U x2 dv xex dx du 2x dx v 2 x3 ex dx 1 2 x2 xe

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Unformatted text preview: y , 1−y √ 2−y dy = 4π 2π (2 1 − y ) √ 1−y −1 0 S= 9 ,S= 9 − y2 0 −1 √ √ 2 − y dy = 8π (3 3 − 2 2)/3 2 dy = 6π dy = 24π −2 Exercise Set 8.5 9. f (x) = 3 S= 1 276 1 2π x1/2 − x3/2 3 2 1 3 1 −1 x+ x 3 4 2π (3 + 2x − x2 )dx = 16π/9 1 1 2 1 + [g (y )]2 = 1 + y 6 − 2 1 √ 1 15 1 x + x + x−3 dx = 515π/64 3 3 16 1 16 − y 2π = 2 1 y 3 + y −3 4 1 π y 3 + y −3 dy = 4 16 , 2 (8y 7 + 6y + y −5 )dy = 16, 911π/1024 1 1 65 − 4y , , 1 + [g (y )]2 = 16 − y ; g (y ) = − √ 4(16 − y ) 2 16 − y 15 S= 1 1 + y −6 2 16 1 4 1 −2 y+ y 4 8 2π 12. x = g (y ) = 1 x2 + x−2 , 4 = 1 1 4 1 −2 y + y , g (y ) = y 3 − y −3 , 4 8 4 11. x = g (y ) = S= 2 1 x2 + x−2 dx = 2π 4 1 −1/2 1 1/2 x +x , 2 2 3 1 −1/2 1 1/2 π x +x dx = 2 2 3 1 1 1 10. f (x) = x2 − x−2 , 1 + [f (x)]2 = 1 + x4 − + x−4 4 2 16 S= 2 1 −1/2 1 1/2 1 11 x − x , 1 + [f (x)]2 = 1 + x−1 − + x = 2 2 4 24 0 65 − 4y dy = π 4(16 − y ) √ √π 65 − 4y dy = (65 65 − 5 5) 6 15 0 1 13. f (x) = ex , 1 + e2x dx ≈ 22.94 2πex 1 + [f (x)]2 = 1 + e2x , S = 0 π 2π sin x 1 + cos2 x dx ≈ 14.42 14. f (x) = cos x, 1 + [f (x)]2 = 1 + cos2 x, S = 0 e 15. x = g (y ) = ln y, g (y ) = 1/y, 1 + [g (y )]2 = 1 + 1/y 2 ; S = 2π 1 + 1/y 2 ln y dy ≈ 7.05 1 16. x = g (y ) = tan y, g (y ) = sec2 y, 1 + [g (y )]2 = 1 + sec4 y ; π /4 2π tan y S= 1 + sec4 y dy ≈ 3.84 0 17. Revolve the line segment joining the points (0, 0) and (h, r) about the x-axis. An equation of the line segment is y = (r/h)x for 0 ≤ x ≤ h so h 2π (r/h)x 1 + r2 /h2 dx = S= 0 18. f (x) = √ 2πr h2 h r 2 + h2 r 2 + h2 √ r2 − x2 , f (x) = −x/ r2 − x2 , 1 + [f (x)]2 = r2 /(r2 − x2 ), r S= x dx = πr 0 r r2 − x2 (r/ r2 − x2 )dx = 2πr 2π −r 19. g (y ) = r2 − y 2 , g (y ) = −y/ dx = 4πr2 −r r2 − y 2 , 1 + [g (y )]2 = r2 /(r2 − y 2 ), r r 2π (a) S = r −h r2 − y2 r2 /(r2 − y 2 ) dy = 2πr dy = 2πrh r −h (b) From part (a), the surface area common to two polar caps of height h1 > h2 is 2πrh1 − 2πrh2 = 2πr(h1 − h2 ). 277 Chapter 8 20. For (4), express the curve y = f (x) in the parametric form x = t, y = f (t) so dx/dt = 1 and dy/dt = f (t) = f (x) = dy/dx. For (5), express x = g (y ) as x = g (t), y = t so dx/dt = g (t) = g (y ) = dx/dy and dy/dt = 1. 21. x = 2t, y = 2, (x )2 + (y )2 = 4t2 + 4 4 √ 8π (17 17 − 1) 3 4 (2t) 4t2 + 4dt = 8π S = 2π t 0 t2 + 1dt = 0 22. x = et (cos t − sin t), y = et (cos t + sin t), (x )2 + (y )2 = 2e2t π /2 S = 2π √ √ (et sin t) 2e2t dt = 2 2π 0 π /2 e2t sin t dt 0 √ 1 = 2 2π e2t (2 sin t − cos t) 5 π /2 0 √ 22 π (2eπ + 1) = 5 1 23. x = 1, y = 4t, (x )2 + (y )2 = 1 + 16t2 , S = 2π t 1 + 16t2 dt = 0 √ π (17 17 − 1) 24 24. x = −2 sin t cos t, y = 2 sin t cos t, (x )2 + (y )2 = 8 sin2 t cos2 t √ 8 sin2 t cos2 t dt = 4 2π π /2 cos2 t S = 2π 0 π /2 cos3 t sin t dt = √ 2π 0 25. x = −r sin t, y = r cos t, (x )2 + (y )2 = r2 , π π √ S = 2π r sin t r2 dt = 2πr2 sin t dt = 4πr2 0 26. 