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Review Exercises for Chapter 14

215

Review Exercises for Chapter 14
1. F x, y, z
z 3 2

xi

j

2k

3. f x, y, z F x, y, z

8x 2 16x

xy yi

z2 xj 2z k

3 4 x 4 y

5. Since M y

1 y2

N

x, F is not conservative. x 6xy 2 y3 7y 3x 2 and N U y 6x 2 y 3y 2 7, g x which suggests h y y3 7y,

N x, F is conservative. From M U 7. Since M y 12xy partial integration yieldsU 3x 2 y 2 x3 h y and U 3x 2 y 2 gx x3, and U x, y 3x 2 y 2 x3 y3 7y C. 9. Since M y 4x N , x

P M 1 . z x F is not conservative. 11. Since M y 1 y 2z N , x M z 1 yz2 P , x N z x y 2z2 P , y

F is conservative. From M we obtain U x yz x2i 2x P y f y, z , U x yz g x, z , U x yz h x, y ⇒ f x, y, z x yz K U x 1 , N yz U y x , P y 2z U z x yz2

13. Since F (a) div F (b) curl F

y 2j 2y

z2k: 2z N i z P x M j z N x M k y 0i 0j 0k 0

15. Since F (a) div F (b) curl F

cos y

y cos x i x cos y cos x

sin x xy sin y

x sin y j

x yz k:

y sin x xz i yz j

sin y

cos x k

xz i

yz j

216

Chapter 14

Vector Analysis ln x 2 2x x2 2x x2 (b) curl F 2x x2 y2 2y y2 2y k y2 x2 1 y2 i ln x 2 2y y2 y2 j 1

17. Since F

arcsin x i xy 2 j yz2 k: 1 (a) div F 2xy 2yz1 x2 (b) curl F z2 i y2k

19. Since F (a) div F

z k:

21. (a) Let x x2
C

t, y

t, y 2 ds

1 ≤ t ≤ 2, then ds
2

2 dt. 2 2 t3 3
2

2t 2 2 dt
1

6 2
1

(b) Let x x2
C

4 cos t, y y 2 ds

4 sin t, 0 ≤ t ≤ 2 , then ds
2

4 dt.

16 4 dt
0

128

23. x

cos t x2
C

t sin t, y y 2 ds

sin t
2

t cos t, 0 ≤ t ≤ 2 , cos t t sin t
2 2

dx dt

t cos t,t cos t
2

dy dt

t sin t
2

sin t

t 2 cos2 t

t 2 sin2 t dt
0

t3

t dt

0

2 25. (a) Let x 2t, y 2x
C

2

1

2

3t, 0 ≤ t ≤ 1
1 1

y dx

x

3y dy
0

7t 2 3 sin t dt, dy
2

7t

3 dt
0

35t dt

35 2

(b) x

3 cos t, y 2x
C

3 sin t, dx x 3y dy

3 cos t dt, 0 ≤ t ≤ 2 9 sin t cos t dt 18

y dx

9
0

27.
C

2x x t y t 2x
C

yds, r t 3a 3a

a cos3 t i

a sin3 t j, 0 ≤ t ≤

2

cos2 t sin t sin2 t cos t
2

y ds
0

2a

cos3 t

a

sin3 t

x t

2

y t 2 dt

9 a2 5

29. f x, y C: y rt r t r t

5

sin x

y

3x from 0, 0 to 2, 6 ti i 3t j, 0 ≤ t ≤ 2 3j 10

Lateral surface area:
2 2

f x, y ds
C2 0

5

sin t

3t

10 dt

10
0

5

sin 4t dt

10 41 4

cos 8

32.528Review Exercises for Chapter 14 31. d r F F
C

217

2t i t5 i dr

3t 2 j dt t 4 j, 0 ≤ t ≤ 1
1

33. dr F F
C

2 sin t i 2 cos t i
2

2 cos t j 2 sin t j

k dt

t k, 0 ≤ t ≤ 2
2

5t 6 dt
0

5 7

dr
0

t dt

2

35. Let x F F
C

t, y t dr
2

t, z 2t 2 i
2

2t 2, 2t 2 tj 4t 3

2 ≤ t ≤ 2, dr 2t k 64 3

i

j

4t k dt.

3 2 2

4t 2 dt

37. For yFor y xy dx
C

x 2, r1 t 2x, r2 t x2

ti 2 y 2 dy

t 2 j, 0 ≤ t ≤ 2 ti 4 xy dx
C1

y

y = 2x
4 3

2t j, 0 ≤ t ≤ 2 x2 32 4 3 y 2 dy
C2

(2, 4) C2 y = x2 C1
x
1 2 3 4

xy dx

x2

y 2 dy

2

100 3

1

39. F

xi

y j is conservative. 1 2 x 2 2 3 y 3
4, 8 2 0, 0

Work

1 16 2 x 2 yz

2 3 8 3
1, 4, 3

2

8 3 3 12

4 2

41.
C

2xyz dx

x 2z dyx 2 y dz

0, 0, 0

1

43. (a)
C

y 2 dx

2xy dy
0 1

1 3 t2
0 1

t

2

3 1

21

3t 1 2 3t 2 4t

t dt 1 dt

2t 14t 7t 2

9t 2
0

5 dt
1

3t 3
4

5t
0

15 t 1 2 t dt

(b)
C

y 2 dx

2xy dy
1 4

t1 t
1 4

2t t dt 15

t2
1

(c) F x, y Hence, F
C

y 2i

2xy j

f where f x, y

xy 2.

dr

42

2

11

2

15

2

2

245.
C

y dx

2x dy
0 0

2

1 dy dx
0

2 dx

4

47.
C

xy 2 dx

x 2y dy
R

2xy

2xy dA

0

218

Chapter 14

Vector Analysis
1 x 1

49.
C

xy dx

x 2 dy
0 x
2

x dy dx
0

x2

x3 dx

1 12

51. r u, v 0 ≤ u ≤

sec u cos v i 3

1

2 tan u sin v j

2u k
6

z

, 0 ≤ v ≤ 2
−4 2 4 x −2 2 4 y

53. (a)
3 −4 4 x −2 −3

z

(b)
3...