Solucionario zill 4edicion capitulo 15

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Chapter 15

Vector Integral Calculus

15.1 Line Integrals

1. Z C 2xy dx = Z π/4 0

2(5 cos t)(5 sin t)(−5 sin t) dt = −250 Z π/4 0 sin2t cos t dt = −250 1 3sin 3_t π/4 0 = −125 √ 2 6 Z C 2xy dy = Z π/4 0

2(5 cos t)(5 sin t)(5 cos t) dt = 250 Z π/4 0 cos2t sin t dt = 250 −1 3cos 3_t π/4 0 = 250 3 1 − √ 2 4 ! =125 6 (4 − √ 2) Z C 2xy ds = Z π/4 0

2(5 cos t)(5 sin t)p25 sin2t + 25 cos2_{t dt = 250}

Z π/4 0 sin t cos t dt = 250 1 2sin 2_t π/4 0 =125 2 2. Z C (x3+ 2xy2+ 2x) dx = Z 1 0 [8t3+ 2(2t)(t4) + 2(2t)]2 dt = 2 Z 1 0 (8t3+ 4t5+ 4t) dt = 2 Z 1 0 (8t3+ 4t5+ 4t) dt = 2 2t4+2 3t 6_{+ 2t}2 1 0 = 28 3 Z C (x3+ 2xy2+ 2x) dy = Z 1 0 [8t3+ 2(2t)(t4) + 2(2t)]2t dt = 2 Z 1 0 (8t4+ 4t6+ 4t2) dt = 2 8 5t 5₊4 7t 7₊4 3t 3 1 0 = 736 105 218

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15.1. LINE INTEGRALS 219 Z C (x3+ 2xy2+ 2x) ds = Z 1 0 [8t3+ 2(2t)(t4) + 2(2t)]p4 + 4t2 _{dt = 8} Z 1 0 t(1 + t2)5/2 dt = 1 7(1 + t 2₎7/2 1 0 =8 7(2 7/2_{− 1)} 3. Z C (3x2+ 6y2) dx = Z 0 −1 [3x2+ 6(2x + 1)2] dx = Z 0 −1 (27x2+ 24x + 6) dx = (9x3+ 12x2+ 6x) 0 −1 = −(−9 + 12 − 6) = 3 Z C (3x2+ 6y2) dy = Z 0 −1 [3x2+ 6(2x + 1)2]2 dx = 6 Z C (3x2+ 6y2) ds = Z 0 −1 [3x2+ 6(2x + 1)2]√1 + 4 dx = 3√5 4. Z C x2 y3 dx = Z 8 1 x2 27x2_/8 dx = 8 27 Z 8 1 dx = 56 27 Z C x2 y3 dy = Z 8 1 x2 27x2_/8x −1/3 _{dx =} 8 27 Z 8 1 x−1/3 dx = 4 9x 2/3 8 1 =4 3 Z C x2 y3 ds = Z 8 1 x2 27x2_/8 p 1 + x−2/3 _{dx =} Z 8 1 x−1/3p1 + x2/3 _{dx =} 8 27(1 + x 2/3₎3/2 8 1 = 8 27(5 3/2_{− 2}3/2₎ 5. Z C (x2+ y2)ds = Z 2π 0

(25 cost−25 sin2_t)p_{25 sin}2_{t + 25 cos}2_{tdt = 125}

Z 2π 0 (cos2t − sin2t)dt = 125 Z 2π 0 cos 2tdt = 125 2 sin 2t 2π 0 = 0 6. Z C (2x + 3y) d = Z π 0

(6 sin 2t + 6 cos 2t)(−4 sin 2t) dt =

Z π

0

−24 sin22t − 24 sin 2t cos 2t dt = Z π 0 −24 1 2(1 − cos 2t) − 24 sin 2t cos 2t dt = −12t + 6 sin22t − 12 sin22t π 0 = −12π 7. Z C z dx = Z π/2 0

t(− sin t) dt Integration by parts = (t cos t − sin t)|π/2₀ = −1 Z C z dy = Z π/2 0

t cos t dt Integration by parts = (t sin t + cos t)|π/2₀ = π

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Z C z dz = Z π/2 0 t dt = 1 2t 2 π/2 0 =π 2 8 Z C z dx = Z π/2 0 tpsin2t + cos2_{t + 1 dt =}√₂ Z π/2 0 t dt = π 2√₂ 8 8. Z C 4xyz dx = Z 1 0 4 1 3t 3 (t2)(2t)t2 dt = 8 3 Z 1 0 t8 dt = 8 27t 9 1 0 = 8 27 Z C 4xyz dy = Z 1 0 4 1 3t 3 (t2)(2t)2t dt = 16 3 Z 1 0 t7 dt = 2 3t 8 1 0 =2 3 Z C 4xyz dz = Z 1 0 4 1 3t 3 (t2)(2t)2 dt = 16 21 Z 1 0 t6 dt = 16 21 1 0 = 16 21 Z C 4xyz ds = Z 1 0 4 1 3t 3 (2t)pt4_{+ 4t}2_{+ 4 dt =} 8 3 Z 1 0 t6(t2+ 2) dt = 8 3 1 9t 9₊2 7t 7 1 0 =200 189 9. Using x as the parameter, dy = dx and

Z C = (2x + y) dx + xy dy = Z 2 −1 (2x + x + 3 + x2+ 3x) dx = Z 2 −1 (x2+ 6x + 3) dx = 1 3x 3_{+ 3x}2_{+ 3x} 2 −1 = 21.

10. Using x as the parameter, dy = 2x dx and Z C (2x + y) dx + xy dy = Z 2 −1 (2x + x2+ 1) dx + Z 2 −1 x(x2+ 1)2x dx = Z 2 −1 (2x4+ 3x2+ 2x + 1) dx = 2 5x 5_{+ x}3_{+ x}2_{+ x} 2 −1 =141 5 .

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 C (2x + y) d + xy dy = Z 2 −1 (2x + 2) dx + Z 5 2 2y dy = (x2+ 2x) 2 −1+ y 2 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 C (2x + y) d + xy dy = Z 0 2 (−1)y dy + Z 2 −1 2x dx + Z 5 0 2y dy = −1 2y 2 0 2 + x2 2 −1+ y 2 2 0 = 2 + 3 + 25 = 30

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15.1. LINE INTEGRALS 221 13. Using x as a the parameter, dy = 2xdx.

Z C y dx + x dy = Z 1 0 x2 dx + Z 1 0 x(2x) dx = Z 1 0 3x2 dx = x3 1 0= 1

14. Using x as a the parameter, dydx.

Z C y dx + x dy = Z 1 0 x dx + Z 1 0 x dx = Z 1 0 2x dx = x2 1 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 C y dx + x dy = 0 + Z 1 0 1 dx = 1

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 C y dx + x dy = 0 + Z 1 0 1 dy = 1 17. Z C (6x2+ 2y + 2) dx + 4xy dy = Z 9 4 (6t + 2t2)1 2t −1/2_{dt +}Z 9 4 4√tt dt + Z 9 4 (3t1/2+ 5t3/2) dt = (2t3/2+ 2t3/2) 9 4 = 460 18. R C(−y 2_{) dx + xy dy =}R2 0(−t 6_{)2 dt +}R2 0(2t)(t 3_)3t2 _{dt =}R2 0 4t 6 _{dt =} 4 7t 7 2 0 = 512 7 19. Z C 2x3y d + (3x + y) dy = Z 1 −1 2(y6)y2y dy + Z 1 −1 (3y2+ y) dy = Z 1 −1 (4y8+ 3y2+ y) dy = 4 9y 9_{+ y}3₊1 2y 2 1 −1 =26 9 20. Z C 4x dx + 2y dy = Z 2 −1 4(y3+ 1)3y2 dy + Z 2 −1 2y dy = Z 2 −1

(12y5+ 12y2+ 2y) dy = 2y6+ 4y3+ y2

2

−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.

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Z C (x2+ y2) dx − 2xy dy = Z 2 −2 x2 dx + Z π 0 4(−2 sin θ dθ) − Z π 0

8 cos θ sin θ(2 cos θ dθ)

= 1 3x 3 2 −2 − 8 Z π 0

(sin θ + 2 cos2θ sin θ) dθ

=16 3 − 8 − cos θ −2 3cos 3_θ π 0 =16 3 − 80 3 = − 64 3 22. We start at (0, 0) and use x as a parameter.

Z C (x2+ y2) dx − 2xy dy = Z 1 0 (x2+ x4) dx − 2 Z 1 0 xx2(2x dx) + Z 0 1 (x2+ x) dx − 2 Z 0 1 x√x 1 2x −1/2 dx = Z 1 0 (x2− 3x4_{) dx +}Z 0 1 x2 dx = Z 1 0 (−3x4) dx = −3 5x 5 1 0 = −3 5 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 C x2y3 dx − xy2 dy = Z −1 1 x2(1) dx + Z −1 1 −(−1)y2 _{dy +} Z 1 −1 x2(−1)3 dx + Z 1 −1 −(1)y2 _dy = 1 3x 3 −1 1 + 1 3y 3 1 −1 − 1 3x 3 1 −1 − 1 3y 3 1 −1 = −8 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 C x2y3 dx − xy2 dy = Z 0 2 x2(64) dx − Z 0 4 0 dy + Z 2 0 x2(8x3) dx − Z 2 0 x(4x2)2 dx = 64 3 x 3 0 2 + 4 3x 6 2 0 − 2x4 2 0= − 512 3 + 256 3 − 32 = − 352 3 25. Z C y dx − x dy = Z π 0 3 sin t(−2 sin t) dt − Z π 0 2 cos t(3 cos t) dt = −6 Z π 0 (sin2t + cos2) dt = −6 Z π 0 dt = −6π Thus, Z −C y dx − x dy = 6π.

