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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online