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Unformatted text preview: 2 dy = (27 − 5 5)/12 0 41. ∂ r/∂u = cos v i + sin v j + 2uk, ∂ r/∂v = −u sin v i + u cos v j, √ ∂ r/∂u × ∂ r/∂v = u 4u2 + 1; S = 2π 2 u 0 √ √ 4u2 + 1 du dv = (17 17 − 5 5)π/6 1 42. ∂ r/∂u = cos v i + sin v j + k, ∂ r/∂v = −u sin v i + u cos v j, √ π /2 2v √ √ 23 π 2 u du dv = ∂ r/∂u × ∂ r/∂v = 2u; S = 12 0 0 √ 5/6 585 Chapter 16 2 2 43. zx = y , zy = x, zx + zy + 1 = x2 + y 2 + 1, π /6 r2 + 1 dr dθ = r 0 R √ 1 (10 10 − 1) 3 3 x2 + y 2 + 1 dA = S= 0 π /6 √ dθ = (10 10 − 1)π/18 0 2 2 44. zx = x, zy = y , zx + zy + 1 = x2 + y 2 + 1, √ 8 2π x2 S= + y2 + 1 dA = r2 + 1 dr dθ = r 0 R 0 26 3 2π dθ = 52π/3 0 2 2 45. On the sphere, zx = −x/z and zy = −y/z so zx + zy + 1 = (x2 + y 2 + z 2 )/z 2 = 16/(16 − x2 − y 2 ); the planes z = 1 and z = 2 intersect the sphere along the circles x2 + y 2 = 15 and x2 + y 2 = 12; S= 16 − x2 − y 2 R √ 15 2π 4 dA = √ 12 0 √ 4r dr dθ = 4 16 − r2 2π dθ = 8π 0 2 2 46. On the sphere, zx = −x/z and zy = −y/z so zx + zy + 1 = (x2 + y 2 + z 2 )/z 2 = 8/(8 − x2 − y 2 ); the cone cuts the sphere in the circle x2 + y 2 = 4; √ 2π 2 2π √ √ 2 2r √ S= dr dθ = (8 − 4 2) dθ = 8(2 − 2)π 2 8−r 0 0 0 47. r(u, v ) = a cos u sin v i + a sin u sin v j + a cos v k, ru × rv = a2 sin v, π 2π π a2 sin v du dv = 2πa2 S= 0 0 sin v dv = 4πa2 0 h 2π 48. r = r cos ui + r sin uj + v k, ru × rv = r; S = r du dv = 2πrh 0 49. zx = h a x 2π S= 0 , zy = h a y 0 2 2 , zx + zy + 1 = x2 + y 2 x2 + y 2 √ a 1 a2 + h2 r dr dθ = a a2 + h2 a 2 0 h2 x2 + h2 y 2 + 1 = (a2 + h2 )/a2 , a2 (x2 + y 2 ) 2π dθ = πa a2 + h2 0 50. Revolving a point (a0 , 0, b0 ) of the xz -plane around the z -axis generates a circle, an equation of which is r = a0 cos ui + a0 sin uj + b0 k, 0 ≤ u ≤ 2π . A point on the circle (x − a)2 + z 2 = b2 which generates the torus can be written r = (a + b cos v )i + b sin v k, 0 ≤ v ≤ 2π . Set a0 = a + b cos v and b0 = a + b sin v and use the first result: any point on the torus can thus be written in the form r = (a + b cos v ) cos ui + (a + b cos v ) sin uj + b sin v k, which yields the result. 51. ∂ r/∂u = −(a + b cos v ) sin ui + (a + b cos v ) cos uj, ∂ r/∂v = −b sin v cos ui − b sin v sin uj + b cos v k, ∂ r/∂u × ∂ r/∂v = b(a + b cos v ); 2π 2π b(a + b cos v )du dv = 4π 2 ab S= 0 52. 