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15.1. LINE INTEGRALS 223 26. Z C x2y3+ x3y2dy = Z 1 −1 x2(x12) dx + Z 1 −1 x3(x8)(4x3) dx = Z 1 −1 x14dx + Z 1 −1 4x14dx = Z 1 −1 5x14dx = 5 15x 15 1 −1 = 5 15+ 5 15= 2 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 C y dx + z dy + x dz = Z 1 0 3t(2 dt) + Z 1 0 4t(3 dt) + Z 1 0 2t(4 dt) + Z 1 0 (3 + 5t)(4 dt) Z 1 0 (4 + t)(5 dt) + Z 1 0 (2 + 4t) dt = Z 1 0 (55t + 34) dt = 55 2 t 2_{+ 34t} 1 0 = 123 2 28. Z C y dx + z dy + x dz = Z 2 0 t3(3 dt) + Z 2 0 5 4t 2 (3t2dt) + Z 2 0 (3t) 5 2t dt = Z 2 0 3t3+15 4 t 4₊15 2 t 2 dt = 3 4t 4₊3 4t 5₊5 2t 3 2 0 = 56

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 C y dx + z dy + z dz = Z 6 0 0 dx + Z 5 0 6 dz + Z 8 0 5 dy = 70

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 C y dx + z dy + z dz = Z 1 0 8t(6 dt) + Z 5 0 6 dz = 24t2 1 0+ 30 = 54 31. Z C 10x dx − 2xy2dy + 6xz dz = Z 1 0 10(t) dt − Z 1 0 2(t)(t2)2(2t) dt + Z 1 0 6(t)(t3)(3t2) dt = 5t2 1 0− 4t 6 1 0+ 18t 6 1 0 = 5 − 4 + 18 = 19

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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 then have Z C1 3x dx − y2dy + z2dz = Z 1 0 3t dt − Z 1 0 t2dt = 3 2− 1 3 = 7 6 Z C2 3x dx − y2dy + z2dz = Z 1 0 3(1)(0) dt − Z 1 0 (1)2(0) dt + Z 1 0 t2dt = 1 3 Z C3 3x dx − y2dy + z2dz = Z 1 0 3(1 − t)(−1) dt − Z 1 0 (1 − t)2(−1) dt + Z 1 0 (1 − t)2(−1) dt = −3 2 − −1 3 + −1 3 = −3 2 33. Z C1 y2dx + xydy = Z 1 0 (4t + 2)22dt + Z 1 0 (2t + 1)(4t + 2)4dt = Z 1 0 (64t2+ 64t + 16)dt = 64 3 t 3_{+ 32t}2_{+ 16t} 1 0 = 64 3 + 32 + 16 = 208 3 Z C2 y2dx + xydy = Z √ 3 1 4y4(2t)dt + Z √ 3 1 2t4(4t)dt = Z √ 3 1 16t5dt = 8 3t 6 √ 3 1 = 72 −8 3 = 208 3 Z C3 y2dx + xydy = Z e3 e 4(ln t)21 tdt + Z e3 e 2(ln t)22 tdt = Z e3 e 8 t(ln t) 2_{dt =} 8 3(ln t) 3 e3 e = 8 3(27 − 1) = 208 3 34. R C1xyds = R2 0 t(2t) √ 1 + 4dt = 2√5R2 0 t 2_{dt = 2}√₅ 1 3t 3 2 0 = 16 √ 5 3 Z C2 xyds = Z 2 0 t(t2)p1 + 4t2_{dt =} Z 2 0 t3p1 + 4t2_dt _{u = 1 + 4t}2_{, du = 8tdt; t}2₌1 4(u − 1) = Z 17 1 1 4(u − 1)u 1/21 8du = 1 32 Z 17 1 (u3/2− u1/2)du = 1 32 2 5u 5/2₋2 3u 3/2 17 1 = 391 √ 17 + 1 120 Z C3 xyds = Z 3 2 (2t − 4)(4t − 8)√4 + 16dt = 16√5 Z 3 2 (t − 2)2dt 16√5 1 3(t − 2) 3 3 2 =16 √ 5 3 C1 and C3 are different parameterization of the same curve, while C1 and C2 are different

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15.2. LINE INTEGRALS OF VECTOR FIELDS 225 curves.

35. We are given ρ = kx. Then

m = Z C ρ dx = Z π 0 kx ds = k Z π 0

(1 + cos t)psin2t + cos2_{t dt = k}

Z π

0

(1 + cos t) dt = k (t + sin t)|π₀ = kπ.

36. From Problem 35, m = kπ and ds = dt. Mx= Z C yρds = Z C kxyds = k Z π 0 (1 + cos t) sin tdtk − cos t +1 2sin 2_t π 0 = 2k My= intCxρds = intCkx2ds = k Z π 0 (1 + cos t)2dt = k Z π 0 (1 + 2 cos t + cos2t)dt = k t + 2 sin t +1 2t + 1 4sin 2t π 0 = 3 2kπ x = My/m = 3kπ/2 kπ = 3 2; y = Mx/m = 2k kπ = 2

π. The center of mass is (3/2, 2/π).

15.2 Line Integrals of Vector Fields

1.

x

y

2.

y

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3.

x

y

4.

x

y

5.

x

y

6.

x

y

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).

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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).

15. F = e3ti − (e−4t)etj = e3ti − e−3tj; dr = (−2e−2ti + etj)dt; F · dr = (−2et− e−2t)dt; Z c F · dr = Z ln 2 0 (−2et− e−2t)dt = (−2et+1 2e −2t₎ ln 2 0 = −31 8 − (− 3 2) = − 19 8 16. F = 2(t)(t2_{)i + t}2_{j = 2t}3_{i + t}2_{j; dr = (i + 2tj)dt;} F · dr = 4t3_dt; R CF · dr = R2 0 4t 2_{dt = t}4 2 0= 16

17. F = 2(2t − 1)i − 2(6t + 1)j = (4t − 2)i + (12t + 2)j; dr = (2i + 6j)dt; F · dr = −64t − 16; R CF · dr = R1 −1(−64t − 16) dt = −32t 2_{− 16t} 1 −1= −32

18. F = cos2_{ti + sin tj; dr = (− sin ti + cos tj);}

F · dr = (− cos2_{t sin t + sin t cos t)dt;}

Z

C

F · dr = Z π/6

0

(− cos2t sin t + sin t cos t) dt

= cos 3_t 3 + sin2t 2 π/6 0 = 1 3 √ 3 2 !3 +1 2 1 2 2 − 1 3 + 0 = √ 3 8 + 1 8− 1 3 = √ 3 8 − 5 24

19. F = −3 sin ti + 2 cos tj + 6tk; dr = (−2 sin ti + 3 cos tj + 3k)dt; F · dr = (−6 sin2t + 6 cos2_{t + 18t)dt;} Z C F · dr = Z π 0 (−6 sin2t + 6 cos2t + 18t) dt = Z π 0 −6 1 2(1 − cos 2t) + 6 1 2(1 + cos 2t) + 18t dt = 3 sin 2t + 9t2 π 0 = 9π2

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20. F = eti + tet3j + t3et6k; dr = (i + 2tj + 3t2k)dt; Z C F · dr = Z 1 0 (et+ 2t2et3+ 3t5et6)dt = (et+2 3e t3₊1 2e t6₎ 1 0 = 13 6 (e − 1) 21. Using x as a parameter, r(x) = xi + ln xj. Then F = ln xi + xj, dr = (i + 1

xj)dx, and W = Z c F · dr = Z e 1 (ln x + 1)dx = (x ln x)|e₁= e. x y c1 c₂

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

Then dr1= 2i − 2j, dr2= 2i + 3j, F1= 2(−2 + 2t)(2 − 2t)i + 4(2 − 2t)2j = (−8t2+ 16t − 8)i + (16t2− 32t + 16)j, F2= 2(2t)(3t)i + 4(3t)2j = 12t2i + 36t2j, and W = Z C1 F1· dr1+ Z cC2 F2· dr2 = Z 1 0 (−16t2+ 32t − 16 − 32t2+ 64t − 32)dt + Z 1 0 (24t2+ 108t2)dt = Z 1 0 (84t2+ 96t − 48)dt = (28t3+ 48t2− 48t) 1 0= 28. x y c3 c2 c1

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, 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 W = Z C1 F1· dr1+ Z c2 F2· dr2+ Z c3 F3· dr3 = Z 1 0 (6 + 4t)dt + Z 1 0 6tdt + Z 1 0 (−14 + 8t − 6 + 2t)dt = Z 1 0 (−14 + 20t)dt = (−14t + 10t2) 1 0= −4. 24. F = t3i + t4j + t5k; dr = 3t2i + 2tj + k; W =R CF · dr = R3 1(3t 5_{+ 2t}5_{= t}5_{)dt =}R3 1 6t 5_{dt = t}6 3 1= 728

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15.2. LINE INTEGRALS OF VECTOR FIELDS 229 25. r = 3 cos ti + 3 sin tj, 0 ≤ t ≤ 2π; dr = −3 sin ti + 3 cos tj; F = ai + bj;

W =R_CF · dr =R₀2π(−3a sin t + 3b cos t)dt = (3a cos t + 3b sin t)|2π₀ = 0 26. Let r = ti + tj + tk for 1 ≤ t ≤ 3. Then dr = i + j + k, and

F = c |r|3(ti + tj + tk) = ct (√3t2₎3(i + j + k) = c 3√3t2(i + j + k), W = Z C F · dr = Z 3 1 c 3√3t2(1 + 1 + 1)dt = c √ 3 Z 3 1 1 t2dt = c √ 3 −1 t 3 1 = √c 3(− 1 3+ 1) = 2c 3√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;

R

CF · dr =

R2π

0 (−100 cos t sin t) dt = 50 cos 2_t

2π 0 = 0

28. F = 10 cos ti + 2 sin tj; dr = (2 cos ti − 10 sin tj)dt; F · dr = (10 cos2t − 20 sin2t)dt; Z C F · dr = Z 2π 0 (10 cos2t − 20 sin2t) dt = Z 2π 0 10 1 2(1 + cos 2t) − 20 1 2(1 − cos 2t) dt = −5t +15 2 sin 2t π 0 = −10π

29. On C1, T = i and F · TcompTF ≈ 1. On C2, T = −j and F · T = compTF ≈ 2. OnC3, T = −i

and F · T = comp_TF ≈ 1.5. Using the fact that the lengths of C1, C2, and C3 are 4, 5, and

5, respectively, we have W =R_CF · Tds =R_C 1F · Tds + R C2F · Tds + R C3F · Tds ≈ 1(4)+2(5)+1.5(5)=21.5 ft-lb. 30. W = Z C F · dr = Z b a (ma · r0(t)) dt = Z b a m(a · v) dt = Z b a m dv dt · v dt = Z b a m 2 dv dt · v + v · dv dt dt = Z b a m 2 d dt(v · v) dt = Z b a m 2 d dt(v 2_{) dt =} m 2v 2 b a = 1 2m[v(b)] 2₋1 2m[v(a)] 2 = K(B) − K(A) 31. ∇f (x, y) = 1 3(3x − 6y)3i + 1 3(3x − 6y)(−6)j = (3x − 6y)i + (−6x + 12y)j

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32. ∇f (x, y) = (1 + 2 cos 5xy − 10xy sin 5xy)i + (−1 − 10x2_{sin 5xy)j} 33. ∇f (x, y, z) = tan−1zyi + xz y2_z2_{+ 1}j + xy y2_z2_{+ 1}k 34. ∇f (x, y, z) = (1 − 2xyz4)i − x2z4j − 4x2yz3k

35. ∇f (x, y, z) = e−y2i +1 + 2xye−y2j + k

36. ∇f (x, y, z) = 2x x2_{+ 2y}4_{+ 3z}6i + 8y3 x2_{+ 2y}4_{+ 3z}6j + 18z5 x2_{+ 2y}4_{+ 3z}6k

37. ∇ x2+1₂y2 = 2xi + yj = F(x, y). Therefore, the answer is (b).

