Let y sech1 x then x sech y 1 cosh y cosh y 1x

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Unformatted text preview: = 4 × πa = πab. a2 − x2 dx = a2 − x2 dx = a a0 a 4 0 37. Let A be the area between the curve and the x-axis and AR the area of the rectangle, then b b k kbm+1 A= xm+1 = , AR = b(kbm ) = kbm+1 , so A/AR = 1/(m + 1). kxm dx = m+1 m+1 0 0 EXERCISE SET 8.2 3 1 1. V = π −1 (3 − x)dx = 8π [(2 − x2 )2 − x2 ]dx 2. V = π 0 1 (4 − 5x2 + x4 )dx =π 0 = 38π/15 2 2 0 1 (3 − y )2 dy = 13π/6 4 4. V = π x4 dx = 32π/5 3. V = π 6. V = π (4 − 1/y 2 )dy = 9π/2 1/2 √ sec2 x dx = π ( 3 − 1) π /3 2 5. V = π π/4 0 y y y = sec x y = x2 2 1 x x 2 3 -1 4 -2 π /2 cos x dx = (1 − 7. V = π √ 2/2)π y y = √cos x 1 π/4 x 3 6 -1 y 1 [(x2 )2 − (x3 )2 ]dx 8. V = π 0 1 (x4 − x6 )dx = 2π/35 =π 1 (1, 1) y = x2 y = x3 0 1 x Exercise Set 8.2 262 4 3 9. V = π −4 [(25 − x2 ) − 9]dx 10. V = π −3 4 3 (16 − x2 )dx = 256π/3 = 2π (9 − x2 )2 dx =π −3 0 5 y y= √25 – x 2 (81 − 18x2 + x4 )dx = 1296π/5 y 9 y = 9 – x2 y=3 -3 x ln 3 π e dx = e2x 2 0 1 12. V = π e−4x dx = 0 x ln 3 2x 11. V = π 3 = 4π 0 π (1 − e−4 ) 4 4 [(4x)2 − (x2 )2 ]dx 13. V = π 0 y 4 (16x2 − x4 )dx = 2048π/15 =π 1 0 y (4, 16) 16 x y = 4x 1 y = x2 x 4 -1 π /4 1 (cos2 x − sin2 x)dx 14. V = π y 2/3 dy = 3π/5 15. V = π 0 0 y π /4 cos 2x dx = π/2 =π 0 y 1 1 x y = sin x 3 -1 y = x3 y = cos x x -1 1 263 Chapter 8 1 3 16. V = π −1 (1 − y 2 )2 dy 17. V = π (1 + y )dy = 8π −1 1 =π −1 y (1 − 2y 2 + y 4 )dy = 16π/15 3 y 1 x = 1 – y2 x= √1 + y x 2 x -1 1 -1 3π/4 3 [22 − (y + 1)]dy 18. V = π csc2 y dy = 2π 19. V = π π/4 0 3 y (3 − y )dy = 9π/2 =π 9 0 y = x2 – 1 y 6 x = csc y (2, 3) 3 3 -2 x -1 1 2 x 1 2 (y − y 4 )dy = 3π/10 20. V = π 0 y 21. V = π −1 [(y + 2)2 − y 4 ]dy = 72π/5 y x = y2 1 -1 (1, 1) y = x2 x = y2 (4, 2) x =y+2 x 1 x -1 (1, –1) 1 22. V = π −1 (2 + y 2 )2 − (1 − y 2 )2 dy x = 2 + y2 y x = 1 – y2 1 1 (3 + 6y 2 )dy = 10π =π x −1 1 -1 2 Exercise Set 8.2 264 a 23. V = π −a y b2 2 (a − x2 )dx = 4πab2 /3 a2 b y= b a √a 2 – x 2 x –a 2 a 1 dx = π (1/b − 1/2); π (1/b − 1/2) = 3, b = 2π/(π + 6) x2 24. V = π b 0 4 (x + 1)dx 25. V = π 6 26. V = π −1 0 1 [(x + 1) − 2x]dx +π 0 = π/2 + π/2 = π 4 = 8π + 8π/3 = 32π/3 y y = √x y 1 -1 y = √ 2x 4 6 x 1 3 9 (9 − y 2 )2 dy 27. V = π [32 − (3 − 28. V = π 0 √ x)2 ]dx 0 3 9 (81 − 18y 2 + y 4 )dy =π =π 0 √ (6 x − x)dx 0 = 648π/5 = 135π/2 y 3 y =6–x x (1, √ 2) y = √x + 1 (6 − x)2 dx x dx + π y x = y2 x y=3 y = √x x 9 9 1 29. V = π √ [( x + 1)2 − (x + 1)2 ]dx 0 1 =π 0 √ (2 x − x − x2 )dx = π/2 y x=y