (u)f (u)] + [f (u)] cos v + [f (u)] sin v p = [f (u)]2 [f 0 (u)]2 + [f (u)]2 p = f (u) 1 + [f 0 (u)]2 p p R b R 2π Rb By (11), A = a 0 f (u) 1 + [f 0 (u)]2 dvdu = 2π a f (u) 1 + [f 0 (u)]2 du

48. (a) x = u, y = sin u cos v, z = sin u sin v, −2π ≤ u ≤ 2π 0 ≤ v ≤ 2π (b)

z

y

x (c) Let S1 be the surface corresponding to the parameter domain 0 ≤ u ≤ pi, 0 ≤ v ≤ 2π. Then Z π p A(S1 ) = 2π sin(x) 1 + cos2 xdx 0  √ √ √  = ln( 2 + 1) − ln( 2 − 1) = 2 2 π √ √
√ Finding the area of the entire surface S, we have A(S) = 4A(S1 ) = ln( 2 + 1) − ln( 2 − 1) = 2 2 4π 49. The surface is a plane passing through the point (x0 , y0 , z0 ) with a normal vector n = v1 × v2 . 50. x = 5 sin φ cos θ + 2, y = 5 sin φ sin θ + 3, z = 5 cos φ + 4

15.6

Surface Integrals √

z

4x2 dA

1. zx = −2x, zy = 0; dS = 1 + Z Z Z 4 Z √2 p Z xdS = x 1 + 4x2 dxdy = S

0

Z = 0

0 4

13 26 dy = 6 3

0

4

√2 1 2 3/2 (1 + 4x ) dy 12 0

2

z=2-x2 R

2 x

4

y

15.6. SURFACE INTEGRALS

253

2. See Problem 1. Z Z Z Z Z xy(9 − 4z)dS = xy(1 + 4x2 )dS = S

S

= 0

3. zx = p

x x2

y2

0

, zy = p

y x2

y2

; dS =

√

+ + Using Z Z polar coordinates, Z Z √ xz 3 dS = x(x2 + y 2 )3/2 2dA S

4

242 121 ydy = 20 10

Z

4

0

2dA.

121 ydy = 10



2π

1

√ Z 2

2π

2π

0 2π

0

x

R

1

Z

r7/2 cos θdrdθ

y2

y

x

1 2 9/2 r cos θ dθ 9 0 2π √ 2 2 2 cos θdθ = sin θ = 0. 9 9 0

y

√

z

2dA. + + Using Z Z polar coordinates, Z Z p √ (x + y + z)dS = R(x + y + x2 + y 2 ) 2dA x2

1

1

0

0

√ Z = 2 √ Z = 2

(r cos θ)r3/2 rdrdθ

0

0

=

z=Mx2+y2

1

Z

 4 484 1 2 y = 2 5 0

z

R

√ Z = 2

4. zx = p

xy(1 + 4x2 )3/2 dxdy

0

0

0

√ 2

Z

√2 Z y 2 5/2 (1 + 4x ) dy = 20

4

Z

4

, zy = p

x2

y2

; dS =

z=Mx2+y2

4

S

=

√ Z 2 0

√ Z = 2 0

2π

Z

4

(r cos θ + r sin θ + r)rdrdθ 1 2π Z

R

4

r2 (1 + cos θ + sin θ)drdθ

4

y

4 x

1

4 1 3 r (1 + cos θ + sin θ) dθ 3 1

√ Z 2π 2 0 √ Z 2π 2π √ √ 63 2 = (1 + cos θ + sin θ)dθ = 21 2(θ + sin θ − cos θ) = 42 2π. 3 0 0 =

254

CHAPTER 15. VECTOR INTEGRAL CALCULUS x 36 − x2 − y 2 , zx = − p , 36 − x2 − y 2 y ; zy = − 2 − y2 36 − x s y2 x2 dS = 1 + + dA 36 − x2 − y 2 36 − x2 − y 2 6 =p dA. 36 − x2 − y 2 Using polar coordinates,

5. z =

p

z 6 z=M36-x2-y2

6

R

y

6 x

Z Z

2

Z Z

2

(x + y )zdS = S

p (x + y ) 36 − x2 − y 2 p 2

R

Z

2π

=6 0

2

−1

√ Z = 2 =

√

36 − x2 = y 2

z 1

−1

0 1 2

1−x2 2 y(x + 1) dx

z=x+1

2

2

(1 + 2x − 2x3 − x4 )dx

1

1

√  1 √ 8 2 1 4 1 5 2 = 2 x+x − x − x = 2 5 5 −1 p 7. zx = −x, zy = −y; dS = 1 + x2 + y 2 dA Z Z Z 1Z 1 p xydS = xy 1 + x2 + y 2 dxdy

y

R

x

z 2

0

1 1 2 2 3/2 = y(1 + x + y ) dy 3 0 0  Z 1 1 1 = y(2 + y 2 )3/2 − y(1 + y 2 )3/2 dy 3 3 0   1 1 1 2 5/2 2 5/2 = (2 + y ) − (1 + y ) 15 15 1

0

=

2

0

(1 − x )(x + 1) dx

−1 Z 1

0

r2 rdrdθ

0

0



Z

6

Z

dA = 6

−1

S

2π

6 Z 2π 1 4 r dθ = 6 324dθ = 972π. 4 0 0

√ 6. zx = 1, zy = 0; dS = 2dA Z Z Z 1 Z 1−x2 √ Z √ z 2 dS = (x + 1)2 2dydx = 2 S

Z

6

1 5/2 (3 − 27/2 + 1) 15

z=2-x2/2-y2/2

1

1 x

y

15.6. SURFACE INTEGRALS

255 z

1 1 1 8. z = + x2 + y 2 , zx = x, zy = y; 2 2 2 p 1 + x2 + y 2 dA. Using Z Z polar coordinates, Z Z p 2zdS = (1 + x2 + y 2 ) 1 + x2 + y 2 dA S

Z

R π/2 Z 1

=

1 z=1/2+x2/2+y2/2

1

p (1 + r ) 1 + r2 rdrdθ 2

1

R

y

x

0

π/3

Z

dS =

π/2

Z

=

1

(1 + r2 )3/2 rdrdθ

0

π/3

1 Z 1 π/2 5/2 1 2 5/2 (2 − 1)dθ = (1 + r ) dθ = 5 π/3 π/3 5 0 √ √ 4 2 − 1  π π  (4 2 − 1)π = − = . 5 2 3 30 Z

π/2

z

√

3

9. yZx = = 1 + 4x2 dA Z 2x, yz = 0; ZdS3 Z 2 p √ 24 yzdS = 24xz 1 + 4x2 dxdz S

0

Z =

y=x2

R

0

2 dz

3

4

2 3/2

2z(1 + 4x ) 0

y

2 x

0

= 2(173/2 − 1)

3

Z

zdz = 2(173/2 − 1)



0

 3 1 2 z 2 0

= 9(173/2 − 1) z

p 10. xy = −2y, xz = −2z; dS = 1 + 4y 2 + 4z 2 dA Using polar coordinates, Z Z Z π/2 Z 2 (1 + 4y 2 + 4z 2 )1/2 dS = (1 + 4r2 )rdrdθ S

0

Z

π/2

0

1 16

R

1

= =

2

2 1 2 2 (1 + 4r ) dθ 16

2

1

Z

π/2

12dθ = 0

3π . 8

1 (6 − x − 3z). 2 √ p 1 3 14 yz = − , yz = − ; dS = 1 + 1/4 + 9/4 = . 2 2 2

11. Write the equation of the surface as y =

3 x

z=4-y2-z2

y

256

CHAPTER 15. VECTOR INTEGRAL CALCULUS Z Z

 √ 14 1 2 3z + 4z (6 − x − 3z) dxdz 2 2 0 0 6−3z √ Z 2  2  14 2 dz 3z x − z(6 − x − 3z) 2 0 0 √ Z 2 14 [3z 2 (6 − 3z) − 0] − [0 − z(6 − 3z)2 ]dz 2 0 2 √ √ √ Z 2 √ 14 14 14 2 2 3 (36z − 18z )dz = (18z − 6z ) = (72 − 48) = 12 14 2 2 2 0

Z

2

(3z + 4yz)dS = S

= = =

2

Z

6−3z

0

12. Write the equation of the surface as x = 6 − 2y − 3z. Then xy = −2, xz = −3; dS = √ √ 1 + 4 + 9 = 14. 3−3z/2 Z Z Z 2 Z 3−3z/2 √ Z 2 √ (3z 2 + 4yz)dS = dz (3yz + 2y 2 z) (3z 2 + 4yz) 14dydz = 14 S

0

0

0

0

      √ Z 2 z 2 z 45 2 9 3 + 18z 1 − dz = 14 9z 1 − 27z − z + z dz = 14 2 2 2 2 0 0  2  √ √ √ 27 2 15 3 9 4 z − z + z = 14(54 − 60 + 18) = 2 14 = 14 2 2 8 √

Z

2

0

z 1

z=1-x-y 2

13. The density is ρ = kx . The √ surface is z = 1 − x − y. Then zx = −1, zy = −1; dS = 3dA. Z Z Z 1 Z 1−x √ √ Z 1 1 3 1−x 2 2 m= kx dS = k x 3dydx = 3k x dx S 0 0 0 3 1 0 √ Z 1 √  √  1 x 3 3 1 3 3 4 = k (1 − x) dx = k − (1 − x) = k 3 3 4 12 0

1

y

0

z

2 z=M4-x2-y2

R 2 x

2

y

15.6. SURFACE INTEGRALS

257

y x , zy = − p ; dS = 14. zx = − p 2 2 4−x −y 4 − x2 − y 2 s y2 2 x2 + dA = p dA. 1+ 4 − x2 − y 2 4 − x2 − y 2 4 − x2 − y 2 Using symmetry and polar coordinates, Z Z Z π/2 Z 2 2 m=4 (r2 cos θ sin θ) √ rdrdθ |xy|dS = 4 4 − r2 0 S 0 Z π/2 Z 2 r2 (4 − r2 )−1/2 sin 2θ(rdr)dθ =4 0