38. ∇ 1₂x2_{+ y}2_{− 4 = xi + 2yj = F(x, y). Therefore, the answer is (c).}

39. ∇ 2x +1 2y

2_{+ 1 = 2i + yj = F(x, y). Therefore, the answer is (d).}

40. ∇ 1₂x2+1₃y3− 5 = xi + y2j = 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 + y2_{− 4z}3

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15.2. LINE INTEGRALS OF VECTOR FIELDS 231 45. x y 46. x y 47. x y 48. x y

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49. x y 50. x y 51. Let φ(x, y, z) = −c(x2+ y2+ z2)−1/2. Then ∇φ(x, y, z) = cx (x2_{+ y}2_{+ z}2₎3/2i + cy (x2_{+ y}2_{+ z}2₎3/2j + cz (x2_{+ y}2_{+ z}2₎3/2k = c(xi + yj + zk) (x2_{+ y}2_{+ z}2₎3/2 = cr |r|3 = F

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

15.3 Independence of the Path

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

1 3x 3_{+ g(y), φ} y= g0(y) = y2_{, g(y) =} 1 3y 3_{, φ =} 1 3x 3₊1 3y 3_, R(2,2) (0,0) x 2_{dx + y}2_{dy =} 1 3(x 3_{+ y}3₎ (2,2) (0,0) = 16 3 (b) Use y = x for 0 ≤ x ≤ 2. R(2,2) (0,0) x 2_{dx + y}2_{dy =}R2 0(x 2_{+ x}2_{)dx =} 2 3x 3 2 0 =16 3

2. (a) Py= 2x = Qxand the integral is independent of path. φx= 2xy, φ = x2y + g(y), φy=

x2+ g0(y) = x2, g(y) = 0, φ = x2y, R_(1,1)(2,4)2xydx + x2dy = x2y

(2,4) (1,1)= 16 − 1 = 15 (b) Use y = 3x − 2 for 1 ≤ x ≤ 2. R(2,4) (1,1)2xydx + x 2_{dy =}R2 1[2x(3x − 2) + x 2_{(3)]dx =}R2 1(9x 2_{− 4x)dx = (3x}3_{− 2x}2₎ 2 1= 15

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15.3. INDEPENDENCE OF THE PATH 233

3. (a) Py = 2 = Qx and the integral is independent of path. φx= x + 2y, φ =

1 2x

2_{+ 2xy +}

g(y), φy = 2x + g0(y) = 2x − y, g(y) = −

1 2y 2_{, φ =} 1 2x 2_{+ 2xy −}1 2y 2_, R(3,2) (1,0)(x + 2y)dx + (2x − y)dy = 1 2x 2_{+ 2xy −}1 2y 2 (3,2) (1,0) = 14 (b) Use y = x − 1 for 1 ≤ x ≤ 3. Z (3,2) (1,0) (x + 2y)dx + (2x − y)dy = Z 3 1 [x + 2(x − 1) + 2x − (x − 1)]dx = Z 3 1 (4x − 1)dx = (2x2− x) 3 1= 14

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 + g0(y) = 1 − sin x sin y, g(y) = y, φ =

sin x cos y + y, R(π/2,0)

(0,0) cos x cos ydx + (1 − sin x sin y)dy = (sin x cos y + y)| (π/2,0) (0,0) = 1

(b) Use y = 0 for 0 ≤ x ≤ π/2. Z (π/2,0)

(0,0)

cos x cos ydx + (1 − sin x sin y)dy = Z π/2

0

cos xdx = sin x|π/2₀ = 1

5. (a) Py = 1/y2 = Qx and the integral is independent of path. φx = −

1 y, φ = − x y + g(y), φy = x y2 + g 0_{(x) =} x y2, g(y) = 0, φ = − x y, R(4,4) (4,1)− 1 ydx + x y2dy = (− x y) (4,4) (4,1) = 3 (b) Use x = 4 for 1 ≤ y ≤ 4. Z (4,4) (4,1) −1 ydx + x y2 dy = Z 4 1 4 y2dy = − 4 y 4 1 = 3

6. (a) Py= −xy(x2+y2)−3/2= Qxand the integral is independent of path. φx=

x p x2_{+ y}2, φ = p x2_{+ y}2_{+ g(y), φ} y= y p x2_{+ y}2 + g 0_{(y) =} y p x2_{+ y}2, g(y) = 0, φ = p x2_{+ y}2_, Z (3,4) (1,0) xdx + ydy p x2_{+ y}2 = p x2_{+ y}2 (3,4) (1,0) = 4 (b) Use y = 2x − 2 for 1 ≤ x ≤ 3. Z (3,4) (1,0) xdx + ydy p x2_{+ y}2 = Z 3 1 x + (2x − 2)2 px2_{+ (2x − 2)}2 dx = Z 3 1 5x − 4 √ 5x2_{− 8x + 4} =p5x2_{− 8x + 4} 3 1= 4

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7. (a) Py = 4xy = Qx and the integral is independent of path. φx = 2y2x − 3, φ =

x2_y2_{− 3x + g(y),} _φ

y = 2x2y + g0(y) = 2x2y + 4, g(y) = 4y, φ = x2y2 − 3x +

4y, R_(1,2)(3,6)(2y2x − 3)dx + (2yx2+ 4)dy = x2y2− 3x + 4y) (3,6) (1,2)= 330 (b) Use y = 2x for 1 ≤ x ≤ 3. Z (3,6) (1,2)

(2y2x − 3)dx + (2yx2+ 4)dy = Z 3 1 ([2(2x)2x − 3] + [2(2x)x2+ 4]2)dx = Z 3 1 (16x3+ 5)dx = (4x4+ 5x) 3 1= 330

8. (a) Py = 4 = Qx and the integral is independent of path. φx= 5x + 4y, φ =

5 2x

2_{+ 4xy +}

g(y), φy = 4x + g0(y) = 4x − 8y3, g(y) = −2y4, φ =

5 2x 2_{+ 4xy − 2y}4_, R(0,0) (−1,1)(5x + 4y)dx + (4x − 8y 3_{)dy =} 5 2x 2_{+ 4xy − 2y}4 (0,0) (−1,1) = 7 2 (b) Use y = −x for −1 ≤ x ≤ 0. Z (−1, 1)(0,0)(5x + 4y)dx + (4x − 8y3)dy = Z 0 −1 [(5x − 4x) + (4x + 8x3)(−1)]dx = Z 0 −1 (−3x − 8x3)dx = (−3 2x 2 − 2x4) 0 −1 = 7 2 9. (a) Py= 3y2+ 3x2= Qxand the integral is independent of path. φx= y3+ 3x2y,

; φ = xy3_{+ x}3_{y + g(y),} _φ

y = 3xy2+ x3+ g0(y) = x3+ 3y2x + 1, g(y) = y, φ =

xy3_{+ x}3_{y + y,} R(2,8)

(0,0)(y

3_{+ 3x}2_{y)dx + (x}3_{+ 3y}2_{x + 1)dy = .(xy}3_{+ x}3_{y + y)} (2,8) (0,0) = 1096 (b) Use y = 4x for 0 ≤ x ≤ 2. Z (2,8) (0,0) (y3+ 3x2y)dx + (x3+ 3y2x + 1)dy = Z 2 0 [(64x3+ 12x3) + (x3+ 48x3+ 1)(4)]dx = Z 2 0 (272x3+ 4)dx = (68x4+ 4x) 2 0= 1096 10.

11. Py = 12x3y2 = Qx throughout the plane and the vector field is a conservative field. φx =

4x3_y3_{+3, φ = x}4_y3_{+3x+g(y), φ}

y = 3x4y2+g0(y) = 3x4y2+1, g(y) = y, φ = x4y3+3x+y

12. Py = 6xy2 = Qx throughout the plane and the vector field is a conservative field. φx =

2xy3_{, φ = x}2_y3_{+ g(y), φ}

y= 3x2y2+ g0(y) = 3x2y2+ 3y2, g(y) = y3, φ = x2y3+ y3

13. Py = −2xy3sin xy2+ 2y cos xy2, Qx= −2xy3cos xy2− 2y sin xy2 throughout the plane and

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15.3. INDEPENDENCE OF THE PATH 235 14. Py = −4xy(x2+ y2+ 1)−3= Qx throughout the plane and the vector field is a conservative

field. φx= x(x2+ y2+ 1)−2, φ = − 1 2(x 2_{+ y}2_{+ 1)}−1_{+ g(y), φ} y= y(x2+ y2+ 1)−2+ g0(y) = y(x2_{+ y}2_{+ 1)}−2_{, g(y) = 0, φ = −}1 2(x 2_{+ y}2_{+ 1)}−1

15. Py = 1 = Qx throughout the plane and the vector field is a conservative field. φx = x3+

y, φ = 1 4x 4_{+ xy + g(y), φ} y = x + g0(y) = x + y3, g(y) = − 1 4y 4_{, φ =} 1 4x 4_{+ xy +}1 4y 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. φx= 2x, φ = x2+ g(y, z)φy = ∂g ∂y = 3y 2_{− x,} g(y, z) = y3− yz + h(z), φ = x2_{+ y}3_{− yz + h(z),} φz= −y + h0(z) = −y, h(z) = 0, φ = x2+ y3− yz

18. Py = 2x = Qx, Pz= 0 = Rx, Qz = −e−y = Ry throughout 3-space and the vector field is a

conservative field. φx= 2xy, φ = x2y + g(y, z), φy= x2+ ∂g ∂y = x 2_{− ze}−y_, g = ze−y+ h(z), φ = x2_{y + ze}−y_{+ h(z),} φz= e−y+ h0(z) = ey− 1, h(z) = −z, φ = x2y + ze−y− z

19. Since Py = −e−y = Qx, F is conservative and R_CF · 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

W = Z

C1

(2x + e−y)dx + (4y − xe−y)dy

= Z 1 0 (2x + e−x)dx + Z 1 0 (4x − e−x)dx Integration by parts = (x2− e−x) 1 0+ (2x 2_{+ xe}−x₎ 1 0 = [(1 − e−1) − (−1)] + [(2 + e−1+ e−1) − (1)] = 3 + e−1. 20. Since Py = −e−y = Qx, F is conservative and

R

CF · 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 W =R_C

1(2x+e

−y_{)dx+(4y −xe}−y_{)dy =}R−2

2 (2x+1)dx = (x 2_{+ x)}

−2

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)

(1,1,1)

yzdx + xzdy + xydz = Z 1 0 [(1 + 3t)(1 + 7t) + (1 + t)(1 + 7t)(3) + (1 + t)(1 + 3t)(7)]dt = Z 1 0 (11 + 62t + 63t2)dt = (11t + 31t2+ 21t3) 1 0= 63.

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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 R(1,1,1) (0,0,0) 2xdx + 3y 2_{dy + 4z}3_{dz =}R1 0(2t + 3t 2_{+ 4t}3_{)dt = (t}2_{+ t}3_{+ t}4₎ 1 0= 3.

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 = x2sin y + xe3z+ g(y, z). Then φy =

x2cos y + gy = Q = x2cos y, so gy = 0, g(y, z) = h(z), and φ = x2sin y + xe3z+ h(z). Now

φz= 3xe3z+ h0(z) = R = 3xe3z+ 5, so h0(z) = 5 and h(z) = 5z. Thus φ = x2sin y + xe3z+ 5z

and

Z (2,π/2,1)

(1,0,0)

(2x sin y + e3z)dx + x2cos ydy(3xe3z+ 5)dz

= (x2sin y + xe3z+ 5z)

(2,π/2,1)

(1,0,0) = [4(1) + 2e

3_{+ 5] − [0 + 1 + 0] = 8 + 2e}3_.

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 Z (3,4,1) (1,2,1) (2x + 1)dx + 3y2dy +1 zdz = Z 1 0 [(2 + 4t + 1)2 + 3(2 + 2t)22]dt = Z 1 0 (24t2+ 56t + 30)dt = (8t3+ 28t2+ 30t) 1 0= 66.

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) (1,1,ln 3) e2zdx + 3y2dy + 2xe2zdz = Z 1 0 [e2 ln 3+ 3(1 + t)2]dt = [9t + (1 + t)3] 1 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) (−2,3,1) 2xzdx + 2yzdy + (x2+ y2)dz = Z 1 0 [−4(1 − t)2(2) + 6(1 − t)2(−3) + 4(1 − t)2(−1) + 9(1 − t)2(−1)]dt = Z 1 0 −39(1 − t)2_{dt = 13(1 − t)}3 1 0 = −13.