x = y2 1 x 1 y = −1 265 Chapter 8 y 1 [(y + 1)2 − (y 2 + 1)2 ]dy 30. V = π x=y 0 1 1 (2y − y 2 − y 4 )dy = 7π/15 =π 0 x 1 x = y2 x = –1 1 31. A(x) = π (x2 /4)2 = πx4 /16, (x − x4 )dx = 3π/10 32. V = π 0 20 (πx4 /16)dx = 40, 000π ft3 V= 0 1 (x − x2 )2 dx 33. V = 0 1 4 (x2 − 2x3 + x4 )dx = 1/30 = V= 0 0 Square 1√ x 2 1 π 2 34. A(x) = = 1 πx, 8 1 πx dx = π 8 y y y = √x y = x (1, 1) y = x2 1 2 4 x 35. On the upper half of the circle, y = x √ 1 − x2 , so: (a) A(x) is the area of a semicircle of radius y , so A(x) = πy 2 /2 = π (1 − x2 )/2; V = π 2 1 −1 1 (1 − x2 ) dx = π (1 − x2 ) dx = 2π/3 0 y y -1 1 x y = √1 – x 2 (b) A(x) is the area of a square of side 2y , so 1 A(x) = 4y 2 = 4(1 − x2 ); V = 4 −1 1 (1 − x2 ) dx = 8 (1 − x2 ) dx = 16/3 0 y -1 2y y = √1 – x 2 1 x Exercise Set 8.2 266 (c) A(x) is the area of an equilateral triangle with sides 2y , so √ √ √ 3 (2y )2 = 3y 2 = 3(1 − x2 ); A(x) = 4 1√ 1 √ √ 3(1 − x2 ) dx = 2 3 (1 − x2 ) dx = 4 3/3 V= −1 0 2y 2y y -1 2y y = √1 – x 2 1 x 36. By similar triangles, R/r = y/h so r R = ry/h and A(y ) = πr2 y 2 /h2 . h V = (πr2 /h2 ) R y 2 dy = πr2 h/3 0 h y 37. The two curves cross at x = b ≈ 1.403288534, so b π /2 ((2x/π )2 − sin16 x) dx + π V =π (sin16 x − (2x/π )2 ) dx = 0.710172176. b 0 e (1 − (ln y )2 ) dy = π 38. V = π 1 r 39. (a) V = π r −h (r2 − y 2 ) dy = π (rh2 − h3 /3) = 12 πh (3r − h) 3 y (b) By the Pythagorean Theorem, r2 = (r − h)2 + ρ2 , 2hr = h2 + ρ2 ; from part (a), πh πh (3hr − h2 ) = 3 3 1 = πh(h2 + 3ρ2 ). 6 V= h r 32 (h + ρ2 ) − h2 ) 2 −10+h −10 (100 − y 2 )dy = y π2 h (30 − h) 3 Find dh/dt when h = 5 given that dV /dt = 1/2. π π dV dh V = (30h2 − h3 ), = (60h − 3h2 ) , 3 dt 3 dt π dh dh 1 = (300 − 75) , = 1/(150π ) ft/min 2 3 dt dt x r 40. Find the volume generated by revolving the shaded region about the y -axis. V =π x2 + y2 = r 2 h – 10 10 x h −10 x= √ 100 – y2 267 Chapter 8 41. (b) ∆x = 5 = 0.5; {y0 , y1 , · · · , y10 } = {0, 2.00, 2.45, 2.45, 2.00, 1.46, 1.26, 1.25, 1.25, 1.25, 1.25}; 10 9 2 yi 2 left = π i=0 10 2 yi 2 right = π i=1 ∆x ≈ 11.157; ∆x ≈ 11.771; V ≈ average = 11.464 cm3 √ 42. If x = r/2 then from y 2 = r2 − x2 we get y = ± 3r/2 √ √ as limits of integration; for − 3 ≤ y ≤ 3, y √ 3r 2 x= A(y ) = π [(r2 − y 2 ) − r2 /4] = π (3r2 /4 − y 2 ), thus x √ 3r/2 V =π √ r 2 – y2 (3r √ − 3r/2 √ 3r/2 2 /4 − y 2 )dy (3r2 /4 − y 2 )dy = = 2π r 2 √ 3πr3 /2. – 0 y 43. (a) √ 3r 2 y (b) h –4 x h –4 -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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