A cosh 3x cosh2x x cosh 2x cosh x sinh 2x sinh x

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Unformatted text preview: x h h -4 -4 2≤h≤4 0≤h<2 If the cherry is partially submerged then 0 ≤ h < 2 as shown in Figure (a); if it is totally submerged then 2 ≤ h ≤ 4 as shown in Figure (b). The radius of the glass is 4 cm and that of the cherry is 1 cm so points on the sections shown in the ﬁgures satisfy the equations x2 + y 2 = 16 and x2 + (y + 3)2 = 1. We will ﬁnd the volumes of the solids that are generated when the shaded regions are revolved about the y -axis. For 0 ≤ h < 2, h−4 V =π −4 [(16 − y 2 ) − (1 − (y + 3)2 )]dy = 6π h−4 (y + 4)dy = 3πh2 ; −4 for 2 ≤ h ≤ 4, −2 V =π −4 [(16 − y 2 ) − (1 − (y + 3)2 )]dy + π −2 h−4 (y + 4)dy + π = 6π −4 = so −2 h−4 −2 (16 − y 2 )dy 1 (16 − y 2 )dy = 12π + π (12h2 − h3 − 40) 3 1 π (12h2 − h3 − 4) 3 2 3πh V= 1 π (12h2 − h3 − 4) 3 if 0 ≤ h < 2 if 2 ≤ h ≤ 4 Exercise Set 8.3 44. 268 x=h± r2 − y2 , y r V =π r2 − y 2 )2 − (h − (h + −r (x – h 2) + y2 = r 2 r2 − y 2 )2 dy r −r 12 πr 2 = 4πh x r2 − y 2 dy = 4πh = 2π 2 r2 h 45. tan θ = h/x so h = x tan θ, A(y ) = 1 1 1 hx = x2 tan θ = (r2 − y 2 ) tan θ 2 2 2 h because x2 = r2 − y 2 , V= u x r 1 tan θ 2 −r (r2 − y 2 )dy r 23 r tan θ 3 (r2 − y 2 )dy = = tan θ 0 r2 − x2 ) 46. A(x) = (x tan θ)(2 47. Each cross section perpendicular to the y -axis is a square so = 2(tan θ)x r2 − x2 , A(y ) = x2 = r2 − y 2 , r x r2 − x2 dx V = 2 tan θ 1 V= 8 0 = 23 r tan θ 3 r (r2 − y 2 )dy 0 V = 8(2r3 /3) = 16r3 /3 y x tan u y x x= √ r 2 – y2 √r 2 – x 2 r x 48. The regular cylinder of radius r and height h has the same circular cross sections as do those of the oblique clinder, so by Cavalieri’s Principle, they have the same volume: πr2 h. EXERCISE SET 8.3 2 2 2πx(x2 )dx = 2π 1. V = 1 x3 dx = 15π/2 1 √ 2 √ 2 2πx( 2. V = 4− x2 − x)dx = 2π 0 0 1 1 2πy (2y − 2y 2 )dy = 4π 3. V = 0 (x 4 − x2 − x2 )dx = (y 2 − y 3 )dy = π/3 0 √ 8π (2 − 2) 3 269 Chapter 8 2 2 2πy [y − (y 2 − 2)]dy = 2π 4. V = 0 (y 2 − y 3 + 2y )dy = 16π/3 0 1 9 2π (x)(x3 )dx 5. V = 6. V = 0 √ 2πx( x)dx 4 1 9 x4 dx = 2π/5 = 2π x3/2 dx = 844π/5 = 2π 0 4 y y y = x3 3 1 2 x -1 y = √x 1 1 x -1 -9 3 7. V = -4 √ π/2 3 2πx(1/x)dx = 2π 1 4 dx = 4π √ 2πx cos(x2 )dx = π/ 2 8. V = 1 9 0 y y y = cos (x 2) 1 y= x x -3 -1 1 x 3 √p 2 2 1 2πx[(2x − 1) − (−2x + 3)]dx 9. V = 10. V = 2π 1 0 2 1 = π ln(x2 + 1) (x2 − x)dx = 20π/3 = 8π x dx x2 + 1 = π ln 2 0 1 y y 1 (2, 