27. Py = 1 − z sin x = Qx, Qz = cos x = Ry, Rx = −y sin x = Pz and the integral is

in-dependent of path. Integrating θx = y − yz sin x we find θ = xy + yz cos x + g(y, z).

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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

C

F · dr = (xy + yz cos x)|π,1,4)_(0,4,0)= (π − 4) − (0 + 0) = π − 4.

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) = y2_{− y + h(z) and φ = 2x − xe}z_{+ y}2_{− y + h(z). Now φ}

z = −xez+ h0(z) = R = 2 − xez_{, so h}0_{(z) = 2, h(z) = 2z, and φ = 2x − xe}z_{+ y}2 _{− y + 2z. Thus} Z C F · dr = (2x − xez+ y2− y + 2z) (2,4,8) (−1,1,−1) = (4 − 2e8+ 16 − 4 + 16) − (−2 + e−1+ 1 − 1 − 2) = 36 − 2e8− e−1

29. Since Py = Gm1m2(2xy/|r|5) = Qx, Qz= Gm1m2(2yz/|r|5) = Ry, and Rx= Gm1m2(2xz/|r|5) =

Pz, the force field is conservative.

θx= −Gm1m2 x (x2_{+ y}2_{+ z}2₎3/2, θ = Gm1m2(x 2_{+ y}2_{+ z}2₎−1/2_{+ g(y, z),} θy= −Gm1m2 y (x2_{+ y}2_{+ z}2₎3/2 + gy(y, z) = −Gm1m2 y (x2_{+ y}2_{+ z}2₎3/2, g(y, z) = h(z), θ = Gm1m2(x2+ y2+ z2)−1/2+ h(z), θz= −Gm1m2 z (x2_{+ y}2_{+ z}2₎3/2 + h 0_{(z) = −Gm} 1m2 z (x2_{+ y}2_{+ z}2₎3/2, h(z) = 0, θ = _p Gm1m2 x2_{+ y}2_{+ z}2 = Gm1m2 |r|

30. Since Py = 24xy2z = Qx, Qz = 12x2y2 = Ry, and Rx = 8xy3 = Pz, F is conservative.

Thus, the work done between two points is independent of the path. From θx = 8xy3z we

obtain θ = 4x2_y3_{z which is a potential function for F. Then}

W = Z (1, √ 3,π/3) (2,0,0) F · dr = 4x2y3z (1,√3,π/3) (2,0,0) = 4 √ 3π and W =R(0,2,π/2) (2,0,0) F · dr = 0. c1 c₂ 31. Since F is conservative, R C1F · dr = R

−C2F · dr. Then, since the simply

closed curve C is composed of C1and C2,

Z C F · dr = Z C1 F · dr + Z C2 F · dr = Z C1 F · dr − Z −C2 F · dr = 0.

32. From F = (x2_{+ y}2₎n/2_{(xi + yj) we obtain P}

y= nxy(x2+ y2)n/2−1= Qx, so that F is

conser-vative. From θx= x(x2+ y2)n/2 we obtain the potential function θ + (x2+ y2)(n+2)/2/(n + 2).

Then W = Z (x2,y2) (x1,y1) F·dr = ((x 2_{+ y}2₎(n+2)/2 n + 2 ) (x2,y2) (x1,y1) = 1 n + 2 h (x2₂+ y2₂)(n+2)/2− (x2 1+ y 2 1) (n+2)/2i_.

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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 C F · dr = Z 2 1 (2x cos(x − 1) − x2sin(x − 1))dx = x2cos(x − 1) 2 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 C F · dr = Z 1 0 (sin x + x cos x + x2)dx = x sin x +x 3 3 1 0 = sin 1 +1 3 35. F cannot be a conservative field in the region.

36. (a) Py=

y2_{− x}2

(y2_{+ x}2₎2 = Qx. Using the hint, we have

Z C F · dr = Z 2π 0 (sin2t + cos2t)dt = Z 2π 0 dt = 2π. 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 Qxare not defined at the origin. Therefore, the theorem

does not apply.

37. From Problem 45 in Exercises 15.2, dv dt · dr dt = dv dt · v = 1 2 d dtv 2_{. Then, using} dp dt = ∂p ∂x dx dt + ∂p ∂y dy dt = ∇p · dr dt, we have Z mdv dt · drdtdt + Z ∇p · dr dt = Z 0dt 1 2m Z _d dtv 2_{dt +}Z dp dtdt = constant 1 2mv 2_{+ p = constant.}

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).

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15.4. GREEN’S THEOREM 239

15.4 Green’s Theorem

x y x=1 1

1. The sides of the triangle are C1: y = 0, 0 ≤ x ≤ 1; C2:

x = 1, 0 ≤ y ≤ 3; C3: y = 3x, 0 ≤ −x ≤ 1. Z C (x − y)dx + xydy = Z 1 0 xdx + Z 3 0 ydy + Z 0 1 (x − 3x)dx + Z 0 1 x(3x)dx = 1 2x 2 1 0 + 1 2y 2 3 0 + (−x2) 1 0+ (3x 2₎ 0 1 = 1 2 + 9 2 + 1 − 3 = 3 Z Z R (y + 1)dA = Z 1 0 Z 3x 0 (y + 1)dydx = Z 1 0 1 2y 2_{+ y} 3x 0 dx = Z 1 0 9 2x 2_{+ 3x} dx = 3 2x 3₊3 2x 2 1 0 = 3

2. The sides of the rectangle are C1: y = 0, −1 ≤ x ≤ 1; C2: x = 1, 0 ≤ y ≤ 1; C3:

y = 1, 1 ≥ x ≥ −1; C4: x = −1, 1 ≥ y ≥ 0. Z C 3x2ydx+(x2− 5y)dy = Z 1 −1 0dx + Z 1 0 (1 − 5y)dy = Z 1 −1 0dx + Z 1 0 (1 − 5y)dy = Z −1 1 3x2dx + Z 0 1 (1 − 5y)dy = y −5 2y 2 1 0 + x3 −1 1 + y −5 2y 2 1 0 = −2 R R R(2x − 3x 2_{)dA =}R1 0 R1 −1(2x − 3x 2_{)dxdy =} R1 0(x 2_{− x}3₎ 1 −1dy = R1 0(−2)dy = −2 3. Z C −y2_{dx + x}2_{dy =} Z 2π 0 (−9 sin2t)(−3 sin t)dt + Z 2π 0 9 cos2t(3 cos t)dt = 27 Z 2π 0

[(1 − cos2t) sin t + (1 − sin2t) cos t]dt

= 27 − cos t +1 3cos 3_{t + sin t −}1 3sin 3_t 2π 0 = 27(0) = 0 Z Z R (2x + 2y)dA = 2 Z 2π 0 Z 3 0 (r cos θ + r sin θ)rdrdθ = 2 Z 2π 0 Z 3 0 r2(cos θ + sin θ)drdθ = 2 Z 2π 0 1 3r 3_{(cos θ + sin θ)} 3 0 dθ = 18 Z 2π 0 (cos θ + sin θ)dθ = 18(sin θ − cos θ)|2π₀ = 18(0) = 0

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x y

R

2

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 C −2y2_{dx + 4xydy =}Z 2 0 0dx + Z 1 2 −2(−x + 2)2_dx + Z 1 2 4x(−x + 2)(−dx) + Z 0 1 −2xdx + Z 0 1 4x√x ₁ 2√x dx = 0 +2 3 + 8 3 + 1 − 1 = 10 3 R R R8ydA = R1 0 R2−y y2 8ydxdy = R1 0 8y(2 − y − y 2_{)dy =} 8y2₋8 3y 3_{− 2y}4 1 0 = 10 3 x y R 5. P = 2y, Py= 2, Q = 5x, Qx= 5 Z C 2ydx + 5xdy = Z Z R (5 − 2)dA = 3 Z Z R dA = 3(25π) = 75π x y R 2 4 6. P = x + y2_{, P} y= 2y, Q = 2x2− y, Qx= 4x Z C (x + y2)dx + (2x2− y)dy = Z Z R (4x − 2y)dA = Z 2 −2 Z 4 x2 (4x − 2y)dydx = Z 2 −2 (4xy − y2) 4 x2 dx = Z 2 −2 (16x − 16 − 4x3+ x4)dx = 8x2− 16x − x4₊1 5x 5 2 −2 = −96 5 x y 2 2 r=2 R 7. P = x4− 2y3_, _P y = −6y2, Q = 2x3− y4, Qx = 6x2.

Using polar coordinates, Z C (x4− 2y3_{)dx + (2x}3_{− y}4_{)dy =}Z Z R (6x2+ 6y2)dA = Z 2π 0 Z 2 0 6r2rdrdθ = Z 2π 0 3 2r 4 2 0 dθ = Z 2π 0 24dθ = 48π.

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15.4. GREEN’S THEOREM 241 x y R 3 3 8. P = x − 3y, Py= −3, Q = 4x + y, Qx= 4 R C(x − 3y)dx + 4(x + y)dy = R R R(4 + 3)dA = 7(10) = 70 x y R y=2x 2 4 9. P = 2xy, Py= 2x, Q = 3xy2, Qx= 3y2 Z C 2xydx + 3xy2dy = Z Z R (3y2− 2x)dA = Z 2 1 Z 2x 2 (3y2− 2x)dydx = Z 2 1 (y3− 2xy) 2x 2 dx = Z 2 1 (8x3− 4x2− 8 + 4x)dx = 2x4−4 3x 3_{− 8x + 2x}2 2 1 =40 3 − −16 3 =56 3 10. P = e2x_{sin 2y, P}

y= 2e2xcos 2y, Q = e2xcos 2y, Qx= 2e2xcos 2y

R

C= e

2x_{sin 2ydx + e}2x_{cos 2ydy =}R R

R0dA = 0 x y R r=1 1 1

11. P = xy, Py = x, Q = x2, Qx = 2x. Using polar

coordinates, Z C xydx + x2dy = Z Z R (2x − x)dA = Z π/2 −π/2 Z 1 0 r cos θrdrdθ = Z π/2 −π/2 1 3r 3_{cos θ} 1 0 dθ = Z π/2 −π/2 1 3cos θdθ = 1 3sin θ π/2 −π/2 =2 3 x y R 1 y=-x -1 12. P = ex2, Py= 0, Q = 2 tan−1x, Qx= 2 1 + x2 Z C ex2dx + 2 tan−1xdy = Z Z R 2 1 + x2dA = Z 0 −1 Z 1 −x 2 1 + x2dydx = Z 0 −1 _2y 1 + x2 1 −x dx = Z 0 −1 ₂ 1 + x2 + 2x 1 + x2 dx = [2 tan−1x + ln(1 + x2)] 0 −1= 0 − −π 2 + ln 2 =π 2 − ln 2

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x y R x=y2 x=1-y2 1 1 13. P = 1 3y 3_{, P} y = y2, Q = xy + xy2, Qx= y + y2 Z C 1 3y 3_{dx + (xy + xy}2_{)dy =}Z Z R ydA = Z 1/ √ 2 0 Z 1−y2 y2 ydxdy = Z 1/ √ 2 0 (xy) 1−y2 y2 dy = Z 1/ √ 2 0 (y − y2− y3_)dy = 1 2y 2₋1 2y 4 1/√2 0 = 1 4− 1 8 = 1 8 x y R y=x2 y=x3 14. P = xy2, Py= 2xy, Q = 3 cos y, Qx= 0 Z C xy2dx + 3 cos ydy = Z Z R (−2xy)dA = − Z 1 0 Z x2 x3 2xydydx = − Z 1 0 (xy) x2 x3 dx = − Z 1 0 (x3− x4_)dx = 1 4x 4₋1 5x 5 1 0 = −1 20 15. P = ay, Py = a, Q = bx, Qx= b. R Caydx + bxdy = R R

R(b − a)dA = (b − a) × (area bounded by C)

16. P = P (x), Py= 0, Q = Q(y), Qx= 0.

R

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

R R

R0dA = 0

17. For the first integral: P = 0, Py = 0, Q = x, Qx= 1;

R

Cxdy = −

R R

R1dA =area of R.