0

u = 4 − r2 , du = −2rdr, r2 = 4 − u   Z π/2 Z 0 Z π/2 Z 0 1 −1/2 (4 − u)u sin 2θ − du dθ = −2 (4u−1/2 − u1/2 ) sin 2θdudθ =4 2 4 4 0 0  0  Z π/2  Z π/2  2 3/2 32 1/2 = −2 8u − u − sin 2θ dθ sin 2θdθ = −2 3 3 0 0 4   π/2 64 64 1 = = − cos 2θ . 3 2 3 0 15. The surface is g(x, y, z) = y 2 + z 2 − 4 = 0. ∇g = 2yj + 2zk, p yi + zk |∇g| = 2 y 2 + z 2 ; ∇ p ; y2 + z2 yz 3yz 2yz + p = p ;; z = F·∇ = p 2 2 2 2 y +z y +z y2 + z2 p 4 − y 2 , zx = 0, s y y2 2 zy = − p dA = p ; dS = 1 + dA 2 2 4 − y 4 − y2 Z 4Z− y Z Z 3yz 2 p p Flux = F · ndS = dA 2 + z2 y 4 − y2 S R p Z Z 3y 4 − y 2 2 p p = dA y2 + 4 − y2 4 − y2 R 2 Z 3Z 2 Z 3 Z 3 3 2 = y dx = 6dx = 18 3ydydx = 0 0 0 0 2 0

z 2 z=M4-y2

R 3 x

2

y

258

CHAPTER 15. VECTOR INTEGRAL CALCULUS

16. The surface is g(x, y, z) = x2 +py 2 + z − 5 = 0. ∇g = 2xi + 2yj + k, |∇g| = 1 + 4x2 + 4y 2 ; n = z 2xi + 2yj + k p ; F · n = p ; zx = 1 + 4x2 + 4y 2 1 + 4x2 + 4y 2 p −2x, zy = −2y, dS = 1 + 4x2 + 4y 2 dA. Using polar coordinates, Z Z Z Z p z p Flux = 1 + 4x2 + 4y 2 dA F · ndS = 1 + 4x2 + 4y 2 S R Z Z = (5 − x2 − y 2 )dA R 2π Z 2

Z

5 z=5-x2-y2

2

2 2

(5 − r )rdrdθ

=

z

x

y

R

0

0 2π

Z

 2 Z 2π 5 2 1 4 6dθ = 12π. r − r dθ = 2 4 0 0



= 0

2xi + 2yj + k 2x2 + 2y 2 + z 17. From Problem 16, n = p . Then F · n = p . Also, from Problem 1 + 4x2 + 4y 2 1 + 4x2 + 4y 2 p 16, dS = 1 + 4x2 + 4y 2 dA. Using polar coordinates, Z Z Z Z Z Z 2x2 + 2y 2 + z p 2 2 p Flux = F · ndS = 1 + 4x + 4y dA = (2x2 + 2y 2 + 5 − x2 − y 2 )dA 1 + 4x2 + 4y 2 S R R  2 Z 2π Z 2 Z 2π  Z 2π 1 4 5 2 2 = (r + 5)rdrdθ = 14dθ = 28π. r + r dθ = 4 2 0 0 0 0 0 18. The surface is g(x, y, z) = z − x − 3 = 0. ∇g = √ −i + k −i + k, |∇g| = 2; n = √ ; 2 √ 1 1 F · n = √ x3 y + √ xy 3 zx = 1, zy = 0, dS = 2dA. 2 2 Using polar Z Z coordinates,Z Z Z Z √ 1 3 3 √ (x y + xy ) 2dA = Flux = F · ndS = xy(x2 + y 2 )dA 2 S R R Z π/2 Z 2 cos θ = (r2 cos θ sin θ)r2 rdrdθ 0

5 z=x+3

1 y

0 π/2

Z

z

Z

=

5

r cos θ sin θdrdθ 0

Z = 0

r=2cos θ

2

2 cos θ

x R

0 π/2

2 cos θ   π/2 Z 1 6 1 π/2 32 1 4 7 8 r cos θ sin θ dθ = 64 cos θ sin θdθ = − cos θ = . 6 6 0 3 8 3 0 0

15.6. SURFACE INTEGRALS

259 z

19. The surfacep is g(x, y, z) = x2 + y 2 + z − 4. ∇g = 2xi + 2yj + k, |∇g| = 4x2 + 4y 2 + 1; 2xi + 2yj + k x3 + y 3 + z n = p ; F·n = p ; zx = 4x2 + 4y 2 + 1 4x2 + 4y 2 + 1 −2x, p zy = −2y, 1 +Z4x2 + 4y 2 dA.Z Using polar coordinates, dS = Z Z 3 Flux = F · ndS = (x + y 3 + z)dA R Z ZS 2 2 = (4 − x − y + x3 + y 3 )dA R 2π Z 2

Z =

4 z=4-x2-y2

2

2

y

R

x

(4 − r2 + r3 cos3 θ + r3 sin3 θ)rdrdθ

0

0

 2 1 5 3 1 4 1 5 3 2r − r + r cos θ + r sin θ dθ = 4 5 5 0 0  Z 2π  32 32 2π cos3 θ + sin3 θ dθ = 4θ|0 + 0 + 0 = 8π. = 4+ 5 5 0 Z

2π



2

z

20. The surface is g(x,√ y, z) = x + y + z −√6. ∇g = i + j + k, |∇g| √ = 3; n = (i + j + k)/ 3; F · n = y x (e zx = −1, zy = −1, dS = √ + e + 18y)/√ 3; 1 + 1 + 1dA = 3dA. Z Z Z Z Flux = F · ndS = (ey + ex + 18y)dA S 6

Z

Z

= 0

Z

6

0 6

=

z=6-x-y

R

6

y

6 x

r 6−x

(ey + ex + 18y)dydx

0

= Z

6

6−x dx (ey + yex + 9y 2 ) 0

[e6−x + (6 − x)ex + 9(6 − x)2 − 1]dx

0

6 = [−e6−x + 6ex − xex + ex − 3(6 − x)3 − x] 0 = (−1 + 6e6 − 6e6 + e6 − 6) − (−e6 + 6 + 1 − 648) = 2e6 + 634 ≈ 1440.86

p 21. For S1 : g(x, y, z) = x2 + y 2 − z, ∇g = 2xi + 2yj − k, |∇g| = 4x2 + 4y 2 + 1; n1 = p 2xy 2 + 2x2 y − 5z 2xi + 2yj − k p ; F · n1 = p ; zx = 2x, zy = 2y, dS1 = 1 + 4x2 + 4y 2 dA. 2 2 2 2 4x + 4y + 1 4x + 4y + 1 For S2 : g(x, y, z) = z − 1, ∇g = k; |∇g| = 1; n2 = k; F · n2 = 5z; zx = 0, zy = 0, dS2 = dA. Using polar coordinates and R : x2 + y 2 ≤ 1 we have

260

CHAPTER 15. VECTOR INTEGRAL CALCULUS Z Z

Z Z

Z Z

2

2

Z Z

F · n1 dS1 + F · n2 dS2 = (2xy + 2x y − 5z)dA + S2 R Z ZS1 = [2xy 2 + 2x2 y − 5(x2 + y 2 ) + 5(1)]dA

Flux =

R 2π Z 1

Z =

5zdA R

(2r3 cos θ sin2 θ + 2r3 cos2 θ sin θ − 5r2 + 5)rdrdθ

0

0

 1 2 5 2 5 5 r cos θ sin2 θ + r5 cos2 θ sin θ − r4 + r2 dθ 5 5 4 2 0 0 2π    2π Z 2π  2 2 1 1 5 5 2 3 2 3 = (cos θ sin θ + cos θ sin θ) + dθ = sin θ − cos θ + θ 5 4 5 3 3 4 0 0 0    2 1 1 5 5 = − − − + π = π. 5 3 3 2 2 p 22. For S1 : g(x, y, z) = x2 + y 2 + z − 4, ∇g = 2xi + 2yj + k, |∇g| = 4x2 + 4y 2 + 1; n1 = p p 2xi + 2yj + k p ; F·n1 = 6z 2 / 4x2 + 4y 2 + 1; zx = −2x, zy = −2y, dS1 = 1 + 4x2 + 4y 2 dA. 4x2 + 4y 2 + 1 p For S2 : g(x, y, z) = x2 + y 2 − z, ∇g = 2xi + 2yj − k, |∇g| = 4x2 + 4y 2 + 1; n2 = p p 2xi + 2yj − k p ; F·n2 = −6z 2 / 4x2 + 4y 2 + 1; zx = 2x, zy = 2y, dS2 = 1 + 4x2 + 4y 2 dA. 4x2 + y 2 + 1 2 2 Using polar Z Z Z R Z : x + y ≤ 2 ZweZhave Z Z coordinates and F · n2 dS2 = 6z 2 dA + −6z 2 dA F · n1 dS1 + Flux = Z

2π



=

R

S1

S1

Z Z

2

2 2

2

2 2

Z

2π

√

Z

[6(4 − x − y ) − 6(x + y ) ]dA = 6

= R

Z =6 0

0

2π

2

[(4 − r2 )2 − r4 ]rdrdθ

0

  √2 Z 2π Z 2π √ 6 1 1 2 3 6 3 3 − (4 − r ) − r dθ = − [(2 − 4 ) + ( 2) ]dθ = 48dθ = 96π. 6 6 0 0 0

23. The surface is g(x, y, z) = x2 + y 2 + z 2 − a2 = 0. ∇g = 2xi + 2yj + 2zk, |∇g| = p xi + yj + zk xi + yj + zk 2 x2 + y 2 + z 2 ; n = p ; F · n = −(2xi + 2yj + 2zk) · p = 2 2 2 x +y +z x2 + y 2 + z 2 p 2x2 + 2y 2 + 2z 2 −p = −2 x2 + y 2 + z 2 = −2a. 2 2 x2 + R Ry + z Flux = −2adS = −2a × area = −2a(4πa2 ) = −8πa3 S 24. n1 = k, n2 = −i, n3 = j, n4 = −k, n5 = i, n6 = −j; F · n1 = z = 1, F · n2 = −x = 0, F · nR3 R= y = 1, RF R· n4 = −z R=R0, F · n5 = x = 1, F · n6 = −y = 0; Flux = 1dS + 1dS + 1dS = 3 S1 S3 S5 a xi + yj + zk and dS = p dA. 25. Refering to the solution to Problem 23, we find n = p x2 + y 2 + z 2 a2 − x2 − y 2

Now

F · n = kq

r r kq kq kq kq · = 4 |r|2 = 2 = 2 = 2 |r|3 |r| |r| |r| x + y2 + z2 a

15.6. SURFACE INTEGRALS

261

Z Z and

Z Z F · ndS =

Flux = S

S

26. We are given σ = kz. Now zx − p y

kq kq kq dS = 2 × area = 2 (4πa2 ) = 4πkq. a2 a a

x 16 − x2 − y 2

,

z

zy = − p ; 2 2 s 16 − x − y x2 y2 + dA dS = 1 + 16 − x2 − y 2 16 − x2 − y 2 4 =p dA 16 − x2 − y 2 UsingZpolar Z coordinates, Z Z p 4 Q= 16 − x2 − y 2 p dA kzdS = k 16 − x2 − y 2 S R Z 2π Z 3 = 4k rdrdθ 0

Z

4

z=M16-x2-y2

3

R

y

3 x

0 2π

= 4k 0

3 Z 2π 9 1 2 r dθ = 4k dθ = 36πk. 2 0 2 0

27. The surface√ is z = 6 − √ 2x − 3y. Then zx = −2, zy = −3, dS = 1 + 4 + 9 = 14dA. The area of the surface is Z Z

Z

A(s) =

3

Z

dS = S

0

0

2−2x/3

√

14dydx =

√

Z 14 0

3



z 6



2 2 − x dx 3

z=6-2x-3y 3 √ 1 2 = 14(2x − x ) = 3 14. 3 0 Z Z Z 3 Z 2−2x/3 √ 1 1 2 y x= √ 14xdydx xdS = √ 3 R 3 14 3 14 0 0 s 2−2x/3  Z Z  x 1 3 1 3 2 2 = dx = xy 2x − x dx 3 0 3 0 3 0  3  2 1 x2 − x3 = 1 = 3 9 0 2−2x/3 Z Z Z 3 Z 2−2x/3 √ Z 1 1 1 3 1 2 √ √ ydS = y= 14ydydx = y dx 3 0 2 0 3 14 3 14 0 0 S 2   3 Z  1 3 2 1 1 2 2 = 2 − x dx = − (2 − x)3 = 6 0 3 6 2 3 3 0 Z Z Z 3 Z 2−2x/3 √ 1 1 z= √ zdS = √ (6 − 2x − 3y) 14dydx 3 14 3 14 0 0 S  2−2x/3    3 Z 3 Z  1 3 2 1 3 2 2 1 2 3 2 = 6y − 2xy − y dx = 6 − 4x + x dx = 6x − 2x + x = 2 3 0 2 3 0 3 3 9 0 0 The centroid is (1, 2/3, 2).