0 dy dx = a(1 − cos φ), = a sin φ, dφ dφ 2π S = 2π dx dφ 2 + dy dφ 2 = 2a2 (1 − cos φ) √ a(1 − cos φ) 2a2 (1 − cos φ) dφ = 2 2πa2 0 2π (1 − cos φ)3/2 dφ, 0 √ φ φ so (1 − cos φ)3/2 = 2 2 sin3 for 0 ≤ φ ≤ π and, taking advantage of the 2 2 π 3φ 2 symmetry of the cycloid, S = 16πa sin dφ = 64πa2 /3. 2 0 but 1 − cos φ = 2 sin2 27. (a) length of arc of sector = circumference of base of cone, 1 θ = 2πr, θ = 2πr/ ; S = area of sector = 2 (2πr/ ) = πr 2 (b) S = πr2 2 − πr1 1 = πr2 ( 1 + ) − πr1 1 = π [(r2 − r1 ) 1 + r2 ]; Using similar triangles 2 /r2 = 1 /r1 , r1 2 = r2 1 , r1 ( 1 + ) = r2 1 , (r2 − r1 ) so S = π (r1 + r2 ) = π (r1 + r2 ) . l1 r1 l2 l r2 1 = r1 Exercise Set 8.6 28. 2πk 278 1 + [f (x)]2 2πf (x) 1 + [f (x)]2 ≤ 2πK b 1 + [f (x)]2 , so b 1 + [f (x)]2 dx ≤ 2πk b 2πf (x) 1 + [f (x)]2 dx ≤ a 1 + [f (x)]2 dx, 2πK a a b b 1 + [f (x)]2 dx ≤ S ≤ 2πK 2πk a 1 + [f (x)]2 dx, 2πkL ≤ S ≤ 2πKL a 1 + [f (x)]2 so 2πf (x) ≤ 2πf (x) 1 + [f (x)]2 , 29. (a) 1 ≤ b b 2πf (x)dx ≤ b f (x)dx ≤ S, 2πA ≤ S 2πf (x) 1 + [f (x)]2 dx, 2π a a a (b) 2πA = S if f (x) = 0 for all x in [a, b] so f (x) is constant on [a, b]. EXERCISE SET 8.6 1. (a) W = F · d = 30(7) = 210 ft·lb 6 (b) W = 6 F (x) dx = 1 1 5 2 0 5 40 dx − F (x) dx = 2. W = x−2 dx = − 0 2 5 v (t) dt = 0 work done is 10 · 10 = 100 ft·lb. 6 = 5/6 ft·lb 1 40 (x − 5) dx = 80 + 60 = 140 J 3 5 3. distance traveled = 1 x 0 2 4t dt = t2 5 5 5 = 10 ft. The force is a constant 10 lb, so the 0 4. (a) F (x) = kx, F (0.05) = 0.05k = 45, k = 900 N/m 0.03 0.10 900x dx = 0.405 J (b) W = (c) W = 900x dx = 3.375 J 0.05 0 0.8 500xdx = 160 J 5. F (x) = kx, F (0.2) = 0.2k = 100, k = 500 N/m, W = 0 2 12x dx = 24 J 6. F (x) = kx, F (1/2) = k/2 = 6, k = 12 N/m, W = 0 1 kx dx = k/2 = 10, k = 20 lb/ft 7. W = 0 6 (9 − x)62.4(25π )dx 8. W = 0 5 9 6 (9 − x)dx = 56, 160π ft·lb = 1560π 0 6 x 0 6 (9 − x)ρ(25π )dx = 900πρ ft·lb 9. W = 0 9-x 279 Chapter 8 10. r/10 = x/15, r = 2x/3, 10 10 15 (15 − x)62.4(4πx2 /9)dx W= 0 = 15 - x 10 83.2 π 3 10 x (15x2 − x3 )dx r 0 = 208, 000π/3 ft·lb 0 11. w/4 = x/3, w = 4x/3, 4 3 2 (3 − x)(9810)(4x/3)(6)dx W= 0 3-x 2 2 x (3x − x2 )dx = 78480 w(x) 0 = 261, 600 J 0 12. w = 2 4 − x2 3 w(x) 2 W= −2 2 (3 − x)(50)(2 4 − x2 )(10)dx 2 = 3000 −2 2 4 − x2 dx − 1000 −2 3-x x 2 x 4 − x2 dx = 3000[π (2)2 /2] ...
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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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