For the second integral: P = y, Py = 1, Q = 0, Q = 0; −

R Cydx = − R R R−dA =area of R. Thus,R Cxdy = − R Cydx. 18. P = −y, Py = −1, Q + x, Qx= 1. 1 2 R C−ydx + xdy = 1 2 R R R2dA = R R RdA =area of R 19. A = Z Z R dA = Z C xdy = Z 2π 0

a cos3t(3a sin2t cos tdt) = 3a2 Z 2π 0 sin2t cos4tdt = 3a2 1 16t − 1 64sin 4t + 1 48sin 3 2t 2π 0 = 3 8πa 2 20. A = Z Z R dA = Z C xdy = Z 2π 0 a cos t(b cos tdt) = ab Z 2π 0 cos2tdt = ab 1 2t + 1 4sin 2t 2π 0 = πab

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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 C −ydx + xdy = Z 1 0 −[y1+ (y2− y1)t](x2− x1)dt + Z 1 0 [x1+ (x2− x1)t](y2− y1)dt = −(x2− x1)[y1t + 1 2(y2− y1)t 2_] 1 0 + (y2− y1)[x1t + 1 2(x2− x1)t 2_] 1 0 = −(x2− x1) y1+ 1 2(y2− y1) + (y2− y1) x1+ 1 2(x2− x1) = x1y2− x2y1.

(b) Let Cibe the line segment from (xi, yi) to (xi+1, yi+1) for i = 1, 2, · · · , n − 1, and C2

the line segment from (xn, yn)to (x1, y1). Then

A = 1 2

Z

C

−ydx + xdy Problem 18

= 1 2 " Z C1 −ydx + xdy + Z C2 −ydx + xdy + · · · + Z Cn−1 −ydx + xdy + Z Cn −ydx + xdy # = 1 2(x1y2− x2y1) + 1 2(x2y3− x3y2) + 1 2(xn−1yn− xnyn−1) + 1 2(xny1− x1yn).

22. From part (b) of Problem 21 A = 1 2 (−1)(1) − (1)(3)] +1 2[(1)(2) − (4)(1) +1 2 (4)(5) − (3)(2)] +1 2[(3)(3) − (−1)(5) = 1 2(−4 − 2 + 14 + 14) = 11. 23. P = 4x2_{− y}3_{, P} y = −3y2; Q = x3+ y2, Qx= 3x2. Z C (4x2− y3_{)dx + (x}3_{+ y}2_{)dy =} Z Z R (3x2+ 3y2)dA = Z 2π 0 Z 2 1 3r2(rdrdθ) = Z 2π 0 3 4r 4 2 1 dθ = Z 2π 0 45 4 dθ = 45π 2 24. P = cos2x − y, Py= −1; Q = p y2_{+ 1, Q} x= 0 I C

(cos2x − y)dx +py2_{+ 1dy =}

Z Z R (0 + 1)dA = Z Z R dA = (6√2)2− π(2)(4) = 72 − 8π

25. We first observe that Py+ (y4− 3x2y2)/(x2+ y2)3= Qx. Letting C0be the circle x2+ y2=

1 4 we have

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Z C −y3_{dx + xy}2_dy (x2_{+ y}2₎2 = Z C0 −y3_{dx + xy}2_dy (x2+ y2)2 x = 1 4cos t, dx = − 1 4sin tdt, y = 1 4sin t, dy = 1 4cos tdt = Z 2π 0 −1 64sin 3_t −1 4sin tdt +1 4cos t 1 16sin 2_t 1 4cos tdt 1/256 = Z 2π 0

(sin4t + sin2t cos2t)dt = Z 2π

0

(sin4t + (sin2t − sin4t)dt

= Z 2π 0 sin2tdt = 1 2t − 1 4sin 2t 2π 0 = π

26. We first observe that Py= [4y2− (x + 1)2]/[(x + 1)2+ 4y2]2= Qx. Letting C0 be the ellipse

(x + 1)2_{+ 4y}2_{= 4 we have} Z C −y (x + 1)2_{+ 4y}2dx+ x + 1 (x + 1)2_{+ 4y}2dy = Z C0 −y (x + 1)2_{+ 4y}2dx + x + 1 (x + 1)2_{+ 4y}2dy

x + 1 = 2 cos t, dx = −2 sin tdt, y = sin t, dy = cos tdt = Z 2π 0 − sin t 4 (−2 sin t) + 2 cos t 4 cos t dt =1 2 Z 2π 0 (sin2t + cos2t)dt = π. 27. Writing R R Rx 2_{dA =} R R

R(Qx− Py)dA we identify Q = 0 and P = −x

2_{y. Then, with}

C : x = 3 cos t, y = 2 sin t, 0 ≤ t ≤ 2π, we have Z Z R x2dA = Z C P dx + Qdy = Z C −x2_{ydx = −}Z 2π 0

9 cos2t(2 sin t)(−3 sin t)dt

= 54 4 Z 2π 0 4 sin2t cos2tdt = 27 2 Z 2π 0 sin22tdt = 27 4 Z 2π 0 (1 − cos 4t)dt = 27 4 t −1 4sin 4t 2π 0 =27π 2 . 28. WritingR R R[1 − 2(y − 1)]dA = R R

R(Qx− Py)dA we identify Q = x and P = (y − 1)

2_{. Then,}

with C1: x = cos t, y − 1 = sin t, −π/2 ≤ t ≤ π/2, and C2: x = 0, 2 ≥ y ≥ 0,

Z Z R [1 − 2(y − 1)]dA = Z C1 P dx + Qdy + Z C2 P dx + Qdy = Z C1 (y − 1)2dx + xdy + Z C2 0dy = Z π/2 −π/2

[sin2t(− sin t) + cos t cos t]dt = Z π/2

−π/2

[cos2t − (1 − cos2t) sin t]dt

= Z π/2

−π/2

1

2(1 + cos 2t) − sin t + cos

2_{t sin t} dt = 1 2t + 1 4sin 2t + cos t − 1 3cos 3_t π/2 −π/2 = π 4 − −π 4 = π 2.

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15.5. PARAMETRIC SURFACES AND AREA 245 29. P = x − y, Py = −1, Q = x + y, Qx= 1; W =R_CF · dr =R R_R2dA = 2 × area = 2(3π 4 ) = 3 2π 30. P = −xy2_{, P}

y = −2xy, Q = x2y, Qx= 2xy. Using polar coordinates,

W = Z C F · dr = Z Z R 4xydA = Z π/2 0 Z 2 1 4(r cos θ)(r sin θ)rdrdθ = Z π/2 0 (r4cos θ sin θ) 2 1 dθ = 15 Z π/2 0 sin θ cos θdθ = 15 2 sin 2_θ π/2 0 = 15 2 . 31. Let P = 0 and W = x2_{. Then Q}

x− Py= 2x and 1 2A I C x2dy = 1 2A Z Z R 2xdA = R R RxdA A = x.

Let P = y2_{and Q = 0. Then Q}

x− Py= −2y and − 1 2A I C y2dx = − 1 2A I C Z Z R −2ydA = R R RydA A = y.

32. Using Green’s Theorem, W = Z C F · dr = Z C −ydx + xdy = Z Z R 2dA = 2 Z 2π 0 Z 1+cos θ 0 rdrdθ = 2 Z 2π 0 1 2r 2 1+cos θ 0 dθ = Z 2π 0 (1 + 2 cos θ + cos2θ)dθ = θ + 2 sin θ +1 2θ + 1 4sin 2θ 2π 0 = 3π.

33. SinceR_ABP dx + Qdy is independent of path, Py = Qx by Theorem 17.3. Then, by Green’s

Theorem Z C P dx + Qdy = Z Z R (Qx− Py)dA = Z Z R 0dA = 0. 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 < ∞}

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6. r(u, v) = 2 cos ui + 3 sin uj + vk, 0 ≤ u ≤ 2π, −∞ < v < ∞ 7. x2+ y2= cos2u + sin2u = 1, circular cylinder

8. z = x2_{+ y}2_{, paraboloid}

9. x = sin u, y = sin u cos v, z = sin u sin v

y2_{+ z}2_{= sin}2_{u cos}2_{v + sin}2_{u sin}2_{v = sin}2_(cos2_{v + sin}2_{v) = sin}2_{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, x2 4 + y2 9 + z2 16 = sin

2_{φ cos}2_{θ + sin}2_{φ sin}2_{θ + cos}2_φ

= sin2(cos2θ + sin2θ) + cos2φ = sin2φ + cos2φ = 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, y = sin v, z = cos v so R is defined by −2 ≤ u ≤ 2, −π₂ ≤ v ≤ π

2

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, π

2 ≤ 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 ∂u π 6, 2 = 5 √ 3, ∂y ∂u π 6, 2 = −5, ∂z ∂u π 6, 2 = 0 ∂x ∂v π 6, 2 = 0, ∂y ∂v π 6, 2 = 0, ∂z ∂v π 6, 2 = 1.

A normal vector is given by n = i j k 5√3 −5 0 0 0 1 = −5i − 5√3j. 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. ∂x ∂u(1, 0) = 1, ∂y ∂u(1, 0) = 0, ∂z ∂u(1, 0) = 2 ∂x ∂v(1, 0) = 0, ∂y ∂v(1, 0) = 1, ∂z ∂v(1, 0) = 0. A normal vector is given by n =

i j k 1 0 2 0 1 0 = −2i + k. The tangent plane is −2(x − 1) + (z − 1) = 0 or −2x + z = −1.

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15.5. PARAMETRIC SURFACES AND AREA 247 17. At u = 1, v = 2, we have x = 3, y = 3, z = −3. ∂x ∂u(1, 2) = 2, ∂y ∂u(1, 2) = 1, ∂z ∂u(1, 2) = 2 ∂x ∂v(1, 2) = 1, ∂y ∂v(1, 2) = 1, ∂z ∂v(1, 2) = −4. A normal vector is given by n =

i j k 2 1 2 1 1 −4 = −6i + 10j + k.

The tangent plane is −6(x − 3) + 10(y − 3) + (z + 3) = 0 or −6x + 10y + z = 9. 18. At u = −1, v = π₃, we have x = −4, y = 3₂, z = 3 √ 3 2 . ∂x ∂u −1, π 3 = 4, ∂y ∂u −1, π 3 = −3, ∂z ∂u −1, π 3 = −3 √ 3 ∂x ∂v −1, π 3 = 0, ∂y ∂v −1, π 3 = −3√3 2 , ∂z ∂v −1, π 3 = 3 2. A normal vector is given by n =

i j k 4 −3 −3√3 0 −3 √ 3 2 3 2 = −18i − 6j − 6√3k.