√

262

CHAPTER 15. VECTOR INTEGRAL CALCULUS

28. The area of the hemisphere is A(s) = 2πa2 . By symmetry, s x = y = 0. y x2 x y2 , zy = − p ; dS = 1 + 2 zx = − p + 2 dA = 2 2 a −x −z a − x2 − y 2 a2 − x2 − y 2 a 2 − x2 − y 2 a p dA a2 − x2 − y 2 UsingZ polar coordinates,Z Z Z Z 2π Z a p 1 a 1 zdS 2 2 2 rdrdθ z= = a −x −y p dA = 2 2πa2 2πa 0 a2 − x2 − y 2 0 S 2πa R a Z 2π Z 2π 1 1 2 1 1 2 a = r dθ = s dθ = . 2πa 0 2 0 2πa 0 2 2 The centroid is (0, 0, a/2). p 2 2 2 2 29. (a) The region pin the xy-plane is x + y ≤ 16. From zx = −x/ x + y and 2 2 zy = −y/ x + y we see that dS =

p √ 1 + x2 /(x2 + y 2 ) + y 2 /(x2 + y 2 )dA = 2da Z Z

and

A(S) =

dS = S

Z Z √

2dA =

√

√ 2π42 = 16 2π.

R

Then

Z Z √ Z 2π Z 4 1 1 x= dS = √ 2xdA = r cos θrdrdθ 16π 0 16 2π 16 2π S R 0 4 Z 2π Z 2π 4 1 3 1 r cos θ dθ = cos θdθ = 0 = 16π 0 3 3π 0 0 Z Z Z Z √ Z 2π Z 4 1 1 1 ydS = √ y= √ 2ydA = r sin θrdrdθ 16π 0 16 2π 16 2π S R 0 4 Z 2π Z 2π 4 1 1 3 = r cos θ dθ = sin θdθ = 0 16π 0 3 3π 0 0 Z 2π Z 4 Z Z Z Z √ p 1 1 1 z= √ zdS = √ 2(4 − x2 + y 2 )dA = (4 − r)rdrdθ 16π 0 16 2π 16 2π S R 0  4 Z 2π  Z 2π 2 1 3 4 1 2 2r − r dθ = dθ = . = 16π 0 3 3π 3 0 0 The centroid is (0,0,4/3). Z Z

1 √

√ Z Z √ Z 2 2 (x + y )kdS = k 2 (x + y )dA = k 2

Z Z

2

(b) Iz =

2

S

R

4 √ Z √ Z 2π √ k 2 2π 4 = r dθ = 64k 2 dθ = 128kπ 2 4 0 0

0

2π

Z

4

r2 rdrdθ

0

0

30. The surface is g(x, y, z) = z − f (x, y) = 0. ∇g = −fx i − fy j + k, |∇g| =

q

fx2 + fy2 + 1; n =

15.7. CURL AND DIVERGENCE

263

−fx i − fy j + k −P fx − Qfy + R q ; F·n= q ; 1 + fx2 + fy2 1 + fx2 + fy2 q dS = 1 + fx2 + fy2 dA RR R R −P fx − Qfy + R q RR q F · ndS = 1 + fx2 + fy2 dA = (−P fx − Qfy + R)dA s R R 1 + fx2 + fy2

15.7

Curl and Divergence

1. curlF = (x − y)i + (x − y)j; divF = 2z 2. curlF = −2x2 i + (10y − 18x2 )j + (4xz − 10z)k; divF = 0 3. curlF = 0; divF = 4y + 8z 4. curlF = (xe2y + ye−yz + 2xye2y )i − ye2y j + 3(x − y)2 k; divF = 3(x − y)2 − ze−yz 5. curlF = (4y 3 − 6xz 2 )i + (2x3 − 3x2 )k; divF = 6xy 3 6. curlF = −x3 zi + (3x2 yz − z)j + ( x2 y 2 − y − 15y 2 )k; divF = (x3 y − x) − (x3 y − x) = 0 2 7. curlF = (3e−z − 8yz)i − xe−z j; divF = e−z + 4z 2 − 3ye−z 8. curlF = (2xyz 3 + 3y)i + (y ln x − y 2 z 3 )j + (2 − z ln x)k; divF =

yz − 3z + 3xy 2 z 2 x

9. curlF = (xy 2 ey + 2xyey + x3 yez + x3 yzez )i − y 2 ey j + (−3x2 yzez − xex )k; divF = xyex + yex − x3 zez 10. curlF = (5xye5xy + e5xy + 3xz 3 sin xz 3 − cos xz 3 )i + (x2 y cos yz − 5y 2 e5xy )j + (−z 4 sin xz 3 − x2 z cos yz)k; divF = 2x sin yz 11. div r = 1 + 1 + 1 = 3 i j k 12. curl r = ∂/∂x ∂/∂y ∂/∂z = 0i − 0j + 0k = 0 x y z i      j k  ∂ ∂ ∂ ∂ ∂ ∂ a a a 13. a×∇ = 1 2 3 = a2 ∂z − a3 ∂y i+ a3 ∂x − a1 ∂z j+ a1 ∂y − a2 ∂x k ∂/∂x ∂/∂y ∂/∂z i j k ∂ ∂ ∂ ∂ ∂ ∂ − a3 a3 − a1 a1 − a2 (a × ∇) × r = a2 ∂z ∂y ∂x ∂z ∂y ∂x x y z = (−a1 − a1 )i − (a2 + a2 )j + (−a3 − a3 )k = −2a   ∂ ∂ ∂ + a2 + a3 r 14. ∇ × (a × r) = (∇ · r)a − (∇ · a)r = (1 + 1 + 1)a − a1 ∂x ∂y ∂z = 3i − (a1 i + a2 j + a3 k) = 2a

264

CHAPTER 15. VECTOR INTEGRAL CALCULUS

∂/∂x 15. ∇ · (a × r) = a1 x i 16. ∇ × r = ∂/∂x x

∂/∂y a2 y

j ∂/∂y y

∂/∂z a3 z

k ∂/∂z z

= ∂ (a2 z − a3 y) − ∂ (a1 z − a3 x) + ∂ (a1 y − a2 x) = 0 ∂x ∂y ∂z

= 0; a × (∇ × r) = a × 0 = 0

j k y z = (a3 y − a2 z)i − (a3 x − a1 z)j + (a2 x − a1 y)k; r · r = x2 + y 2 + z 2 a2 a3 i j k ∂/∂y ∂/∂z ∇ × [(r · r)a] = ∂/∂x (r · r)a1 (r · r)a2 (r · r)a3

i 17. r × a = x a1

= (2ya3 − 2za2 )i − (2xa3 − 2za1 )j + (2xa3 − 2ya1 )k = 2(r × a) 18. r · a = a1 x + a2 y + a3 z; r · r = x2 + y 2 + z 2 ; ∇ · [(r × r)a] = 2xa1 + 2ya2 + 2za3 = 2(r · a) 19. Let F = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k and G = S(x, y, z)i + T (x, y, z)j + U (x, y, z)k. ∇ · (F + G) = ∇ · [(P + S)i + (Q + T )j + (R + U )k] = Px + Sx + Qy + Ty + Rx + Uz = (Px + Qy + Rz ) + (Sx + Ty + Uz ) = ∇ · F + ∇ · G 20. Let F = P (x, y, z)i and G = S(x, y, z)i + T (x, y, z)j + U (x, y, z)k. + Q(x, y, z)j + R(x, y, z)k j j k ∂/∂z ∇ × (F + G) = ∂/∂x ∂/∂y P +S Q+T R+U = (Ry + Uy − Qz − Tz )i − (Rx + Ux − Pz − Sz )j + (Qx + Tx − Py − Sy )k = (Ry − Qz )i − (Rx − Rz )j + (Qx − Py )k + (Uy − Tz )i − (Ux − Sz )j + (Tx − Sy )k =∇×F+∇×G 21. ∇ · (f F) = ∇ · (f P i + f Qj + f Rk) = f Px + P fx + f Qy + Qfy + f Rz + Rfz = f (Px + Qy + Rz ) + (P fx + Qfy + Rfz ) = f (∇ · F) + F · (∇f ) j 22. ∇ × (f F) = ∂/∂x fP

j ∂/∂y fQ

k ∂/∂z fR



= (f Ry + Rfy − f Qz − Qfz )i − (f Rx + Rfx − f Pz − P fz )j (f Qx + Qfx − f Py − P fy )k = (f Ry − f Qz )i − (f Rx − f Pz )j + (f Qx − f Py )k + (Rfy − Qfz )i − (Rfx − P fz )j + (Qfx − P fy )k i f [(Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k] + fx P

j fy Q

k fz R

= f (∇ × F) + (∇f ) × F

15.7. CURL AND DIVERGENCE

265

23. Assuming continuous second partial derivatives, j curl(gradf ) = ∇ × (fx i + fy j + fz k) = ∂/∂x fx

j ∂/∂y fy

k ∂/∂z fz



= (fzy − fyz )i − (fzx − fxz )j + (fyx − fxy )k = 0. 24. Assuming continuous second partial derivatives, div(curlF) = ∇ · [(Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k] = (Ryx − Qzx − (Rxy − Pzy ) + (Qxz − Pyz ) = 0. 25. Let F = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k and G = S(x, y, z)i + Y (x, y, z)j + U (x, y, z)k. i j k F × G = P Q R = (QU − RT )i − (P U − RS)j + (P T − QS)k S T U div (F × G) = (QUx + Qx U − RTx − Rx T ) − (P Uy + Py U − RSy − Ry S) + (P Tz + Pz T − QSz − Qz S) = S(Ry − Qz ) + T (Pz − Rx ) + U (Qx − Py ) − P (U − y − Tz ) − Q(Sz − Ux ) − R(Rx − Sy ) = G · (curlF) − F · (curlG) 26. Using Problems 20 and 23, curl(curlF + gradf ) = ∇ × (curlF + gradf ) = ∇ × (curlF) + ∇ × (gradf ) = curl(curlF) + curl(gradf ) = curl(curlF) + 0 = curl(curlF). 27. curl F = −8yzi − 2zj − xk; curl (curl F) = 2i − (8y − 1)j + 8zk 28. For F = P i + Qj + Rk, curl (curl F) = (Qxy − Pyy − Pzz + Rxz )i + (Ryz − Qzz − Qxx + Pyx )j + (Pzx − Rxx − Ryy + Qzy )k and −∇2 F + grad(divF) = −(Pxx + Pyy + Pzz )i − (Qxx + Qyy + Qzz )j − (Rxx + Ryy + Rzz )k + grad(Px + Qy + Rz ) = −Pxx i − Qyy j − Rzz k + (−Pyy − Pzz )i + (−Qxx − Qzz )j + (−Rxx − Ryy )k + (Pxx + Qyx + Rzx )i + (Pxy + Qyy + Rzy )i + (Pxz + Qyz + Rzz )k = (−P − P + Q + R)i + (−Qxx − Qzz + Pxy + Rzy )j + (−Rxx − Ryy + Pxz + Qyz )k. Thus, curl(curlF) = −∇2 F + grad(divF).

