EVNREV06 - Review Exercises for Chapter 6 26. From Exercise...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Review Exercises for Chapter 6 26. From Exercise 21: F 64 15 1 2 2 299 753.98 lb 28. h y 3 y 5x 2 x 2 4y 5 2 4y 5 y 3 30. h y 4 for x, you obtain y. Ly F 4y 5 y y 5 y y 12 2 62.4 0 y 7 16 2 4 Solving y x Ly F y2 7 16 7 16 y 16 y2 12 4 y 12 y2 dy y2 dy 21373.7 lb 62.4 2 0 3 3 3 0 y dy dy 546.265 lb 62.4 7 0 y 2 124.8 y 5 4 10 8 6 x 2 1 x −3 −2 −1 −1 1 2 3 −6 −4 −2 2 4 6 32. Fluid pressure is the force per unit of area exerted by a fluid over the surface of a body. 34. The left window experiences the greater fluid force because its centroid is lower. Review Exercises for Chapter 6 5 2. A 12 4 4x y 1 dx x2 1 x 5 12 1 4. A 0 1 y2 y2 0 1 2y 2y 1 2 dy 1 dy 1 dy 81 5 6 y 0 1 ( 2 , 4) 5 (5, 4) y 3 1 31 0 3 2 1 1 2 3 4 1 ( 5, 25 ) 6 1 3 y x 3 2 (−1, 1) 1 1 2 −2 3 −2 1 −2 x 300 Chapter 6 2 Applications of Integration 2 6. A 1 2 y 2 1 3 y 12 y 2 y2 y2 dy 13 y 3 2 1 1 dy 8. A 2 6 2 csc x d x 2 2 2x 9 2 2 2 2 3 ln csc x 0 ln 2 cot x 6 2y y 3 3 3 ln 2 3 1.555 (5, 2) 2 1 x 2 −1 −2 3 4 5 3 y (π , 2) 6 1 ( 56π, 2) (2, −1) π 6 π 3 π 2 2π 3 5π 6 x 5 3 3 10. A y 2 3 1 2 sin y 7 3 cos y d y 5 5 3 3 cos y y 2 7 5 3 3 1 dy 2 12. Point of intersection is given by: x3 A x2 0.783 4x 3 3 4x 0⇒x x2 0.783. sin y 3 x3 dx 14 x 4 0.783 0 23 y 3 0 3x 2x 2 13 x 3 2 ( 21 , 73π ) ( 21 , 53π ) 1 ( 2, π ) 3 1.189 4 −1 x 2 3 − 2 (.7832, .4804) −2 −1 2 14. A 2 0 2 2x 2 4x 2 0 x4 x4 dx 15 x 5 2 0 2x 2 d x 16. y y x x 1 2 2 1⇒x ⇒x 1 1 1 32 y2 2y y2 x 2 1 x 4 1 1 1 2 4 2 x3 3 10 128 15 8.5333 A 0 5 2y x 1 dy dx 5 (−2, 8) (2, 8) 1 2 x 3 −4 −2 4 y 1 2 1 4 3 (0, 0) 3 (5, 2) 2 1 x (1, 0) 2 −1 −2 3 4 5 R eview Exercises for Chapter 6 1 5 301 18. A 0 2 dx 1 2 x 1 dx 5 y x A y2 2 1 y2 1 dy 2 4 3 (5, 2) 0 13 y 3 y 0 14 3 1 (1, 0) x 2 3 4 5 20. (a) R1 t 40 5.2834 1.2701 t 5.2834 e0.2391t (b) R2 t Difference 10 15 5.28 e0.2t R1 t 10 R2 t dt 171.25 billion dollars 0 0 10 22. (a) Shell 2 2 (b) Shell 2 V y 4 3 2 0 y3 dy 2 y4 0 8 V 2 0 2 2 2y 2 0 y y2 dy y3 dy 14 y 4 2 0 2 2 y 23 y 3 8 3 2 1 4 x 1 2 3 4 3 1 x 1 2 3 4 (c) Disk 2 2 (d) Disk y4 dy 0 V y 5 y5 0 32 5 2 V 0 2 y2 y4 0 1 2 12 d y 2y 2 d y 23 y 3 2 0 4 3 15 y 5 y 176 15 2 1 x 1 2 3 4 