The tangent plane is −18(x + 4) − 6(y −3₂) − 6√3z −3

√ 3 2 = 0 or 3x + y +√3z = −6. 19. At u = 3, v = 3, we have x = 3, y = 3, z = 9. ∂x ∂u(3, 3) = 1, ∂y ∂u(3, 3) = 0, ∂z ∂u(3, 3) = 3 ∂x ∂v(3, 3) = 0, ∂y ∂v(3, 3) = 1, ∂z ∂v(3, 3) = 3. A normal vector is given by n =

i j k 1 0 3 0 1 3 = −3i − 3j + k.

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 = √ 2 2 , y = √ 2 2 , z = 1. ∂x ∂u 1, π 4 = √ 2 2 , ∂y ∂u 1, π 4 = √ 2 2 , ∂z ∂u 1, π 4 = 1 ∂x ∂v 1, π 4 = √ 2 2 , ∂y ∂v 1, π 4 = − √ 2 2 , ∂z ∂v 1, π 4 = 0.

A normal vector is given by n = i j k √ 2 2 √ 2 2 1 √ 2 2 − √ 2 2 0 = √ 2 2 i + √ 2 2 j − k.

The tangent plane is √ 2 2 x − √ 2 2 ! + √ 2 2 y − √ 2 2 ! − (z − 1) = 0 or x + y −√2z = 0. 21. At u = −2, v = 1, we have x = −1, y = 3, z = −2. ∂x ∂u(−2, 1) = 1, ∂y ∂u(−2, 1) = −1, ∂z ∂u(−2, 1) = 1 ∂x ∂v(−2, 1) = 1, ∂y ∂v(−2, 1) = 1, ∂z ∂v(−2, 1) = −2.

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A normal vector is given by n = i j k 1 −1 1 1 1 −2 = i + 3j + 2k.

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 ∂u(0, ln 3) = ln 3, ∂y ∂u(0, ln 3) = 1, ∂z ∂u(0, ln 3) = 1 ∂x ∂v(0, ln 3) = 0, ∂y ∂v(0, ln 3) = 1, ∂z ∂v(0, ln 3) = 3. A normal vector is given by n =

i j k ln 3 1 1 0 1 3 = 2i − 3 ln 3j + ln 3k.

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 ∂u(2, 1) = 1, ∂y ∂u(2, 1) = 2, ∂z ∂u(2, 1) = 4 ∂x ∂v(2, 1) = −1, ∂y ∂v(2, 1) = 3, ∂z ∂v(2, 1) = 2. A normal vector is given by n =

i j k 1 2 4 −1 3 2 = −8i − 6 ln 3j + 5k.

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. ∂x ∂u(4, 1) = 0, ∂y ∂u(4, 1) = 1, ∂z ∂u(4, 1) = 8 ∂x ∂v(4, 1) = 2, ∂y ∂v(4, 1) = −1, ∂z ∂v(4, 1) = 0. A normal vector is given by n =

i j k 0 1 8 2 −1 0 = 8i − 16j − 2k.

The tangent plane is 8(x − 1) − 16(y − 3) − 2(z − 16) = 0 or 4x − 8y − z = −36. 25. ∂r ∂u = h2, 1, i, ∂r ∂v = h−1, 1, 0i ∂r ∂u × ∂r ∂v = i j k 2 1 1 −1 1 0 = h−1, −1, 3i ∂r ∂u × ∂r ∂v =√1 + 1 + 9 =√11 A =R2 0 R1 −1 √ 11 dv du = 4√11 26. Let x = u, y = v, z = 1 − u − v. Then A =R1 −1 R √ 1−u2 −√1−u2p1 + (−1)2+ (−1)2dv du = √ 3π 27. ∂r ∂u = h1, 0, 2ui, ∂r ∂v = h0, 1, 2vi

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15.5. PARAMETRIC SURFACES AND AREA 249 ∂r ∂u × ∂r ∂v = i j k 1 0 2u 0 1 2v = h−2u, −2v, 1i ∂r ∂u × ∂r ∂v =√4u2_{+ 4v}2_{+ 1} Since 0 ≤ z ≤ 4, we have 0 ≤ u2_{+ v}2_{≤ 4. So} A = Z 2 −2 Z √ 4−u2 −√4−u2 p 4u2_{+ 4v}2_{+ 1 dv du} = Z 2 −2 v 2 p 4u2_{+ 4v}2_{+ 1 +} (4u 2_{+ 1)} 4 ln |2v + p 4u2_{+ 4v}2_{+ 1|} √ 4−u2 −√4−u2 = 1 4 h

2(4u2+ 1) ln |2p4 − u2₊√_{17| − (4u}2_{+ 1) ln(4u}2_{+ 1) + 4p−17(u}2_{− 4)}i

= Z 2π 0 Z 2 0 p 4r2_{+ 1r drdθ} _{polar transformation} = Z 2π 0 (4r2_{+ 1)}3/2 12 2 0 dθ = Z 2π 0 17√17 − 1 12 dθ = 17 √ 17 − 1 12 θ 2π 0 = (17 √ 17 − 1)π 6 28. ∂r

∂r = hcos θ, sin θ, 1i, ∂r ∂θ = h−r sin θ, r cos θ, 0i ∂r ∂r× ∂r ∂θ = i j k cos θ sin θ 1 r sin θ r cos θ 0 = h−r cos θ, −r sin θ, ri ∂r ∂r× ∂r ∂θ =pr2_cos2_{θ + r}2_sin2_{θ + r}2₌√_r2_{+ r}2_{= r}√_2. A =R₀2R₀2πr√2 dθ dr = 4π√2

29. r = (r cos θ)i + (r sin θ)j + θk ∂r

∂r = hcos θ, sin θ, 0i, ∂r ∂θ = h−r sin θ, r cos θ, f i ∂r ∂r× ∂r ∂θ = i j k cos θ sin θ 0 −r sin θ r cos θ 1 = hsin θ, − cos θ, ri ∂r ∂r× ∂r ∂θ =psin2θ + cos2_{θ + r}2₌√_{1 + r}2_. A =R2π 0 R2 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

∂θ = h−a sin φ sin θ, a sin φ cos θ, 0i, ∂r

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∂r ∂θ× ∂r ∂φ = i j k

− sin φ sin θ a sin φ cos θ 0 a cos φ cos θ a cos φ sin θ −a sin φ

= h−a2sin2φ cos θ, −a sin2φ sin θ, −a2sin φ cos φi ∂r ∂θ× ∂r ∂φ = q

a4_sin4_{φ cos}2_{θ + a}4_sin4_{φ sin}2_{θ + a}4_sin2_{φ cos}2_θ

= q

a4_sin4_{φ + a}4_sin2_{φ cos}2_{φ =}

q a4_sin2_{φ = a}2_{sin φ} A =Rπ 0 R2π 0 a 2_{sin φdθdφ = 4a}2_π

31. We have a = 2, so x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ, π₃ ≤ φ ≤ π, 0 ≤ θ ≤ 2π, A =R_π/3π R₀2π4 sin φ dθ dφ = 12π

32. x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ, π₃ ≤ φ ≤ π

2, 0 ≤ θ ≤ 2π,

A =Rπ

π/3

R2π

0 4 sin φ dθ dφ = 4π

33. x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ, 0 ≤ φ ≤ π

4, 0 ≤ θ ≤ 2π, A =Rπ/4 0 R2π 0 4 sin φ dθ dφ = 4π(2 − √ 2)

34. x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ; the spehre intersects the cylinder when z2_{= 2,}

so the region outside the cylinder is described by π₄ ≤ φ3π

4, 0 ≤ θ ≤ 2π A =R_π/43π/4R₀2π4 sin φdθdφ = 8π√2 35. (a) 1 1 1 y x z (b) y 1 x 1 1 z (c) y 1 x 1 1 z 36. (a) x y z (b) x y z (c) x y z 37. (f )

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15.5. PARAMETRIC SURFACES AND AREA 251 38. (e) 39. (d) 40. (a) 41. (c) 42. (b) 43. x y z 44. ∂r

∂φ = h− cos φ cos θ, − cos φ sin θ, − sin φi, ∂r

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

− cos φ cos θ − cos φ sin θ − sin φ (sin φ − R) sin θ (R − sin φ) cos θ 0

= h(r − sin φ) sin φ cos θ, (R − sin φ) sin φ sin θ, −(R − sin φ) cos φi ∂r ∂φ× ∂r ∂θ = q

(R − sin φ)2_sin2_{φ cos}2_{θ + (R − sin φ)}2_sin2_{φ sin}2_{θ + (R − sin φ)}2_cos2_φ

= q

(R − sin φ)2_sin2_{φ + (R − sin φ)}2_cos2_φ

=p(R − sin φ)2_{= R − sin φ} A =R2π 0 R2π 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 ∂u = h1, f

0_{(u) cos v, f}0_{(u) sin vi,} ∂r

∂v = h0, −f (u) sin v, f (u) cos vi, ∂r ∂u× ∂r ∂v = i j k

1 f0(u) cos v f0(u) sin v 0 −f (u) sin v f (u) cos v

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∂r ∂u× ∂r ∂v = q

[f (u)f0_(u)]2_{+ [f (u)]}2_cos2_{v + [f (u)]}2_sin2_v

=p[f (u)]2_[f0_(u)]2_{+ [f (u)]}2

= f (u)p1 + [f0_(u)]2 By (11), A =Rb a R2π 0 f (u)p1 + [f 0_(u)]2_{dvdu = 2π}Rb af (u)p1 + [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)

x

y

z

(c) Let S1 be the surface corresponding to the parameter domain 0 ≤ u ≤ pi, 0 ≤ v ≤ 2π.

Then A(S1) = 2π Z π 0 sin(x)p1 + cos2_xdx =ln(√2 + 1) − ln(√2 − 1) = 2√2π

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

x y z R z=2-x2 2 2 4 1. zx= −2x, zy= 0; dS = √ 1 + 4x2_dA Z Z S xdS = Z 4 0 Z √ 2 0 xp1 + 4x2_{dxdy =} Z 4 0 1 12(1 + 4x 2₎3/2 √ 2 0 dy = Z 4 0 13 6 dy = 26 3

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15.6. SURFACE INTEGRALS 253 2. See Problem 1. Z Z S xy(9 − 4z)dS = Z Z S xy(1 + 4x2)dS = Z 4 0 Z √ 2 0 xy(1 + 4x2)3/2dxdy = Z 4 0 y 20(1 + 4x 2₎5/2 √ 2 0 dy = Z 4 0 242 20ydy = 121 10 Z 4 0 ydy = 121 10 1 2y 2 4 0 =484 5 x y z R z=Mx2_+y2 1 1 1 3. zx= x p x2_{+ y}2, zy = y p x2_{+ y}2; dS = √ 2dA. Using polar coordinates,

Z Z S xz3dS = Z Z R x(x2+ y2)3/2√2dA =√2 Z 2π 0 Z 1 0 (r cos θ)r3/2rdrdθ =√2 Z 2π 0 Z 1 0 r7/2cos θdrdθ =√2 Z 2π 0 2 9r 9/2_{cos θ} 1 0 dθ =√2 Z 2π 0 2 9cos θdθ = 2√2 9 sin θ 2π 0 = 0. x y z R z=Mx2_+y2 4 4 4 4. zx= x p x2_{+ y}2, zy = y p x2_{+ y}2; dS = √ 2dA. Using polar coordinates,

Z Z S (x + y + z)dS = Z Z R(x + y +px2_{+ y}2₎√_2dA =√2 Z 2π 0 Z 4 1 (r cos θ + r sin θ + r)rdrdθ =√2 Z 2π 0 Z 4 1 r2(1 + cos θ + sin θ)drdθ =√2 Z 2π 0 1 3r 3_{(1 + cos θ + sin θ)} 4 1 dθ =63 √ 2 3 Z 2π 0

(1 + cos θ + sin θ)dθ = 21√2(θ + sin θ − cos θ)

2π

0

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x y z z=M36-x2_-y2 R 6 6 6 5. z =p36 − x2_{− y}2_{, z} x= − x p 36 − x2_{− y}2, zy= − y 36 − x2_{− y}2; dS = s 1 + x 2 36 − x2_{− y}2 + y2 36 − x2_{− y}2dA = _p 6 36 − x2_{− y}2dA.