266

CHAPTER 15. VECTOR INTEGRAL CALCULUS

29. fz = 6x + 4y − 9z; fxx = 6; fy = 10y + 4z; fyy = 10; fz = −9x − 16z; fzz = −16; ∇2 f = fxx + fyy + fzz = 6 + 10 − 16 = 0 30. Using Problem 21, ∇ · (f ∇f ) = f (∇ · ∇f ) + ∇f · ∇f = f (∇2 f ) + |∇f |2 . 31. fx = 6x + 4y − 9z; fxx = 6; fy = 10y + 4x; fyy = 10; fz = −9x − 16z; fzz = −16; ∇2 f + fx x + fy y + zz z = 6 + 10 − 16 = 0 (a − x)A ∂f = 3/2 2 ∂x [(x − a) + (y − b)2 + (z − c)2 ]  2 2 2 A 2(x − a) − (y − b) − (z − c)2 ∂ f = 5/2 ∂x2 [(x − a)2 + (y − b)2 + (z − c)2 ] ∂f (b − y)A = 3/2 2 ∂y [(x − a) + (y − b)2 + (z − c)2 ]  2 2 2 A 2(y − b) − (x − a) − (z − c)2 ∂ f = 5/2 ∂y 2 [(x − a)2 + (y − b)2 + (z − c)2 ] (c − z)A ∂f = 3/2 2 ∂z [(x − a) + (y − b)2 + (z − c)2 ]  A 2(z − c)2 − (x − a)2 − (y − b)2 ∂2f = 5/2 ∂z 2 [(x − a)2 + (y − b)2 + (z − c)2 ] 2 2 2 ∂ f ∂ f ∂ f Now + 2 + 2 = 0. Hence f is harmonic. ∂x2 ∂y ∂z   1 4xy 4xy 33. fx = − 2 =− 2 2 2 2 2 (x + y − 1) (x + y − 1)2 + 4y 2 4y 1+ 2 2 2 (x + y − 1) 12x4 y − 4y 5 + 8x2 y 3 − 8x2 y − 8y 3 − 4y [(x2 + y 2 − 1)2 + 4y 2 ]4y − 4xy[4x(x2 + y 2 − 1)] fxx = − = [(x2+ y 2 − 1) + 4y 2 ]2 [(x2 + y 2 − 1)2 + 4y 2 ]2  2(x2 + y 2 − 1) − 4y 2 1 2(x2 + y 2 − 1)2 fy = = 2 (x2 + y 2 − 1)2 (x + y 2 − 1)2 + 4y 2 4y 2 1+ 2 2 2 (x + y − 1) [(x2 + y 2 − 1)2 + 4y 2 ](−4y) − 2(x2 + y 2 − 1)2 [4y(x2 + y 2 − 1)2 + 8y] fyy = [(x2 + y 2 − 1)2 + 4y 2 ]2 4 5 2 3 −12x y + 4y − 8x y + 8x2 y + 8y 3 + 4y = [(x2 + y 2 − 1)2 + 4y 2 ]2 ∇2 f = fxx + fyy = 0

32.

34.

∂f ∂2f = 4x3 − 12xy 2 , = 12x2 − 12y 2 , ∂x ∂x2 ∂f ∂2f = −12x2 y + 4y 3 , = −12x2 + 12y 2 , ∂y ∂y 2 ∂2f ∂2f Now + 2 = 0. Hence f is harmonic. 2 ∂x ∂y

35. Using Problems 25 and 23, ˙ ∇ · F = div (∇f × ∇g) = ∇g (curl ∇f ) − ∇f · (curl g) = ∇g · 0 − ∇f · 0 = 0.

15.7. CURL AND DIVERGENCE

267

36. Recall that a · (a × b) = 0. Then, using Problmems 25, 23, and 22, ∇ · F = (∇f × f ∇g) = f ∇g · (curl ∇f ) − ∇f · (curl f ∇g) = f ∇g · 0 − ∇f · (∇ × f ∇g) = −∇f · [f (∇ × ∇g) + (∇f × ∇g)] = −∇f · [f curl ∇g + (∇f × ∇g)] = −∇f · [f 0 + (∇f × ∇g)] = −∇f · (∇f × ∇g) = 0.

2 37. The surface is g(x, y) = x2 + y 2 + z p4z − 4 = 0. 2 ∇g = 2xi + 2yj + 8zk, |∇g| = 2 x + y 2 + 16z 2 ; 1 z=M1-x2/4-y2/4 2xi + 2yj + 8zk xi + yj + 4zk n= p =p ; 2 x2 + y 2 + 16z 2 x2 + y 2 + 16z 2 2 y 12z(x2 + y 2 ) 2 2 R ∇ × F = (3x − 3y )k, (∇ × F) · n = p x2 + y 2 + 16z 2 2 r=2 Writing the equation of the surface as z = p 2 2 1 − x /4 − y /4, we have x x zx = − p , zy = 4 1 − x2 /4 − y 2p /4 y 16 − 3x2 − 3y 2 − p , and dS = p dA. 2 2 4 1 − x /4 − y /4 2 4 − x2 − y 2 Then, using polar coordinates, p Z Z Z Z 16 − 3x2 − 3y 2 12z(x2 − y 2 ) p p dA Flux = (∇ × F) · ∇dS = x2 + y 2 + 16z 2 2 4 − x2 − y 2 S R p Z Z p 6 1 − x2 /4 − y 2 /4(x2 − y 2 ) 16 − 3x2 − 3y 2 dA p p = x2 + y 2 + 16 − 4x2 − 4y 2 4 − x2 − y 2 R Z π/4 Z 2 p Z π/4 Z 2 p = 1 − r2 /4(r2 cos2 θ − r2 sin2 θ)rdrdθ 4 − r2 = 3r2 cos 2θdrdθ 0

Z

0 π/4

= 0

0

0

2 Z π/4 3 4 π/4 r cos 2θ dθ = 12 cos 2θdθ = 6 sin 2θ|0 = 6. 4 0 0

i j k 1 1 1 ∂/∂x ∂/∂y ∂/∂z 38. curl v = curl (ω × r) = 2 2 2 ω2 z − ω3 y ω3 x − ω1 z ω1 y − ω2 x 1 = [(ω1 + ω1 )i − (ω2 − ω2 )j + (ω3 + ω3 )k] = ω1 i + ω2 j + ω3 k = ω 2 i j k 39. curl F = −Gm1 m2 ∂/∂x ∂/∂y ∂/∂z x/|r|3 y/|r|3 z/|r|3 = −Gm1 m2 [(−3yz/|r|5 + 3yz/|r|5 )i − (−3xz/|r|5 + 3xz/|r|5 )j + (−3xy/|r|5 + 3xy/|r|5 )k] =0  div F = −Gm1 m2

 −2x2 + y 2 + z 2 x2 − 2y 2 + z 2 x2 + y 2 − 2z 2 + + =0 |r|5/2 |r|5/2 |r|5/2

268

CHAPTER 15. VECTOR INTEGRAL CALCULUS

40. (a) Expressing the vertical component of V in polar coordinates, we have 2xy 2r2 sin θ cos θ sin 2θ = = 2 2 2 4 (x + y ) r r2 Similarly, x2 − y 2 r2 (cos2 θ − sin2 θ) cos 2θ = = . (x2 + y 2 )2 r4 r2 Since lim (sin 2θ)/r2 = lim (cos 2θ)/r2 , V ≈ Ai for r large or (x, y) far from the origin. r→∞ r→∞   x2 − y 2 2Axy (b) Identify P (x, y) = A 1 − 2 , Q(x, y) = − 2 , and R(x, y) = 0, we (x − y 2 )2 (x − y 2 )2 have Py =

2Ay(3x2 − y 2 ) 2Ay(3x2 − y 2 ) , Q = , x (x2 + y 2 )3 (x2 + y 2 )3

and Pz = Qz = Rx = Ry = 0.

Thus, curlV = (Ry − Qz )i + (Pz − Rx )j + (Qx − Py )k = 0 and V is irrotational. 2Ax(3y 2 − x2 ) 2Ax(x2 − 3y 2 ) , Qy = , and Rz = 0, ∇·F = Px +Qy +Rz = 0 2 2 3 (x + y ) (x2 + y 2 )3 and V is incompressible.

(c) Since Px =

41. We first note that curl (∂H/∂t) = ∂(curl H)/∂t and curl (∂E/∂t) = ∂(curl E)/∂t. Then, from Problem 30, −∇2 E = −∇2 E + 0 = −∇2 E + grad 0 = −∇2 E + grad (div E) = curl (curl E)     1 ∂H 1 ∂ 1 ∂ 1 ∂E 1 ∂2E = curl − =− curl H = − =− 2 c ∂t c ∂t c ∂t c ∂t c ∂t and ∇2 E = 2

1 2 2 c2 ∂ E/∂t .

Similarly,

2

−∇ H = −∇ H + grad (div H) = curl (curl H) = curl   1 ∂ 1 ∂H 1 ∂2H = − =− 2 c ∂t c ∂t c ∂t2 and ∇2 H =



1 ∂E c ∂t

 =

1 ∂ curl E c ∂t

1 2 ∂ H/∂t2 . c2

42. We note that div F = 2xyz − 2xyz + 1 = 1 6= 0. If F = curl G, then div (curl G) = div F = 1. But, by problem 24, for any vector field G, div (curl G) = 0. Thus, F cannot be the curl of G.