3 2 1 x 1 2 3 4 5 4 302 Chapter 6 Applications of Integration (b) Disk 24. (a) Shell a V 4 0 x 2b a b a a a a2 a2 x2 x2 x2 dx 12 V 2x d x 42 ab 3 2 0 b2 2 a a2 x2 dx 13 x 3 a 0 0 a 32 0 2 b2 2 ax a2 4 ab2 3 y 4b 2 a 3a y (0, b) 2 x2 + y =1 a2 b2 (0, b) 2 x2 + y =1 a2 b2 x x (a, 0) (a, 0) 26. Disk 1 28. Disk 2 0 V 1 1 1 0 2 1 x2 dx V 0 1 e e 0 x2 dx 1 2 arctan x 2 2 2x dx 1 2 2 e 2x 0 4 0 y 2e 2 2 1 e2 2 y 3 2 x 1 1 −2 −1 −1 x 1 2 30. (a) Disk 0 (b) Shell x2 x 1 0 V 1 dx x2 0 1 u x u2 x 1 1 y x3 1 dx dx V 2u du 0 −1 x x4 4 y x3 3 12 2 1 1 x2 x u2 0 1 1 dx −1 4 x 1 2 u 2du 2u 4 25 u 5 u2 du 13 u 3 1 0 −1 4 0 u6 17 u 7 −1 4 32 105 Review Exercises for Chapter 6 1 bh 2 3 a2 a 303 32. A x 1 2 a2 2 x2 x2 3 a2 x2 34. y y x3 6 12 x 2 12 x 2 3 1 2x 1 2x 2 1 2x 2 2 V 3 a a2 4a3 3 x2 dx 3 a 2x x3 3 a a 1 y 2 3 s 1 12 x 2 1 dx 2x 2 13 x 6 1 2x 3 1 14 3 Since 4 3 a 3 3 a 3 10, we have a 3 1.630 meters. 5 3 2. Thus, 53 2 2 a 2− x 2 2 a 2− x 2 2 a 2− x 2 tan x has f x sec 2 x, this integral 36. Since f x represents the length of the graph of tan x from x 0 to 4. This length is a little over 1 unit. Answers (b). x 38. y 2x y 1 x 1 2 0 1 y 2 1 x 3 x x 2x 1 x x 3 S 1 3 dx 56 3 4 0 x 1 dx 4 2 x 3 1 32 0 40. F 50 F W kx k9 ⇒ k 50 x 9 9 0 42. We know that 50 9 dV dt V 25 2 x 9 9 0 4 gal min 12 gal min 7.481 gal ft3 r 2h dh 9 dt 9 dV dt 9 3.064 t 8 7.481 150 1 h 9 8 ft3 min 7.481 50 x dx 9 lb dV dt lb dh dt 225 in 18.75 ft 3.064 ft min. Depth of water: Time to drain well: t 49 12 150 49 minutes 3.064 588 gallons pumped Volume of water pumped in Exercise 41: 391.7 gallons 391.7 52 x Work 588 x 588 52 391.7 78 ft 78 ton 304 Chapter 6 Applications of Integration 44. (a) Weight of section of cable: 4 x Distance: 200 200 x 200 W 4 0 200 x dx 2 200 x 2 0 80,000 ft lb 40 ft lb ton 30 ft ton (b) Work to move 300 pounds 200 feet vertically: 200 300 Total work work for drawing up the cable 40 ft b 60,000 ft work of lifting the load ton ton 30 ft ton 70 ft 46. W a F x dx 2 9x 4 3x 9 Fx W 0 6, 16, 6 dx 0≤x≤9 9 ≤ x ≤ 12 12 9 2 x 9 12 x 9 9 3 4 x 3 16x 9 16 d x 12 9 6x 0 22 x 3 96 192 54 54 144 13 x 3 51 ft 3 1 lbs 32 3 48. A 1 2x 3 32 3 32 3 3 x2 dx x2 3x 1 A x y x 2x 1 3 3 2x x 2 dx 3 43 x 3 2 3 32 3 3x 1 2x 2 3 x 3 dx 12x 4x 2 3 32 