Using polar coordinates,

Z Z S (x2+ y2)zdS = Z Z R (x2+ y2)p36 − x2_{− y}2 6 p 36 − x2_{= y}2dA = 6 Z 2π 0 Z 6 0 r2rdrdθ = 6 Z 2π 0 1 4r 4 6 0 dθ = 6 Z 2π 0 324dθ = 972π. x y z R 1 2 1 z=x+1 6. zx= 1, zy = 0; dS = √ 2dA Z Z S z2dS = Z 1 −1 Z 1−x2 0 (x + 1)2√2dydx = √2 Z 1 −1 y(x + 1)2 1−x2 0 dx =√2 Z 1 −1 (1 − x2)(x + 1)2dx =√2 Z 1 −1 (1 + 2x − 2x3− x4)dx = √2 x + x2−1 2x 4₋1 5x 5 1 −1 =8 √ 2 5 x y z 1 1 2 z=2-x2_/2-y2_/2 7. zx= −x, zy= −y; dS = p 1 + x2_{+ y}2_dA Z Z S xydS = Z 1 0 Z 1 0 xyp1 + x2_{+ y}2_dxdy = Z 1 0 1 3y(1 + x 2_{+ y}2₎3/2 1 0 dy = Z 1 0 1 3y(2 + y 2 )3/2−1 3y(1 + y 2 )3/2 dy = 1 15(2 + y 2₎5/2 − 1 15(1 + y 2₎5/2 1 0 = 1 15(3 5/2_{− 2}7/2_{+ 1)}

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15.6. SURFACE INTEGRALS 255 x y z 1 1 1 R z=1/2+x2_/2+y2_/2 8. z = 1 2 + 1 2x 2 ₊ 1 2y 2_, _z x = x, zy = y; dS = p 1 + x2_{+ y}2_dA.

Using polar coordinates, Z Z S 2zdS = Z Z R (1 + x2+ y2)p1 + x2_{+ y}2_dA = Z π/2 π/3 Z 1 0 (1 + r2)p1 + r2_rdrdθ = Z π/2 π/3 Z 1 0 (1 + r2)3/2rdrdθ = Z π/2 π/3 1 5(1 + r 2₎5/2 1 0 dθ = 1 5 Z π/2 π/3 (25/2− 1)dθ =4 √ 2 − 1 5 π 2 − π 3 = (4 √ 2 − 1)π 30 . x y z R 4 2 3 _y=x2 9. yx= 2x, yz= 0; dS = √ 1 + 4x2_dA Z Z S 24√yzdS = Z 3 0 Z 2 0 24xzp1 + 4x2_dxdz = Z 3 0 2z(1 + 4x2)3/2 2 0 dz = 2(173/2− 1) Z 3 0 zdz = 2(173/2− 1) 1 2z 2 3 0 = 9(173/2− 1) x y z R 3 2 2 z=4-y2_-z2 10. xy= −2y, xz= −2z; dS = p 1 + 4y2_{+ 4z}2_dA

Using polar coordinates, Z Z S (1 + 4y2+ 4z2)1/2dS = Z π/2 0 Z 2 1 (1 + 4r2)rdrdθ = Z π/2 0 1 16(1 + 4r 2₎2 2 1 dθ = 1 16 Z π/2 0 12dθ = 3π 8 .

11. Write the equation of the surface as y = 1

2(6 − x − 3z). yz= − 1 2, yz= − 3 2; dS =p1 + 1/4 + 9/4 = √ 14 2 .

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Z Z S (3z2+ 4yz)dS = Z 2 0 Z 6−3z 0 3z2+ 4z1 2(6 − x − 3z) √₁₄ 2 dxdz = √ 14 2 Z 2 0 3z2_{x − z(6 − x − 3z)}2 6−3z 0 dz = √ 14 2 Z 2 0 [3z2(6 − 3z) − 0] − [0 − z(6 − 3z)2]dz = √ 14 2 Z 2 0 (36z − 18z2)dz = √ 14 2 (18z 2_{− 6z}3₎ 2 0 = √ 14 2 (72 − 48) = 12 √ 14

12. Write the equation of the surface as x = 6 − 2y − 3z. Then x_√ y = −2, xz = −3; dS =

1 + 4 + 9 =√14. Z Z S (3z2+ 4yz)dS = Z 2 0 Z 3−3z/2 0 (3z2+ 4yz)√14dydz = √14 Z 2 0 (3yz + 2y2z) 3−3z/2 0 dz =√14 Z 2 0 9z1 − z 2 + 18z1 − z 2 2 dz =√14 Z 2 0 27z −45 2 z 2₊9 2z 3 dz = √14 27 2 z 2₋15 2 z 3₊9 8z 4 2 0 =√14(54 − 60 + 18) = 2√14 x y z 1 1 1 z=1-x-y 13. The density is ρ = kx2_{. The surface is z = 1 − x − y. Then}

zx= −1, zy= −1; dS = √ 3dA. m = Z Z S kx2dS = k Z 1 0 Z 1−x 0 x2√3dydx = √3k Z 1 0 1 3x 3 1−x 0 dx = √ 3 3 k Z 1 0 (1 − x)3dx = √ 3 3 k −1 4(1 − x) 4 1 0 = √ 3 12k x y z R z=

M

4-x2_-y2 2 2 2

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15.6. SURFACE INTEGRALS 257 14. zx = − x p 4 − x2_{− y}2, zy = − y p 4 − x2_{− y}2; dS = s 1 + x 2 4 − x2_{− y}2 + y2 4 − x2_{− y}2dA = 2 p 4 − x2_{− y}2dA.

Using symmetry and polar coordinates, m = 4 Z Z S |xy|dS = 4 Z π/2 0 Z 2 0 (r2cos θ sin θ)√ 2 4 − r2rdrdθ = 4 Z π/2 0 Z 2 0 r2(4 − r2)−1/2sin 2θ(rdr)dθ u = 4 − r2, du = −2rdr, r2= 4 − u = 4 Z π/2 0 Z 0 4 (4 − u)u−1/2sin 2θ −1 2du dθ = −2 Z π/2 0 Z 0 4

(4u−1/2− u1/2_{) sin 2θdudθ}

= −2 Z π/2 0 8u1/2−2 3u 3/2 0 4 sin 2θdθ = −2 Z π/2 0 −32 3 sin 2θ dθ = 64 3 −1 2cos 2θ π/2 0 =64 3 . x y z R 2 2 3 z=M4-y2 15. The surface is g(x, y, z) = y2_{+ z}2_{− 4 = 0. ∇g = 2yj + 2zk,}

|∇g| = 2p y2_{+ z}2_{; ∇} yi + zk p y2_{+ z}2; F · ∇ = _p2yz y2_{+ z}2 + yz p y2_{+ z}2 = 3yz p y2_{+ z}2; ; z = p 4 − y2_{, z} x= 0, zy = − y p 4 − y2; dS = s 1 + y 2 4 − y2dA = 2 p 4 − y2dA Flux = Z Z S F · ndS = Z Z R 3yz p y2_{+ z}2 2 p 4 − y2dA = Z Z R 3yp4 − y2 p y2_{+ 4 − y}2 2 p 4 − y2dA = Z 3 0 Z 2 0 3ydydx = Z 3 0 3 2y 2 2 0 dx = Z 3 0 6dx = 18

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x

y

z

R

2

5 z=5-x

2

_-y

2 16. The surface is g(x, y, z) = x2 _{+ y}2_{+ z − 5 = 0.} _{∇g =} 2xi + 2yj + k, |∇g| = p1 + 4x2_{+ 4y}2_; _n ₌ 2xi + 2yj + k p 1 + 4x2_{+ 4y}2; F · n = z p 1 + 4x2_{+ 4y}2; zx = −2x, zy = −2y, dS = p

1 + 4x2_{+ 4y}2_{dA. Using polar}

coordinates, Flux = Z Z S F · ndS = Z Z R z p 1 + 4x2_{+ 4y}2 p 1 + 4x2_{+ 4y}2_dA = Z Z R (5 − x2− y2_)dA = Z 2π 0 Z 2 0 (5 − r2)rdrdθ = Z 2π 0 5 2r 2₋1 4r 4 2 0 dθ = Z 2π 0 6dθ = 12π.

17. From Problem 16, n = _p2xi + 2yj + k

1 + 4x2_{+ 4y}2. Then F · n =

2x2_{+ 2y}2_{+ z}

p

1 + 4x2_{+ 4y}2. Also, from Problem

16, dS =p1 + 4x2_{+ 4y}2_{dA. Using polar coordinates,}

Flux = Z Z S F · ndS = Z Z R 2x2+ 2y2+ z p 1 + 4x2_{+ 4y}2 p 1 + 4x2_{+ 4y}2_{dA =} Z Z R (2x2+ 2y2+ 5 − x2− y2_)dA = Z 2π 0 Z 2 0 (r2+ 5)rdrdθ = Z 2π 0 1 4r 4₊5 2r 2 2 0 dθ = Z 2π 0 14dθ = 28π. x y z R 5 1 2 r=2cosθ z=x+3 18. The surface is g(x, y, z) = z − x − 3 = 0. ∇g = −i + k, |∇g| =√2; n = −i + k√ 2 ; F · n = √1 2x 3_{y +} _√1 2xy 3 _z x = 1, zy = 0, dS = √ 2dA. Using polar coordinates,

Flux = Z Z S F · ndS = Z Z R 1 √ 2(x 3_{y + xy}3₎√_{2dA =} Z Z R xy(x2+ y2)dA = Z π/2 0 Z 2 cos θ 0 (r2cos θ sin θ)r2rdrdθ = Z π/2 0 Z 2 cos θ 0 r5cos θ sin θdrdθ = Z π/2 0 1 6r 6_{cos θ sin θ} 2 cos θ 0 dθ = 1 6 Z π/2 0 64 cos7θ sin θdθ = 32 3 −1 8cos 8_θ π/2 0 = 4 3.