15.8

Stokes’ Theorem

15.8. STOKES’ THEOREM

269

1. Surface Integral: curlF = −10k. Letting g(x, y, z) = −1, we have ∇g = k and n = k. Then Z Z Z Z (curlF)·ndS = (−10)dS = −10×(area of S) = −10(4π) = −40π. S

z

3

z=1

C

3 y R

3

S

x

Line Integral: Parameterize the curve C by x = 2 cos t, y = 2 sin t, z = 1, for 0 ≤ t ≤ 2π. Then I

I

Z

F · dr =

2π

[10 sin t(−2 sin t) − 10 cos t(2 cos t)]dt

5ydx − 5xdy + 3dz = 0 2π

Z

(−20 sin2 t − 20 cos2 t)dt =

=

Z

2π

−20dt = −40π. 0

0

2. Surface Integral: curlF = 4i−2j−3k. Letting g(x, y, z) = x2 + y 2 +p z − 16, ∇g = 2xi + 2yj + k, and n = (2xi + 2yj + k)/ 4x2 + 4y 2 + 1. Thus, Z Z Z Z 8x − 4y − 3 p dS. (curlF) · ndS = 4x2 + 4y 2 + 1 S S

z

16 z=16-x2-y2

Letting the surface be z =p16 − x2 − y 2 , we have zx = −2x, zy = −2y, and dS = 1 + 4x2 + 4y 2 dA. Then, using polar Z Z coordinates, Z Z (curlF) · ndS =

(8x − 4y − 3)dA

S

Z

2π

Z

0

y

C

4

4

(8r cos θ − 4r sin θ − 3)rdrdθ

=

4

R

R

x

r=4

0

 4 4 3 3 2 8 3 r cos θ − r sin θ − r dθ = = 3 3 2 0 0  Z 2π  512 256 cos θ − sin θ − 24 dθ 3 3 0   2π 512 256 = sin θ + cos θ − 24θ = −48π. 3 3 0 Line Integral: Parameterize the curve C by x = 4 cos t, y = 4 sin t, z = 0, for 0 ≤ t ≤ 2π. Then, I I Z Z

2π



2π

F · dr = C

2zdx − 3xdy + 4ydz = c

Z = 0

[−12 cos t(4 cos t)]dt 0

2π

2π

−48 cos2 tdt = (−24t − 12 sin 2t)|0 = −48π.

270

CHAPTER 15. VECTOR INTEGRAL CALCULUS

3. Surface Integral: curlF = i + j + k. Letting z=3-y/2-x g(x, y, z) = 2x + y + 2z − 6, we have ∇g = 2i + 2j + 2k and z RR RR 5 n = (2i + j + 2k)/3. Then (curlF) · ndS = dS. C3 S S 3 1 Letting the surface be z = 3 − y − x we have zx = C1 2 r 1 3 1 R y −1, zy = − , and dS = 1 + (−1)2 + (− )2 dA = dA. 2 2 2 C2 Then x   Z Z Z Z 5 3 5 5 45 (curlF)·ndS = dA = ×(area of R) = (9) = . 2 2 2 2 S R 3 Line Integral: C1 : z = 3 − x, 0 ≤ x ≤ 3, y = 0; C2 : y = 6 − 2x, 3 ≥ x ≥ 0, z = 0; C3 : z = 3I− y/2, 6 ≥ y ≥ 0, x Z= Z0. Z Z zdx + xdy + ydz =

zdx +

C

C1

Z

xdy + C2

3

0

Z (3 − x)dx +

= 0

ydz C3

Z x(−2dx) +

3

0

y(−dy/2) 6

0  3  0 1 9 1 45 1 3x − x2 − x2 3 − y 2 = − (0 − 9) − (0 − 36) = 2 4 2 4 2 0 6 RR 4. Surface Integral: curlF = 0 and (curlF) · ndS = 0. S Line Integral: The curve is x = cos t, y = sin t, z = 0, 0 ≤ t ≤ 2π. =

I

Z xdx + ydy + zdz =

C

2π

[cos t(− sin t) + sin t(cos t)]dt = 0. 0

5. curlF = 2i + j. A √ unit vector normal to the plane is n = (i + j + k)/ 3. Taking the equation of the plane to be √z = 1 − x − y, √ we have zx = zy = −1. Thus, dS = 1 + 1 + 1dA = 3dA and I Z Z Z Z √ √ Z Z √ F · dr = S(curlF) · ndS = 3dS = 3 3dA C

S

z

1

z=1-x-y

R

R

= 3 × (area of R) = 3(1/2) = 3/2.

C

1

y

1

x

√ 2. From z = 1 − y, 6. curlF = −2xzi + x2 k. A unit vector normal to the plane is n = (j + k)/ √ √ we have zx = 0 and zy = −1. Thus, dS = 1 + 1dA = 2dA and Z Z I Z Z Z Z 1 2√ √ z 2dA = (1 − y)2 dA F · dr = (curlF) · ndS = 2 R C S R 1 Z 2Z 1 Z 2 Z 2 1 1 2 = (1 − y)2 dydx = − (1 − y)3 dx = dx = . 3 3 3 0

0

0

0

0

15.8. STOKES’ THEOREM

271

√ 7. curlF = −2yi − zj − xk. A unit vector normal √ is n = (j + k)/ 2. From z = 1 − y √ to the plane we have zx = 0 and zy = −1. Then dS = 1 + 1dA = 2dA and  Z Z Z Z  √ 1 2dA = (y − x − 1)dA − √ (z + x) 2 R S R  1  Z 2 Z 2 Z 2Z 1 1 1 2 −x − (y − x − 1)dydx = y − xy − y dx = dx = 2 2 0 0 0 0 0   2 1 1 = − x2 − x = −3. 2 2 0

I

Z Z

F · dr = c

(curlF) · ndS =

8. curlF = 2i + 2j + 3k. Letting g(x, y, z) = x + 2y√+ z − 4, we have ∇g = i + 2j + k and n = (i + 2j + k)/ 6. From z = 4√− x − 2y we have zx = −1 and zy = −2. Then dS = 6dA and

z 4 z=4-x-2y

C

√ RR RR 1 √ (9) 6dA = F · dr = (curlF) · ndS = C S R 6 RR 9dA = 9x (area of R)= 9(4) = 36. R H

2 y R

4

9. curlF = (−3x2√ −3y 2 )k. A unit vector normal to the plane is 3. From z = 1−x−y, we have zx = zy = −1 n = (i+j+k)/ √ and dS = 3dA. Then, using polar coordinates, I Z Z Z Z √ √ √ F · dr = (curlF) · ndS = (− 3x2 − 3y 2 ) 3dA C

S

z 3 C

z=1-x-y

R

Z Z

2

2

Z

2π

Z

(−x − y )dA = 3

=3 R 2π

Z

x

=3 0

0

R

1

(−r2 )rdrdθ

1 y

0

1 Z 2π 1 3π 1 − dθ = . − r4 dθ = 3 4 0 4 2 0

x

2yj + k 10. curlF = 2xyzi − y 2 zj + (1 − x2 )k. A unit vector normal to the surface is n = p . From 4y 2 + 1 p z = 9 − y 2 we have zx = 0, zy = −2y and dS = 1 + 4y 2 dA. Then I Z Z Z Z Z 3 Z y/2 F · dr = (curlF) · ndS = (−2y 3 z + 1 − x2 )dA = [−2y 3 (9 − y 2 ) + 1 − x2 ]dxdy C

S

R

0

0

 y/2  Z 3 1 3 1 1 3 3 5 4 6 = −18y x + 2y x + x − x dy = −9y + y + y − y dy 3 2 24 0 0 0   3 9 1 1 1 = − y 5 + y 7 + y 2 − y 4 ≈ 123.57. 5 7 4 96 0 Z

3



272

CHAPTER 15. VECTOR INTEGRAL CALCULUS

11. curlF = 3x2 y 2 k. A unit vector normal to the surface is 8xi + 2yj + 2zk 4xi + yj + zk n= p =p . 2 2 2 64x + 4y + 4z 16x2 + y 2 + z 2

From zx = − p

4x 4 − 4x2 − y 2

, zy = − p

s

y 4 − 4x2 − y 2

we obtain dS = 2

1 + 3x2 dA. 4 − 4x2 − y 2

Then I

Z Z

3x2 y 2 z

Z Z

F · dr =

(curlF) · ndS =

C

S

Z Z

R

3x2 y 2 dA

=

p (2 16x2 + y 2 + z 2

1

√ 2 1−x2

Z

x2 y 2 dydx = 12

= 12 0

Z = 32

1 + 3x2 )dA 4 − 4x2 − y 2

Using symmetry

R

Z

s

0

Z

1



0

 2√1−x2 1 2 3 x y dx 3 0

1

x2 (1 − x2 )3/2 dx

x = sin t, dx = cos tdt

0

Z = 32

π/2

sin2 t cos4 tdt = π.

0

12. curlF = i + j + k. A unit vector normal to the surface is 2xi + 2yj + 2zk xi + yj + zk n= p =p 2 2 2 4x + 4y + 4z x2 + y 2 + z 2 = xi + yj + zk. p y x , zy = − p and dS = From z = 1 − x2 − y 2 , we have zx = − p 2 2 1−x −y 1 − x2 − y 2 1 p dA. Then 1 − x2 − y 2 ! I Z Z Z Z 1 dA F · dr = (curl F) · ndS = (x + y + z) p 1 − x2 − y 2 C S R p Z Z Z Z Z Z x + y + 1 − x2 − y 2 x+y p p = dA = 1dA + dA 2 2 1−x −y 1 − x2 − y 2 R R R Z Z = 1dA + 0 Using Symmetry R

π = 2 since R is the disk x2 + y 2 ≤

1 2

with radius

√1 . 2

13. Parameterize C by x = 4 cos t, y = 2 sin t, z = 4, for 0 ≤ t ≤ 2π. Then

15.8. STOKES’ THEOREM

273

Z Z

I

I

(curlF) · ndS =

F · dr =

S

2

6yzdx + 5xdy + yzex dz

C 2π

Z =

[6(2 sin t)(4)(−4 sin t) + 5(4 cos t)(2 cos t) + 0]dt 0 2π

Z

(−24 sin2 t + 5 cos2 t)dt = 8

=8 0

Z

2π

(5 − 29 sin2 t)dt = −152π.