0 ru × rv = √ u2 + 1; S = 4π 5 5 u2 + 1 du dv = 4π 0 0 u2 + 1 du = 174.7199011 0 53. z = −1 when v ≈ 0.27955, z = 1 when v ≈ 2.86204, ru × rv = | cos v |; 2π 2.86204 | cos v | dv du ≈ 9.099 S= 0 0.27955 Exercise Set 16.5 586 54. (a) x = v cos u, y = v sin u, z = f (v ), for example x = v cos u, y = v sin u, z = 1/v 2 (b) z (c) x y 55. (x/a)2 + (y/b)2 + (z/c)2 = cos2 v (cos2 u + sin2 u) + sin2 v = 1, ellipsoid 56. (x/a)2 + (y/b)2 − (z/c)2 = cos2 u cosh2 v + sin2 u cosh2 v − sinh2 v = 1, hyperboloid of one sheet 57. (x/a)2 + (y/b)2 − (z/c)2 = sinh2 v + cosh2 v (sinh2 u − cosh2 u) = −1, hyperboloid of two sheets EXERCISE SET 16.5 1 2 1 1 2 (x2 + y 2 + z 2 )dx dy dz = 1. −1 0 −1 0 1/2 π π zx sin xy dz dy dx = 1/3 0 0 2 y2 1/3 z −1 0 π /4 −1 2 −1 0 0 π /4 0 x3 cos y dx dy = 0 0 0 √ 9−z 2 3 x 0 0 √ 9−z 2 3 xy dy dx dz = 5. 0 0 x2 3 0 ln z 0 3 0 x 1 0 √ 4−x2 2 1 3−x2 −y 2 √ 4−x2 2 x dz dy dx = 2 0 √ 3y 2 8. 1 z 0 y dx dy dz = 2 + y2 x 2 1 1 1 (1 − cos πx)dx = + 2 12 √ 1 cos y dy = 2/8 4 1 (81 − 18z 2 + z 4 )dz = 81/5 8 15 33 x − x + x2 dx = 118/3 2 2 [2x(4 − x2 ) − 2xy 2 ]dy dx 0 0 = 2 0 3 x −5+x2 +y 2 0 0 3 (xz − x)dz dx = 1 7. 13 x dx dz = 2 x2 xey dy dz dx = 6. 1/3 π /4 1 x cos y dz dx dy = 1/2 47 17 15 1 y + y − y dy = 3 2 6 3 (yz 2 + yz )dz dy = x2 1 4. 1 x sin xy dy dx = 2 y2 2 yz dx dz dy = 3. 0 (10/3 + 2z 2 )dz = 8 −1 0 1/2 1 2. 1 (1/3 + y 2 + z 2 )dy dz = 2 z 4 x(4 − x2 )3/2 dx = 128/15 3 π dy dz = 3 2 1 π (2 − z )dz = π/6 3 √ 3−2 4π 587 Chapter 16 π π /6 1 π 9. 0 0 x[1 − cos(πy/6)]dy dx = 0 0 1−x2 1 y 0 1−x2 1 −1 0 −1 0 √ 2 1 11. −1 0 √ 2 2−x2 x x xyz dz dy dx = 0 0 0 π /2 0 π /2 0 xy π/6 y π /2 √ √ 1−x2 1 1−x2 −y 2 14. 8 0 0 e−x 2 √ 2 0 π /2 π /2 y sin x dx dy = π/6 0 1 (1 − x2 )3 dx = 32/105 3 1 xy (2 − x2 )2 dy dx = 2 cos(z/y )dz dx dy = 12. −y 2 −z 2 (1 − 3/π )x dx = π (π − 3)/2 0 y 2 dy dx = y dz dy dx = 10. π 1 xy sin yz dz dy dx = y 13 x (2 − x2 )2 dx = 1/6 4 √ y cos y dy = (5π − 6 3)/12 π/6 dz dy dx ≈ 2.381 0 (4−x)/2 4 (12−3x−6y )/4 (4−x)/2 4 dz dy dx = 15. V = 0 0 4 = 0 0 0 0 1 (12 − 3x − 6y )dy dx 4 3 (4 − x)2 dx = 4 16 √ y 1−x 1 1−x 1 dz dy dx = 16. V = 0 0 0 2 0 0 4−y 4 x2 0 2 0 √ y 1 4 1−y 2 1 8 − 4x2 + x4 dx = 256/15 2 0 y 1 0 0 2 x2 1 1 − y 2 dx dy = dz dx dy = 0 2 (1 − x)3/2 dx = 4/15 3 (4 − y )dy dx = 2 0 18. V = 1 y dy dx = 0 dz dy dx = 2 17. V = 2 √ 0 0 1 − y 2 dy = 1/3 y 0 19. The projection of the curve of intersection onto the...
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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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