thus the smallest distance is 2 5 5 5 and the largest

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Unformatted text preview: lt; π , so π 12. ∂ (x, y ) du dv = ∂ (u, v ) x=2y dy = x=−y √ sin xy dy dx < 0 2 3 1 1/2 0 √ = −1/( 2π ) y 2 ey dy = 0 π 1 dy dx = π 2 0 1 sin(πx2 ) π 3 2 π 1 y3 e 2 2 = 0 18 e −1 2 π 2 ex ey dx dy 13. 2y 0 0 y 15. 1 x sin x dy dx x 14. 0 16. p /2 r = a (1 + cos u) y = sin x r=a y = tan (x /2) x p /6 6 y 1/3 8 x2 sin y 2 dx dy = 17. 2 0 0 π /2 2 3 2 (4 − r2 )r dr dθ = 2π 18. 0 0 8 0 1 y sin y 2 dy = − cos y 2 3 8 = 0 0 1 (1 − cos 64) ≈ 0.20271 3 605 Chapter 16 2xy , and r = 2a sin θ is the circle x2 + (y − a)2 = a2 , so x2 + y 2 19. sin 2θ = 2 sin θ cos θ = √ a+ a2 −x2 a √ a− a2 −x2 0 π /2 a 2xy dy dx = 2 + y2 x a2 − x2 − ln a − x ln a + 4r2 (cos θ sin θ) r dr dθ = −4 cos 2θ π/4 2 =4 0 2π π/4 2π 16 r2 cos2 θ r dz dr dθ = 21. 0 r4 0 π /2 2 π /2 1 0 0 0 π 1 ρ2 sin φ dρ dφ dθ = 1 − 1 + ρ2 4 = 1− 2π r3 (16 − r4 ) dr = 32π cos2 θ dθ 0 22. 0 π /3 π /2 π 2 π 4 sin φ dφ 0 2π π /3 (ρ2 sin2 φ)ρ2 sin φ dρ dφ dθ = 0 0 2π 0 √ 3a/2 √ a2 −r 2 (b) (c) 0 √ 3a/2 √ − √ 0 π 4 π 2 a 0 √ a2 −r 2 √ r/ 3 0 r3 dz dr dθ (x2 + y 2 ) dz dy dx √ x2 +y 2 / 3 4x dz dy dx √ − 4x−x2 π /2 4 cos θ x2 +y 2 4r cos θ (b) r dz dr dθ −π/2 r2 0 2−y/2 2 2 (y/2)1/3 y y − 2 2 2− dx dy = 25. 0 a −x −y 0 = 1− ρ4 sin3 φ dρ dφ dθ 0 0 √ 3a/2 2π r2 dz rdr dθ = √2 2 2 2 (3a2 /4)−x2 √ 4x−x2 4 0 (3a2 /4)−x √ √ − 3a/2 24. (a) √ r/ 3 π /2 π (− cos φ) 2 a 23. (a) 0 dx = a2 π /2 2 20. a2 − x2 0 0 π /6 cos 3θ 0 √ a/ 3 y 2 4/3 2 = 0 3 2 0 27. V = 0 3 y2 − 4 2 cos2 3θ = π/4 0 2π 2y − dy = π /6 r dr dθ = 3 26. A = 6 1/3 √ 3r 0 √ a/ 3 a r(a − r dz dr dθ = 2π 0 √ πa3 3r) dr = 9 28. The intersection of the two surfaces projects onto the yz -plane as 2y 2 + z 2 = 1, so √ √ 2 2 1−2y 1/ 2 1−y dx dz dy V =4 0 0 √ 1/ 2 =4 ru × rv 1−2y 2 √ 1/ 2 (1 − 2y − z ) dz dy = 4 2 2 0 0 29. √ y 2 +z 2 0 8 (1 − 2x2 )3/2 dx = 3 √ 2π 4 √ = 2u2 + 2v 2 + 4, 2π S= u2 +v 2 ≤4 2 2u2 + 2v 2 + 4 dA = 0 0 √ 2 r2 + 2 r dr dθ = 8π √ (3 3 − 1) 3 Chapter 16 Supplementary Exercises 30. 