0

14. Parameterize C by x = 5 cos t, y = 5 sin t, z = 4, for 0 ≤ t ≤ 2π. Then, Z Z I I (curlF) · ndS = F·r= ydx + (y − x)dy + z 2 dz S

C 2π

C

Z

[(5 sin t)(−5 sin t) + (5 sin t − 5 cos t)(5 cos t)]dt

= 0 2π

Z

 (25 sin t cos t − 25)dt =

= 0

 2π 25 2 sin t − 25t = −50π. 2 0

15. Parameterize C by C1 : x = 0, z = 0, 2 ≥ y ≥ 0; C2 : z = x, y = 0, 0 ≤ x ≤ 2; C3 : x = 2, z = 2, 0 ≤ y ≤ 2; Z ZC4 : z = x, y = 2, I 2 ≥ x ≥I0. Then (curlF) · ndS = F·r= 3x2 dx + 8x3 ydy + 3x2 ydz S C ZC Z = 0dx + 0dy + 0dz + 3x2 dx C1 C2 Z Z + 64dy + 3x2 dx + 6x2 dx Z =

C3 2

C4 2

3x2 dx +

Z

0

Z 64dy +

0

z 2

C3 C4

C2 C1

2

y

2

0

9x2 dx

2

x

2 0 2 = x3 0 + 64y|0 + 3x3 2 = 112. 16. Parameterize C by x = cos t, y = sin t, z = sin t, 0 ≤ t ≤ 2π. Then Z Z I I (curlF) · ndS = F·r= 2xy 2 zdx + 2x2 yzdy + (x2 y 2 − 6x)dz S

C 2π

Z

C

[2 cos t sin2 t sin t(− sin t) + 2 cos2 t sin t sin t cos t

= 0

+ (cos2 t sin2 t − 6 cos t) cos t]dt Z =

2π

(−2 cos t sin4 t + 3 cos3 t sin2 t − 6 cos2 t)dt = −6π.

0

17. We take the surface to be z = 0. Then n = k and dS = dA. Since curlF =

1 2 i + 2zex j + 2 1+y

274

CHAPTER 15. VECTOR INTEGRAL CALCULUS y 2 k, I

2 x2

z e dx + xydy + tan

Z Z

−1

Z Z (curlF) · ndS =

ydz =

C

S

Z

2π

Z

3

0

Z

0 2π

0

r2 sin2 θrdrdθ =

y 2 dA

y dS = S

= 81 = 4

Z Z

2

R 2π

Z 0

3 1 4 2 r sin θ dθ 4 0

81π sin2 θdθ = . 4

2xi + 2yj + k and 18. (a) curlF = xzi − yzj. A unit vector normal to the surface is n = p 4x2 + 4y 2 + 1 p dS = 1 + 4x2 + 4y 2 dA. Then, using x = cos t, y = sin t, 0 ≤ t ≤ 2π, we have Z Z Z Z Z Z (curlF) · ndS = (2x2 z − 2y 2 z)dA = (2x2 − 2y 2 )(1 − x2 − y 2 )dA S R R Z Z = (2x2 − 2y 2 − 2x4 + 2y 4 )dA R 2π Z 1

Z = 0

(2r2 cos2 θ − 2r2 sin2 θ − 2r4 cos4 θ + 2r4 cos4 θ)rdrdθ

0 2π

Z

Z

1

[r3 cos 2θ − r5 (cos2 θ − sin2 θ)(cos2 θ + sin2 θ)]drdθ

=2 0

0 2π

Z

Z

0

1 6

3

Z

5

(r cos 2θ − r cos 2θ)drdθ = 2

=2 =

1

Z

0

2π

 cos 2θ

0

2π

cos 2θdθ = 0. 0

(b) RWe R take the surface to be z = 0. Then n = k, curlF · ndS = 0. S

curlF · n = curlF · k = 0 and

(c) By Stoke’s Theorem, using z = 0, we have Z Z I I I curlF · ndS = F · dr = xyzdz = xy(0)dz = 0. S

15.9

 1 1 4 1 6 r − r dθ 4 6 0

Divergence Theorem

C

C

15.9. DIVERGENCE THEOREM

275 z

1

1. divF = y + x + z The Integral: Z Z Triple Z Z 1Z divFdV = D

Z

1

S4

1

(x + y + z)dxdydz 0

0

0

Z

1

1

Z

=

S6

1 1 2 ( x + xy + xz) dydz 2 0

1

y

1 1 S1 x ( + y + z)dydz = 2 0 0 1 Z 1 1 1 2 = ( y + y + yz) dz 2 0 2 0 1 Z 1 1 3 1 2 (1 + z)dz = (1 + z ) = 2 − = = 2 2 2 0 0 The Surface Integral: Let the surfaces be S1 in z = 0, S2 in z = 1, S3 in y = 0, S4 in y = 1, S5 in x = 0, and S6 in x = 1. The unit outward normal vectors are −k, k, −j, j, −i and i, respectively. Then Z Z Z Z Z Z Z Z Z Z F · ndS = F · (−k)dS1 + F · kdS2 + F · (−j)dS3 + F · jdS4 S S S S3 S4 Z Z1 Z Z2 + F · (−i)dS5 + F · idS6 S6 Z Z S5 Z Z Z Z Z Z = (−xz)dS1 + xzdS2 + (−yz)dS3 + yzdS4 S3 S4 Z SZ1 Z SZ2 + (−xy)dS 5 + xydS6 S5 S6 Z Z Z Z Z Z = xdS2 + zdS4 + ydS6 Z

0 1

Z

0 1

S2 1

Z

1

Z

=

Z

S4 1Z 1

xdxdy + 0

0 1

Z = 0

Z

1 dy + 2

0

1

S6

Z

1

Z

zdzdx + 0

0

1 dx + 2

Z

ydydz 0

0

1

1

0

3 1 dz = . 2 2

z 1

2. divF = 6y + 4z

S3

The Integral: Z Z Triple Z Z 1Z divFdV = D

0

Z

1−x

Z

S4

1−x−y

(6y + 4z)dzdydx

0 1Z

= 0

S1

0

1

0

1−x−y 1−x 2 (6yz + 2z ) dydx 0

1 x

S2

y

276

CHAPTER 15. VECTOR INTEGRAL CALCULUS Z

1

Z

=

1−x

(−4y 2 + 2y − 2xy + 2x2 − 4x + 2)dydx

0

0

  1−x 4 3 2 2 2 − y + y − xy + 2x y − 4xy + 2y = dx 3 0 0    1 Z 1 5 5 5 5 5 5 5 = − x3 + 5x2 − 5x + dx = − x4 + x3 − x2 + x = 3 3 12 3 2 3 12 0 0 The Surface Integral: Let the surfaces be S1 in the plane x+y+z = 1, √S2 in z = 0, S3 in x = 0, and S4 in y = 0. The unit outward normal vectors √ are n1 = (i + j + k)/ 3, n2 = −k, n3 = −i, and n4 = −j, respectively. Now on S1 , dS1 = 3dA1 , on S3 , x = 0, and on S4 , y = 0, so Z Z Z Z Z Z Z Z Z Z F · ndS = F · n1 dS1 + F · (−k)dS2 + F · (−j)dS3 + F · (−i)dS4 Z

1

S

S1

Z

=

1

Z

s2

S3

1−x

(6xy + 4y(1 − x − y) + xe−y )dydx + Z Z Z Z + (−6xy)dS3 + (−4yz)dS4 0

0

S3

S4

Z 0

1

Z

1−x

(−xe−y )dydx

0

S4

1−x  1−x Z 1 4 3 −y −y = xy + 2y − y − xe dx + 0 + 0 dx + xe 3 0 0 0 0  Z 1 Z 1 4 = x(1 − x)2 + 2(1 − x)2 − (1 − x)3 − xex−1 + x dx + (xex−1 − x)dx 3 0 0  1  5 1 2 2 3 1 4 2 1 3 4 x − x + x − (1 − x) + (1 − x) = . = 2 3 4 3 3 12 0 Z

1



2

2

3. divF = 3x2 + 3y 2 + 3z 2 . Using spherical coordinates, Z Z Z Z Z Z 2π Z π Z a F · ndS = 3(x2 + y 2 + z 2 )dV = 3ρ2 ρ2 sin φdρdφdθ S D 0 0 0 a Z Z 2π Z π Z 3a5 2π π 3 5 ρ sin φ dφdθ = = sin φdφdθ 5 0 0 0 0 5 0 π Z Z 6a5 2π 3a5 2π 12πa5 . = − cos φ dθ = dθ = 5 5 5 0

0

0

4. divF = 4 + 1 + 4 = 9. Using the formula for the volume of a sphere,   Z Z Z Z Z 4 3 F · ndS = 9dV = 9 π2 = 96π. 3 S D 5. divF = 2(z − 1). Using cylindrical coordinates, Z Z Z Z Z Z 2π Z 4 Z F · ndS = 2(z − 1)V = S

D

Z

2π

Z

=

0

4

Z 16rdrdθ =

0

0

0

2π

0

5

Z

Z

2(z − 1)dzrdrdθ =

1

0

4 Z 2 8r dθ = 128 0

2π

0

2π

dθ = 256π.

0

4

5 (z − 1) rdrdθ 2

1

15.9. DIVERGENCE THEOREM

277

2 6. divF Z Z = 2x + 2z +Z 12z Z Z. Z F · ndS = divFdV = S

Z

0 3

Z

2

= Z

0 2

Z

z

1

3

(2x + 2z + 12z 2 )dxdydz

0

0

1 2 2 (x + 2xz + 12xz ) dydz 0

(1 + 2z + 12z 2 )dydz

= 0

0 3

Z

2

Z

0

D 3

Z

3

= 0

Z Z Z

Z

F · ndS = D

Z

2π

√

Z

Z

0 2π

= 0 Z 2π

= 0

2π Z

3Z

0

0

√

4−r 2

3z 2 rdzdrdθ

0

2

x

2π

0

0

y

r=M3

2

0

√3 Z 1 − (4 − r2 )5/2 dθ = 5

r=M4-r2

2

√4−r2 Z 2π Z √3 3 rz drdθ = r(4 − r2 )3/2 drdθ 0 0

3

= 0

z

√

divFdV =

S

y

x

7. divF = 3z 2 . Using cylindrical coordinates, Z Z

2

1

3 2(1 + 2z + 12z 2 )dz = (2z + 2z 2 + 8z 3 ) 0 = 240

1 − (1 − 32)dθ 5

31 62π dθ = . 5 5

8.

z

divF = 2x. Z Z Z Z Z F · ndS = divFdV S

Z

3

Z

9 z=9-y

D 9 Z 9−y

=

2xdzdydx x2

0

Z

3

Z

0

9

3

Z 2x(9 − y)dydx =

= x2

0

Z

9 −x(9 − y) dx 2 2

0

3

9

4

x

x(9 − x)2 dx

=

x

0

Z =

3

(x3 − 18x2 + 81x)dx =

0

891 = 4



 3 81 1 4 x − 6x3 + x2 4 2 0

y=x2

y

278

CHAPTER 15. VECTOR INTEGRAL CALCULUS

9. divF =

x2

1 . Using spherical coordinates, + y2 + z2

Z Z

Z Z Z

2π

Z

Z

π

Z

y

b

1 2 ρ sin φdρdφdθ 2 a ρ 0 D 0 π Z 2π Z 2π Z π − cos φ dθ (b − a) sin φdφdθ = (b − a) =

F · ndS = S

a

divFdV =

0

0

0

b

x

0

2π

Z = (b − a)

2dθ = 4π(b − a). 0

Z Z

Z Z Z F · ndS =

10. Since divF = 0,

0dV = 0.