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15.6. SURFACE INTEGRALS 259 x y z R 4 2 2 z=4-x2_-y2 19. The surface is g(x, y, z) = x2_{+ y}2_{+ z − 4. ∇g = 2xi + 2yj +}

k, |∇g| =p4x2_{+ 4y}2_{+ 1;} n = _p2xi + 2yj + k 4x2_{+ 4y}2_{+ 1}; F · n = x3_{+ y}3_{+ z} p 4x2_{+ 4y}2_{+ 1}; zx = −2x, zy= −2y,

dS =p1 + 4x2_{+ 4y}2_{dA. Using polar coordinates,}

Flux = Z Z S F · ndS = Z Z R (x3+ y3+ z)dA = Z Z R (4 − x2− y2_{+ x}3_{+ y}3_)dA = Z 2π 0 Z 2 0 (4 − r2+ r3cos3θ + r3sin3θ)rdrdθ = Z 2π 0 2r2−1 4r 4₊1 5r 5_cos3_{θ +}1 5r 5_sin3_θ 2 0 dθ = Z 2π 0 4 +32 5 cos 3_{θ +} 32 5 sin 3_θ dθ = 4θ|2π₀ + 0 + 0 = 8π. x y z R 6 6 6 z=6-x-y 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 = (ey _{+ e}x _{+ 18y)/}√_3; _z x = −1, zy = −1, dS = √ 1 + 1 + 1dA =√3dA. Flux = Z Z S F · ndS = Z Z r (ey+ ex+ 18y)dA = Z 6 0 Z 6−x 0 (ey+ ex+ 18y)dydx = Z 6 0 (ey+ yex+ 9y2) 6−x 0 dx = Z 6 0 [e6−x+ (6 − x)ex+ 9(6 − x)2− 1]dx = [−e6−x+ 6ex− xex_{+ e}x_{− 3(6 − x)}3_{− x]} 6 0 = (−1 + 6e6− 6e6_{+ e}6_{− 6) − (−e}6_{+ 6 + 1 − 648) = 2e}6_{+ 634 ≈ 1440.86}

21. For S1 : g(x, y, z) = x2+ y2− z, ∇g = 2xi + 2yj − k, |∇g| =

p 4x2_{+ 4y}2_{+ 1;} _n 1 = 2xi + 2yj − k p 4x2_{+ 4y}2_{+ 1}; F · n1= 2xy2_{+ 2x}2_{y − 5z} p 4x2_{+ 4y}2_{+ 1} ; zx= 2x, zy= 2y, dS1= p 1 + 4x2_{+ 4y}2_dA. For S2: g(x, y, z) = z − 1, ∇g = k; |∇g| = 1; n2= k; F · n2= 5z; zx= 0, zy= 0, dS2=

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Flux = Z Z S1 F · n1dS1+ Z Z S2 F · n2dS2= Z Z R (2xy2+ 2x2y − 5z)dA + Z Z R 5zdA = Z Z R [2xy2+ 2x2y − 5(x2+ y2) + 5(1)]dA = Z 2π 0 Z 1 0

(2r3cos θ sin2θ + 2r3cos2θ sin θ − 5r2+ 5)rdrdθ

= Z 2π 0 2 5r 5_{cos θ sin}2_{θ +}2 5r 5_cos2_{θ sin θ −}5 4r 4₊5 2r 2 1 0 dθ = Z 2π 0 2 5(cos θ sin 2 θ + cos2θ sin θ) +5 4 dθ = 2 5 1 3sin 3 θ −1 3cos 3_θ 2π 0 + 5 4θ 2π 0 = 2 5 −1 3 − −1 3 +5 2π = 5 2π.

22. For S1: g(x, y, z) = x2+ y2+ z − 4, ∇g = 2xi + 2yj + k, |∇g| =

p 4x2_{+ 4y}2_{+ 1; n} 1= 2xi + 2yj + k p 4x2_{+ 4y}2_{+ 1}; F·n1= 6z 2_/p 4x2_{+ 4y}2_{+ 1; z} x= −2x, zy= −2y, dS1= p 1 + 4x2_{+ 4y}2_dA.

For S2 : g(x, y, z) = x2+ y2− z, ∇g = 2xi + 2yj − k, |∇g| =

p 4x2_{+ 4y}2_{+ 1;} _n 2 = 2xi + 2yj − k p 4x2_{+ y}2_{+ 1}; F·n2= −6z 2_/p 4x2_{+ 4y}2_{+ 1; z} x= 2x, zy = 2y, dS2= p 1 + 4x2_{+ 4y}2_dA.

Using polar coordinates and R : x2+ y2≤ 2 we have Flux = Z Z S1 F · n1dS1+ Z Z S1 F · n2dS2= Z Z R 6z2dA + Z Z −6z2dA = Z Z R [6(4 − x2− y2₎2_{− 6(x}2_{+ y}2₎2_{]dA = 6} Z 2π 0 Z √ 2 0 [(4 − r2)2− r4_]rdrdθ = 6 Z 2π 0 −1 6(4 − r 2₎3₋1 6r 6 √ 2 0 dθ = − Z 2π 0 [(23− 43_{) + (}√₂₎6_{]dθ =} Z 2π 0 48dθ = 96π. 23. The surface is g(x, y, z) = x2 _{+ y}2 _{+ z}2 _{− a}2 _{= 0.} _{∇g = 2xi + 2yj + 2zk,} _{|∇g| =}

2px2_{+ y}2_{+ z}2_; _{n =} xi + yj + zk p x2_{+ y}2_{+ z}2; F · n = −(2xi + 2yj + 2zk) · xi + yj + zk p x2_{+ y}2_{+ z}2 = −2x 2_{+ 2y}2_{+ 2z}2 p x2_{+ y}2_{+ z}2 = −2 p x2_{+ y}2_{+ z}2_{= −2a.} Flux =R R

S−2adS = −2a × area = −2a(4πa

2_{) = −8πa}3 24. n1 = k, n2 = −i, n3 = j, n4 = −k, n5= i, n6 = −j; F · n1 = z = 1, F · n2 = −x = 0, F · n3= y = 1, F · n4= −z = 0, F · n5= x = 1, F · n6= −y = 0; Flux =R R S11dS + R R S31dS + R R S51dS = 3

25. Refering to the solution to Problem 23, we find n = _pxi + yj + zk

x2_{+ y}2_{+ z}2and dS = a p a2_{− x}2_{− y}2dA. Now F · n = kq r |r|3· r |r| = kq |r|4|r| 2₌ kq |r|2 = kq x2_{+ y}2_{+ z}2 = kq a2

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15.6. SURFACE INTEGRALS 261 and Flux = Z Z S F · ndS = Z Z S kq a2dS = kq a2 × area = kq a2(4πa 2_{) = 4πkq.} x y z R 3 4 3 z=M16-x2_-y2 26. We are given σ = kz. Now zx−

x p 16 − x2_{− y}2, zy = − y p 16 − x2_{− y}2; dS = s 1 + x 2 16 − x2_{− y}2 + y2 16 − x2_{− y}2dA = _p 4 16 − x2_{− y}2dA

Using polar coordinates, Q = Z Z S kzdS = k Z Z R p 16 − x2_{− y}2 4 p 16 − x2_{− y}2dA = 4k Z 2π 0 Z 3 0 rdrdθ = 4k Z 2π 0 1 2r 2 3 0 dθ = 4k Z 2π 0 9 2dθ = 36πk. x y z R 2 3 6 z=6-2x-3y

27. The surface is z = 6 − 2x − 3y. Then zx = −2, zy =

−3, dS =√1 + 4 + 9 =√14dA. The area of the surface is A(s) = Z Z S dS = Z 3 0 Z 2−2x/3 0 √ 14dydx =√14 Z 3 0 2 − 2 3x dx = √14(2x −1 3x 2₎ 3 0 = 3√14. x = 1 3√14 Z Z s xdS = 1 3√14 Z 3 0 Z 2−2x/3 0 √ 14xdydx = 1 3 Z 3 0 xy 2−2x/3 0 dx = 1 3 Z 3 0 2x −2 3x 2 dx = 1 3 x2−2 9x 3 3 0 = 1 y = 1 3√14 Z Z S ydS = 1 3√14 Z 3 0 Z 2−2x/3 0 √ 14ydydx = 1 3 Z 3 0 1 2y 2 2−2x/3 0 dx = 1 6 Z 3 0 2 − 2 3x 2 dx = 1 6 −1 2(2 − 2 3x) 3 3 0 =2 3 z = 1 3√14 Z Z S zdS = 1 3√14 Z 3 0 Z 2−2x/3 0 (6 − 2x − 3y)√14dydx = 1 3 Z 3 0 6y − 2xy − 3 2y 2 2−2x/3 0 dx = 1 3 Z 3 0 6 − 4x +2 3x 2 dx = 1 3 6x − 2x2+2 9x 3 3 0 = 2 The centroid is (1, 2/3, 2).

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28. The area of the hemisphere is A(s) = 2πa2_{. By symmetry, x = y = 0.} zx= − x p a2_{− x}2_{− y}2, zy= − y p a2_{− x}2_{− y}2; dS = s 1 + x 2 a2_{− x}2_{− z}2 + y2 a2_{− x}2_{− y}2dA = a p a2_{− x}2_{− y}2dA

Using polar coordinates, z = Z Z S zdS 2πa2 = 1 2πa2 Z Z R p a2_{− x}2_{− y}2 a p a2_{− x}2_{− y}2dA = 1 2πa Z 2π 0 Z a 0 rdrdθ = 1 2πa Z 2π 0 1 2r 2 a 0 dθ = 1 2πa Z 2π 0 1 2s 2_{dθ =} a 2. The centroid is (0, 0, a/2).

29. (a) The region in the xy-plane is x2+ y2≤ 16. From zx= −x/

p x2_{+ y}2 _and zy = −y/ p x2_{+ y}2_{we see that} dS =p1 + x2_/(x2_{+ y}2_{) + y}2_/(x2_{+ y}2_{)dA =}√_2da and A(S) = Z Z S dS = Z Z R √ 2dA =√2π42= 16√2π. Then x = 1 16√2π Z Z S dS = 1 16√2π Z Z R √ 2xdA = 1 16π Z 2π 0 Z 4 0 r cos θrdrdθ = 1 16π Z 2π 0 1 3r 3_{cos θ} 4 0 dθ = 4 3π Z 2π 0 cos θdθ = 0 y = 1 16√2π Z Z S ydS = 1 16√2π Z Z R √ 2ydA = 1 16π Z 2π 0 Z 4 0 r sin θrdrdθ = 1 16π Z 2π 0 1 3r 3_{cos θ} 4 0 dθ = 4 3π Z 2π 0 sin θdθ = 0 z = 1 16√2π Z Z S zdS = 1 16√2π Z Z R √ 2(4 −px2_{+ y}2_{)dA =} 1 16π Z 2π 0 Z 4 0 (4 − r)rdrdθ = 1 16π Z 2π 0 2r2−1 3r 3 4 0 dθ = 2 3π Z 2π 0 dθ = 4 3. The centroid is (0,0,4/3). (b) Iz= Z Z S (x2+ y2)kdS = k√2 Z Z R (x2+ y2)dA = k√2 Z 2π 0 Z 4 0 r2rdrdθ = k √ 2 4 Z 2π 0 r4 4 0 dθ = 64k√2 Z 2π 0 dθ = 128kπ√2 30. The surface is g(x, y, z) = z − f (x, y) = 0. ∇g = −fxi − fyj + k, |∇g| = q f2 x+ fy2+ 1; n =