3) y= 1 x2 + 1 (1, 1) x -1 x 1 (2, –1) √ 3 11. V = x2 2πxe 1 x2 √ 3 dx = πe 1 y = π (e − e) 3 20 y = ex 2 10 x -√3 -1 1 √3 Exercise Set 8.3 270 y 2 2πx(2x − x2 )dx 12. V = y = 2x – x 2 0 2 (2x2 − x3 )dx = = 2π 0 8 π 3 x 2 1 3 2πy 3 dy = π/2 13. V = 14. V = 0 3 y 2 dy = 76π/3 2πy (2y )dy = 4π 2 2 y y x = y2 1 3 2 x = 2y x x 1 2πy (1 − 15. V = √ 4 y )dy 2πy (5 − y − 4/y )dy 16. V = 0 1 1 4 (y − y 3/2 )dy = π/5 = 2π (5y − y 2 − 4)dy = 9π = 2π 0 1 y y (1, 4) x = 5–y y = √x (4, 1) x x = 4/y 1 π π /2 x sin xdx = 2π 2 17. V = 2π x x cos xdx = π 2 − 2π 18. V = 2π 0 0 1 2πx(x3 − 3x2 + 2x)dx = 7π/30 19. (a) V = 0 (b) much easier; the method of slicing would require that x be expressed in terms of y . y y = x3 – 3x2 + 2x x -1 1 271 Chapter 8 y 2 2π (x + 1)(1/x3 )dx 20. V = 1 x+1 2 = 2π (x−2 + x−3 )dx = 7π/4 y = 1/x 3 1 x -1 1x 2 y 1 2π (1 − y )y 1/3 dy 21. V = 0 1 (y 1/3 − y 4/3 )dy = 9π/14 = 2π 0 1 x = y1/3 1–y b d 2πx[f (x) − g (x)]dx 22. (a) x 2πy [f (y ) − g (y )]dy (b) a c y h (r − y ) is an equation of the line r through (0, r) and (h, 0) so 23. x = r h (r − y ) dy r 2πy V= 0 = 2πh r k /4 24. V = (0, r) x (h, 0) r (ry − y 2 )dy = πr2 h/3 0 √ 2π (k/2 − x)2 kxdx y 0 √ = 2π k k/2 – x y = √kx k /4 1/2 (kx − 2x 3/2 3 )dx = 7πk /60 x 0 y = – √kx a 25. V = a 2πx(2 r2 − x2 )dx = 4π 0 =− x(r2 − x2 )1/2 dx 0 4π 2 (r − x2 )3/2 3 a = 0 x = k/2 x = k/4 y y= √ r 2 – x2 4π 3 r − (r2 − a2 )3/2 3 x a y = – √ r 2 – x2 Exercise Set 8.4 272 y a 26. V = −a 2π (b − x)(2 a2 − x2 )dx a b−x a = 4πb −a a2 − x2 dx − 4π −a √ a2 – x2 x a2 − x2 dx –a = 4πb · (area of a semicircle of radius a) − 4π (0) x a – √ a2 – x2 = 2π 2 a2 b x=b b 27. Vx = π 1/2 b 1 dx = π (2 − 1/b), Vy = 2π x2 dx = π (2b − 1); 1/2 Vx = Vy if 2 − 1/b = 2b − 1, 2b2 − 3b + 1 = 0, solve to get b = 1/2 (reject) or b = 1. EXERCISE SET 8.4 1. (a) (b) 2. dy = 2, L = dx 1 dx = ,L = dy 2 2 √ 1 L= 0 √ 5 1 √ √ 1 + 1/4 dy = 2 5/2 = 5 4 2 1 dy dx = 1, = 5, L = dt dt 3. f (x) = 1 + 4dx = 12 + 52 dt = √ 26 0 9 1/2 81 x , 1 + [f (x)]2 = 1 + x, 2 4 8 1 + 81x/4dx = 243 81 1+ x 4 3/2 1 √ = (85 85 − 8)/243 0 4. g (y ) = y (y 2 + 2)1/2 , 1 + [g (y )]2 = 1 + y 2 (y 2 + 2) = y 4 + 2y 2 + 1 = (y 2 + 1)2 , 1 1 (y 2 + 1)dy = 4/3 (y 2 + 1)2 dy = L= 0 0 2 5. 2 4 9x2/3 + 4 dy dy = x−1/3 , 1 + = 1 + x−2/3 = ,...
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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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