S

D

11. divF = 2z + 10y − 2z = 10y. Z Z

Z Z Z D 2

Z

2−x2 /2

Z

0

dx =

D 2−x

Z

0

0

2

Z

0

Z

2 (80 − 5x4 )dx = (80x − x5 ) 0 = 128

3

3

z

0

x+y

dydx

2−x

2

(90xy − 30x2 y − 30xy 2 )dydx

y

0 2

0 Z 2

2

2−x dx (45xy − 15x y − 10xy ) 2

= =

2

x+y

= Z

3 30xyz

0

0

2−x

2−x2 /2

30xydzdydx

0

=

Z

0

Z

Z

(80 − 40z)dzdx 0

z 2−x2 /2

2

2

Z

(80z − 20z 2 )

0

12. divF Z Z = 30xy. Z Z Z Z F · ndS = 30xydV = S

z

dzdx =

2

Z =

0

5y 2

0

Z

4−z

Z

10ydydzdx

0 4−z

=

2

2−x2 /2

Z

10ydV =

S

Z

2

Z

F · ndS =

2 2

x

3

0

(−5x4 + 45x3 − 120x2 + 100x)dx =



−x5 +

0

13. divF = 6xy + 1 − 6xy 2 = 1. Using cylindrical coordinates, Z Z Z Z Z Z π Z 2 sin θ Z 2r sin θ Z F · ndS = dV = dzrdrdθ =

 2 45 4 x − 40x3 + 50x2 = 28 4 0

2

S

D π

0

0

r2  2 sin θ

0

π

Z

2 sin θ

(2r sin θ − r2 )rdrdθ

0

 Z π 2 3 1 16 r sin θ − r4 dθ = sin4 θ − 4 sin4 θ dθ 3 4 3 0 0  0  π Z π 4 4 3 1 1 π 4 = sin θdθ = θ − sin 2θ + sin 4θ = 3 0 3 8 4 32 2 0 Z

=



15.9. DIVERGENCE THEOREM

279

14. divF = y 2 + x2 . Using spherical coordinates, we have x2 + y 2 = ρ2 sin2 ω and z = ρ cos ω or ρ = z sec ω. Then Z Z

Z Z Z

(x2 + y 2 )dS =

F · ndS = S

Z

D 2π

Z

0

π/4

Z

= 0

0 2π

Z

Z

0

=

992 5

Z

0 2π

0

Z

π/4

Z

4 sec φ

ρ2 sin2 φρ2 sin φdρdφdθ

2 sec φ

0

4 sec φ Z 2π Z π/4 992 1 5 3 dφdθ = ρ sin φ sec5 φ sin3 φdφdθ 5 5 0 0 2 sec φ

π/4

=

2π

992 992 tan3 φ sec2 φdφdθ = 5 5

2π

Z 0

π/4 1 4 dθ tan φ 4 0

1 496π dθ = . 4 5

15. Since div a = 0, by the divergence Theorem Z Z Z Z Z Z Z Z (a · n)dS = div adV = 0dV = 0. S

D

D

16. By the Divergence Theorem and Problem 24 in Section 15.7, Z Z Z Z Z Z Z Z (curlF · n)dS = div(curlF)dV = 0dV = 0. S

D

D

 x2 − 2y 2 + z 2 x2 + y 2 − 2z 2 −2x2 + y 2 + z 2 + + 2 =0 17. (a) divE = q (x2 + y 2 + Zz 2 )Z5/2Z (x2 + y 2 +Zz 2Z)5/2 (x + y 2 + z 2 )5/2 Z Z Z 

(E · n)dS =

divEdV =

SU Sa

D

0dV = 0 D

RR RR RR RR (E · n)dS = − Sa (E · n)dS. on (b) From (a), (E · n)dS + Sa (E · n)dS = 0 and S S Sa , |r| = a, n = −(xi + yj + zk)/a = −r/a and E · n = (qr/a3 ) · (−r/a) = −qa2 /a4 = −qa2 . Thus RR RR q q RR q q (E · n)dS = − Sa (− 2 )dS = 2 dS = 2 × (area of Sa ) = 2 (4πa2 ) = 4πq. S S a a a a a RR RRR RR 18. (a) By Gauss’ (E · n)dS R=R R D 4πρdV, Rand (E · S R RLaw R R R by the Divergence R R RTheorem n)dS = divEdV. Thus 4πρdV = divEdV and (4πρ−divE)dV = D D D D 0. Since this holds for all regions D, 4πρ − divE = 0 and divE = 4πρ. (b) Since E is irrotational, E = ∇φ and ∇2 φ = ∇ · ∇φ = ∇E = divE = 4πρ. 19. By the Divergence Theorem and Problem 21 in Section 15.7, Z Z Z Z Z Z Z Z Z Z Z (f ∇g) · ndS = div(f ∇g)dV = ∇ · (f ∇g)dV = [f (∇ · ∇g) + ∇g · ∇f ]dV S D D Z Z ZD = (f ∇2 g + ∇g · ∇f )dV. D

280

CHAPTER 15. VECTOR INTEGRAL CALCULUS

20. By the Divergence Theorem and Problem 19 and 21 in Section 15.7, Z Z Z Z Z Z Z Z (f ∇g − g∇f ) · ndS = div(f ∇g − g∇f )dV = ∇ · (f ∇g − g∇f )dV S D Z Z ZD = [f (∇ · ∇g) + ∇g · ∇f − g(∇ · ∇f ) − ∇f · ∇g]dV Z Z ZD = (f ∇2 g − g∇2 f )dV. D

21. If G(x, y, z) is a vector valued function then we define surface integrals and triple integrals of G component-wise. In this case, if a is a constant vector it is easily shown that Z Z Z Z Z Z Z Z Z Z a · GdS = a · GdSand a · GdV = a · GdV. S

S

Now let F = f a. Then Z Z

D

D

Z Z

Z Z

F · ndS = S

(f a) · ndS =

a · (f n)dS

S

S

and, using Problem 21 in Section 15.7 and the fact that ∇ · a = 0, we have Z Z Z Z Z Z Z Z Z Z Z Z divFdV = ∇ · (f a)dV = [f (∇ · a) + a · ∇f ]dV = a · ∇f dV. D

D

D

D

By the Divergence Theorem, Z Z Z Z Z Z Z Z Z Z a · (f n)dS = F · ndS = divFdV = a · ∇f dV S

S

D

D

and Z Z



a·

f ndS

Z Z Z



=a·

∇f dV

S

Z Z



Z Z Z

or a ·

f ndS −

D

∇f dV

S

D

Since a is arbitrary, Z Z Z Z Z Z Z Z Z Z f ndS − ∇f dV = 0 and f ndS = ∇f dV. S

D

Z Z

S

Z Z Z

pndS + mg = mg − ∇pdV = mg − D Z Z Z  = mg − ρdV g = mg − mg = 0

D

Z Z Z

22. B + W = −

S

D

Chapter 15 in Review A. True/False 1. True; the value is 4/3.

ρgdV D

= 0.

CHAPTER 15 IN REVIEW

281

2. True; since 2xydx − x2 dy is not exact. R 3. False; C xdx + x2 dy = 0 from (−1, 0) to (1, 0) along the x-axis and along the semicircle √ y = 1 − x2 , but since xdx + x2 dy is not exact, the integral is not independent of path. 4. True 5. True; assuming that the first partial derivatives are continuous. 6. True 7. True 8. True; since curlF = 0 when F is a conservative vector field. 9. True 10. True 11. True 12. True

B. Fill in the Blanks 1. F = ∇φ = −x(x2 + y 2 )−3/2 i − y(x2 + y 2 )−3/2 j i j k 2. curlF = ∂/∂x ∂/∂y ∂/∂z = 0 f (x) g(y) h(z) 3. 2xy + 2xy + 2xy = 6xy i j k 4. ∂/∂x ∂/∂y ∂/∂z = 2xzi − 2yzj + (y 2 − x2 )k x2 y xy 2 2xyz 5.

∂ ∂ 2 ∂ (2xz) − (2yz) + (y − x2 ) = 0 ∂x ∂y ∂z

6. ∇(6xy) = 6yi + 6xj 3

7. 0; since (y − 7ex )dx + (x + ln

√

y)dy is exact.

8. Irrotational 9. At u = 1, v = 4, we have r = h1, 4, 4i. ∂r ∂r (1, 4) = h1, 0, 2i, (1, 4) = h0, 1, 1/2i ∂u ∂v i j k A normal vector is given by n = 0 1 2 = h−2, −1/2, 1i. 0 1 1/2 The tangent plane is −2(x − 1) − 12 (y − 4) + (z − 4) = 0 or 4x + y − 2z = 0. 10. r(2, v) = (8 + v)i + (2 + 2v)j + (2 + v)k So the parametric equations are x = 8 + v, y = 2 + 2v, z = 2 + v

282

CHAPTER 15. VECTOR INTEGRAL CALCULUS

C. Exercises Z 1. C

2π

p 4t2 4 sin2 2t + 4 cos2 2t + 4dt = cos2 2t + sin 2t π √ √ 2π 56 2π 3 8 2 3 = t = 3 3

z2 ds = x2 + y 2

Z

Z

2π

√ 8 2t2 dt

π

π

√ √ [x(2 − 2x) + 4x] 1 + 4dx = 5 0 √   0 √ 7 5 2 3 2 = 5 3x − x = − 3 3 1

Z 2.

Z

1

Z

1

(xy + 4x)ds = C

(6x − 2x2 )dx

0

3. Since P − y = 6x2 y = Qx , the integral is independent of path. φx = 3x2 y 2 , φ = x3 y 2 + g(y), φy = 2x3 y + g 0 (y) = 2x3 y − 3y 2 ; g(y) = −y 3 ; φ = x3 y 2 − y 3 ; (−1,2) R (−1,2) 2 2 3x y dx + (2x3 y − 3y 2 )dy = (x3 y 2 − y 3 ) (0,0) = −12 (0,0) 4. By Green’s Theorem, I

(x2 + y 2 )dx+(x2 − y 2 )dy =

C

=2 0

5.