606 √ 1 + u2 , S = ru × rv = u=1 v =2 0 u=−3 v =0 2+y 2 /8 −4 y 2 /4 = −18, 0, −3 , tangent plane 6x + z = −9 y2 8 2− 4 2+y 2 /8 4 −4 y 2 /4 dy = 32 ; y = 0 by symmetry; ¯ 3 3 256 8 34 1 256 y dy = , x= ¯ = ; centroid 2 + y2 − 4 128 15 32 15 5 x dx dy = −4 1 + u2 du = 53/2 − 1 0 4 dx dy = 33. A = 4 3u = −2, −4, 1 , tangent plane 2x + 4y − z = 5 32. u = −3, v = 0, (ru × rv ) 4 2 1 + u2 dv du = 0 31. (ru × rv ) 3u 2 34. A = πab/2, x = 0 by symmetry, ¯ √ 22 a b 1−x /a 1 a2 y dy dx = b (1 − x2 /a2 )dx = 2ab2 /3, centroid 2 −a −a 0 35. V = 0, 8 ,0 5 4b 3π 12 πa h, x = y = 0 by symmetry, ¯¯ 3 2π h−rh/a a a rh2 1 − rz dz dr dθ = π 0 0 0 2 0 4−y 4 2 −2 x2 4 4−y 2 2 x2 2 4 4 z dz dy dx = −2 −2 0 x2 −2 2 x2 2 4−y x2 x2 (4y − y 2 ) dy dx = −2 0 2 4 y dz dy dx = −2 dr = πa2 h2 /12, centroid (0, 0, h/4) (4 − y )dy dx = −2 0 2 4 dz dy dx = 36. V = r a −2 1 (4 − y )2 dy dx = 2 1 256 , 8 − 4x2 + x4 dx = 2 15 32 16 x − 2x4 + 3 3 2 −2 − dx = 32 x6 + 2x4 − 8x2 + 6 3 1024 35 dx = 2048 105 8 0, , 4 7 x = 0 by symmetry, centroid ¯ √ 37. The two quarter-circles with center at the origin and of radius A and 2A lie inside and outside of the square with corners (0, 0), (A, 0), (A, A), (0, A), so the following inequalities hold: π /2 A 0 0 1 rdr dθ ≤ (1 + r2 )2 A 0 A 0 (1 + x2 √ 2A π /2 1 + y 2 )2 dx dy ≤ 0 0 1 rdr dθ (1 + r2 )2 2 πA and the integral on the right equals 4(1 + A2 ) 2 π 2πA . Since both of these quantities tend to , it follows by sandwiching that 2) 4(1 + 2A 4 The integral on the left can be evaluated as +∞ 0 +∞ 0 π 1 dx dy = . (1 + x2 + y 2 )2 4 38. The centroid of the circle which generates the tube travels a distance 4π √ √ √ s= sin2 t + cos2 t + 1/16 dt = 17π , so V = π (1/2)2 17π = 17π 2 /4. 0 607 Chapter 16 39. (a) The values of x in the formula of the astroidal sphere lie between −a and a; and the same is true for x = (a cos u cos v )3 ; similarly for y and z . Moreover, x2/3 + y 2/3 + z 2/3 = a2/3 cos2 u cos2 v + a2/3 sin2 u cos2 v + a2/3 sin2 v = a2/3 cos2 v + a2/3 sin2 v = a2/3 π /2 π /2 ru × rv du dv (b) S = 8 0 0 π /2 π /2 sin2 v + sin2 u cos2 u cos2 vdu dv ≈ 4.451 sin u cos u sin v cos4 v = 72 0 0 (1−x2/3 )3/2 1 1 − x2/3 − y 2/3 dy dx ≈ 0.3590 (c) 8 0 0 (d) Let x = t cos3 u, y = t sin3 u, 0 ≤ t ≤ 1, 0 ≤ u ≤ π/2, then J = π /2 1 (1 − t2/3 )3/2 3t sin2 u cos2 u du dt = V =8 0 40. V = 0 4 3¯ 3 πa , d = 3 4πa3 ρdV = ρ≤a 3 4πa3 π 2π 3π 2 ∂ (x, y ) = 3t sin2 u cos2 u, ∂ (t, u) 1 t(1 − t2/3 )3/2 dt = 0 a ρ3 sin φ dρ dθ dφ = 0 0 0 4π 35 3 3 a4 2π (2) = a 3 4πa 4 4 41. (a) (x/a)2 +(y/b)2 +(z/c)2 = sin2 φ cos2 θ +sin2 φ sin2 θ +cos2 φ = sin2 φ +cos2 φ = 1, an ellipsoid (b) r(φ, θ) = 2 sin φ cos θ, 3 sin φ sin θ...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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