Z

2π



2π

3

Z

(2x − 2y)dA = 2 R

Z

Z

Z Z

(r cos θ − r sin θ)rdrdθ 0

0

 3 Z 2π 27 1 3 1 3 r cos θ − r sin θ dθ = 2 (cos θ − sin θ)dθ = 0. 3 3 3 0 0

y sin πzdx + x2 ey dy + 3xyzdz

C

Z

1

=

2

[t2 sin πt3 + t2 et (2t) + 3tt2 t3 (3t2 )]dt =

Z

0

1

2

(t2 sin πt3 + 2t3 et + 9t8 )dt

0

 1  Z 1 2 1 t3 et dt cos πt3 + t9 + 2 = − 3π 0 0 1 2 2 2 2 = + 1 + (t2 et − et ) = +2 3π 3π 0

Integration by parts

6. Parameterize C by x = cos t, y = sin t; 0 ≤ t ≤ 2π. Then I

Z

2π

F · dr = C

Z

2π

[4 sin t(− sin tdt) + 6 cos t(cos t)dt] = 0

Z

2π

0 2π

(6 cos2 t − 4 sin2 t)dt

5 sin 2t − 4t) = 2π. 2 0 0 H RR Using Green’s Theorem, Qz − Py = 6 − 4 = 2 and C F · dr = 2dA = 2(π · 12 ) = 2π. R =

(10 cos2 t − 4)dt = (5t +

CHAPTER 15 IN REVIEW

283

π π 7. Let r1 = ti and r2 = i + πtj for 0 ≤ t ≤ 1. Then 2 2 π dr1 = i, dr2 = πj, F1 = 0, 2 F2 =

y π

C2

π π π sin πti + πt sin j = sin πti + πtj, 2 2 2

C1

and

π/2 Z

Z F1 ·dr1 +

W = C1

Z F2 ·dr2 =

C2

0

1

x

1 π2 1 2 2 2 . π tdt = π t = 2 2 0

8. Parameterize the line segment from (-1/2,1/2) TO (-1,1) using y = −x as x goes from -1/2 to -1. Parameterize the line segment from (-1,1) √ to (1,1) using y = 1 as x goes from √ -1 to 1. Parameterize the line segment from (1,1) to (1, 3) using x = 1 as y goes from 1 to 3. Then Z

Z

−1

F · dr =

W = C

F · (dxi − dxj) + −1/2

√

1

Z

Z

−1

3

F · (dyj)

F · (dxi) + 1

Z 1 Z √3 1 2 1 2 − 2 )dx + dx + dy = ( 2 2 2 2+1 x + (−x) x + (−x) x 1 + y2 −1 −1/2 1 Z −1 Z 1 Z √3 1 2 1 = dx + dx + dy 2 2 2x 1 + x 1 + y2 −1/2 1 −1 −1 1 √3 1 π π 13π − 6 1 + 2 tan−1 x −1 + tan−1 y 1 = − + 2( ) + = . = − 2x −1/2 2 2 12 12 Z

−1

9. Py = 2x = Qx , Qz = 2y = Ry , Rx = 0 = Pz and the integral is independent of path. Parameterize the line segment between the points by x = 1, y = 1, z = t, 0 ≤ t ≤ π. Then dx = dy = 0, dz = dt, and Z

(1,1,π)

2xydx + (x2 + 2yz)dy + (y 2 + 4)dz =

(1,1,0)

Z

π

[2(0) + (1 + 2t)(0) + (1 + 4)]dt = 5π. 0

10. Py = 0 = Qx , Qz = 0 = Ry , Rx = 2e2x = Pz and the integral is independent of path. From ω = x2 + y 2 − y + ze2x we obtain Z

(3,2,0)

(0,0,1)

(3,2,0) (2x + 2ze2x )dx + (2y − 1)dy + e2x dz = (x2 + y 2 − y + ze2x ) (0,0,1) = 11 − 1 = 10.

11. Using Green’s Theorem, I Z Z Z Z −4ydx + 8xdy = [8 − (−4)]dA = 12 dA = 12 × (area of R) C

R

= 12(16π − π) = 180π.

R

284

CHAPTER 15. VECTOR INTEGRAL CALCULUS

12. Py = [(x − 1)2 − (y − 1)2 ]/[(x − 1)2 + (y − 1)2 ]2 = Qx . When (1,1) is outside C, Green’s Theorem applies and I Z Z Z Z P dx + Qdy = (Qx − Py )dA = 0dA = 0. C

R

R

For (1,1) inside C, let Ca be a circle of radius a centered at (1,1) and lying entirely inside C. Using x − 1 = a cos θ and y − 1 = a sin θ for 0 ≤ θ ≤ 2π we obtain I I I 1 (y − 1)dx + (1 − x)dy P dx + Qdy = P dx + Qdy = 2 a Ca C Ca Z 2π 1 = 2 [a sin θ(−a sin θ) − a cos θ(a cos θ)]dθ a 0 Z 2π =− (sin2 θ + cos2 θ)dθ = −2π. 0

√ 13. zx = 2x, zy = 0; dS = 1 + 4x2 dA  2 Z 3Z 2 2p Z Z Z 3  z x 1 1 2 3/2 2 dS = (1 + 4x ) 1 + 4x dxdy = dy S xy 1 1 xy 1 y 12 1 3 √ √ Z 3 3/2 1 17 − 53/2 17 17 − 5 5 = dy = ln y 12 1 y 12 1 √ √ 2 17 17 − 5 5 x = ln 3 12

z

4 z=x2

3

14. n = k, R F R · n = 3; RR flux = F·ndS = 3 S dS = 3×(area of S) = 3(1) = 3 S

y

z 2

S

1

1

y

x

√ −x −x −x −2x + 1, 15. The surface √is g(x, y, z) = y +e −2 = 0. Then ∇g = −e i+j, n = (−e i+j)/ e and dS = 1 + e−2x dA. 3 Z Z Z Z Z 2Z 3 Z 2 flux = (F · n)dS = (−4e−x + 2 − y)dA = (−3e−x )dxdz = 3e−x dz S

Z =

R

0

0

0

0

2

(3e−3 − 3)dz = 6e−3 − 6.

0

16. Solving y = 2−e−x for x, we obtain x = − ln(2−y). The surface is g(x, y, z) = x+ln(2−y) = 0. p 1 Then ∇g = i − j, and |∇g| = 1 + 1/(2 − y)2 . Due to the orientation of S we 2−y want the j component of the unit normal vector to be positive. Since y < 2 we shall take

CHAPTER 15 IN REVIEW

285

p p n = [−i + (1/2 − y)j]/ 1 + 1/(2 − y)2 . Now dS = 1 + 1/(2 − y)2 dA and the region R in the yz -plane is 0 ≤ z ≤ 2 and 1 ≤ y ≤ 2 − e−3 . Then Z Z Z Z flux = (F · n)dS = (−4 + 1)dA = −3 × (area of R) S R  
= −3 2 (2 − e−3 ) − 1 = −6 + 6e−3 . 17. The surface is g(x, y, z) = x2 + y 2 + z 2 − a2 = 0. ∇g = 2(xi + yj + zk) = 2r n = r/|r|, F = −xi − yj − zk = cr/|r|3 c∇(1/|r|) + c∇(x2 + y 2 + z 2 )−1/2 = c 2 (x + y 2 + z 2 )3/2 r r |r|2 c c r·r F·n=− 3 · = −c 4 = −c 4 = − 2 = − 2 |r| |r| |r| |r| |r| a RR c c c RR dS = − 2 × (area of S) = − 2 (4πa2 ) = −4πc flux = F · ndS = − 2 S S a a a 18. In Problem 17, F is not continuous at (0, 0, 0) which is in any acceptable region containing the sphere. 19. Since F = c∇(1/r), divF = ∇ · (c∇(1/r)) = c∇2 (1/r) = c∇2 [(x2 + y 2 + z 2 )−1/2 ] = 0 by Problem 31 in Section 17.5. Then, by the Divergence Theorem, Z Z Z Z Z Z Z Z fluxF = F · ndS = divFdV = 0dV = 0. S

D

D

20. Parameterize C by x = 2 cos t, y = 2 sin t, z = 5, for 0Z ≤Zt ≤ 2π. Then I I (curlF · n)dS = F · dr = 6xdx + 7zdy + 8ydz S

C 2π

z

9

C

Z =

[12 cos t(−2 sin t) + 35(2 cos t)]dt 0

Z

2π

C

(70 cos t − 24 sin t cos t)dt

= 0

2π = (70 sin t − 12 sin2 t) 0 = 0.

2 y 2

21. Identify F = −2yi + 3xj + 10zk. Then curlF = 5k. The curve C lies in the plane z = 3, so n = k and dS = dA. Thus, I Z Z F · dr = (curlF) · ndS C Z ZS = 5dA = 5 × (area of R) = 5(25π) = 125π. R

22. Since curlF = 0,

H

F · dr =

RR S

(curlF · n)dS =

RR S

0dS = 0.

z

x

C

6

R

6 x

10

y

286

CHAPTER 15. VECTOR INTEGRAL CALCULUS z 1

23. divF Z Z = 1 + 1 + 1Z=Z3; Z F · ndS = divFdV S Z Z ZD = 3dV = 3 × (volume of D) = 3π

1

D

y

1 x

24. divF = x2 + y 2 + z 2 . Using cylindrical coordinates, Z Z

Z Z Z

Z Z Z

F · ndS =

(x2 + y 2 + z 2 )dV =

divFdV =

S

D

Z

D

2π

Z

0

1

Z

0

1

(r2 + z 2 )rdzdrdθ

0

 1   Z 2π Z 1  1 1 r3 + r drdθ = r3 z + rz 3 drdθ = 3 3 0 0 0 0 0  1 Z 2π Z 2π  5 1 4 1 2 5π = r + r dθ = dθ = . 4 6 12 6 0 0 0 Z

2π

Z

1

25. divF Z Z = 2x + 2(x Z+ y) Z − Z 2y = 4x Z Z Z F · ndS = divFdV = 4xdV S

Z

1

Z

D 1−x2

D

Z

=

z

z=1-x2

2−z

4xdydzdx 0

Z

0 1

Z

y=2-z

1

2

1

0

y

x

1−x2

4x(2 − z)dzdx

= 0

Z

0 1

Z

1−x2

Z

1

(8x − 4xz)dzdx =

= 0

Z =

0

0

1−x2 (8xz − 2xz ) dx 2

0

1

[8x(1 − x2 ) − 2x(1 − x2 )2 ]dx

0

  1 1 5 = −2(1 − x2 )2 + (1 − x2 )3 = 3 3 0 S3

z

p 26. For S1 , n = (xi + yj)/ x2 + y 2 ; for S2 , n2 = −k and z = 0; and for S3 , n3 = k and z = c. Then

c S1

a x

S2

y

CHAPTER 15 IN REVIEW

287

Z Z

Z Z

Z Z

F · ndS =

Z Z

F · n1 dS1 +

S

F · n2 dS2 +

S1

S2

F · n3 dS3 S3

Z Z Z Z x2 + y 2 2 p dS1 + (−z − 1)dS2 + (z 2 + 1)dS3 = x2 + y 2 s2 S3 S1 Z Z Z Z Z Z p 2 2 x + y dS1 + (−1)dS2 + (c2 + 1)dS3 = S2 S3 S1 Z Z Z Z Z Z 2 =a dS1 − dS2 + (c + 1) dS3 Z Z

S1

S2

S3

= a(2πac) − πa2 + (c2 + 1)πa2 = 2πa2 c + πa2 c2 . 27. x2 − y 2 = u2 (cosh v)2 − u2 (sinh v)2 hyperbolic paraboloid   = u2 (cosh v)2 − (sinh v)2 u2 = z; 28. z = x2 + y 2 ; paraboloid 29. y = x2 ; parabolic cylinder 30. x2 + y 2 − z 2 = (cos u cosh v)2 + (sin u cosh v)2 − (sinh v)2 = (cosh v)2 − (sinh v)2 = 1; z 2 = x2 + y 2 − 1 frustum of a cone