x 32 2 x4 dx 23 x 3 14 x 4 3 1 1 y 31 32 2 3 9x 64 x4 dx 15 x 5 3 1 1 3 64 17 5 9 1 9 6x 2 6 ( x, y ) 3 x, y 17 1, 5 −3 x 3 6 8 50. A 0 x2 3 5 16 5 16 8 1 x dx 2 3 53 x 5 12 x 4 8 0 16 5 y 6 4 2 1 A x ( x, y ) x x2 3 0 1 x dx 2 13 x 6 8 0 x 2 −2 4 6 8 5 3 83 x 16 8 y 51 16 2 15 2 16 x, y 10 40 , 3 21 8 10 3 x4 3 0 12 x dx 4 13 x 12 8 0 3 73 x 7 40 21 Problem Solving for Chapter 6 52. Wall at shallow end: 5 305 54. F 1248 y2 2 5 62.4 16 5 4992 lb F 62.4 0 y 20 d y 15,600 lb 0 Wall at deep end: 10 10 F 62.4 0 y 20 d y 624 y 2 0 62,400 lb Side wall: 5 5 F1 62.4 0 5 y 40 d y 1248 y 2 0 5 31,200 lb F2 F y 20 15 10 5 62.4 0 10 F2 y 8y d y 72,800 lb 62.4 0 80y 8y 2 d y F1 −5 x 5 10 15 20 25 30 35 40 45 Problem Solving for Chapter 6 1 2. R 0 x1 x dx x2 2 x3 3 1 0 1 2 1 3 1 6 4. 8y2 y ± x2 1 x y 0.5 0.25 x2 1 x2 Let c, mc be the intersection of the line and the parabola. Then, mc 11 26 1 12 c1 1 0 c ⇒m m 22 1 c or c 1 m. x x2 2 1 2 1 61 1 1 1 2 1 1 m m m m m 1 2 13 2 2 2 x2 x2 m 2 1 mx dx 1 0 m −1.5 x3 3 m 2 − 0.5 − 0.25 x 0.5 1.5 m 3 3 m m 3 1 2 6m 1 6m m 2 − 0.5 41 41 2m m 2 6 2 m For x > 0, y 1 1 2x2 2 2 1 x2 x 1 1 2x2 2 2 1 x2 2 3 S 22 0 dx 1 2 13 52 3 0.2063 m 1 306 Chapter 6 Applications of Integration 8. f x fx 1 2 w2 l2 lw fx f0 fx 2 6. By the Theorem of Pappus, V 2 rA 2 d ex ex 2 2 2e x C 2 0⇒C 2e x 2 2 10. Let f be the density of the fluid and 0 fg h 0 the density of the iceberg. The buoyant force is F A y dy where A y is a typical cross section and g is the acceleration due to gravity. The weight of the object is L h W F fg h 0 f 0g h A y dy. W 0 L h A y dy 0g h A y dy 0.92 1.03 103 103 0.893 or 89.3% submerged volume total volume 0 by symmetry 6 12. (a) y My m x (b) My m x y 2 1 6 x 1 dx x4 6 2 1 1 dx x3 35 36 −1 3 2 1 y= 1 x4 2 1 1 dx x4 215 324 63 43 b2 b2 2 b3 1 3b3 3b b 1 2 b2 b 1 3 ,0 2 160 0 28 160 0 38 x, y 1 x, y 63 ,0 43 x −1 −2 −3 2 3 4 5 35 36 215 324 b y=−1 x4 2 1 b 1 dx x3 1 dx x4 2 1 b2 2 b3 3 2 1 b2 1 3b3 x, y 3b b 1 ,0 2 b2 b 1 b→ lim x 14. (a) Trapezoidal: Area (b) Simpson’s: Area 2 50 4 50 2 54 2 54 2 82 4 82 2 82 2 82 2 73 4 73 2 75 2 75 2 80 4 80 0 0 9920 sq ft 10,4131 sq ft 3 16. Point of equilibrium: 1000 x P0, x0 840, 20 20 0.4x2 840 42x 20, p Consumer surplus 0 20 1000 840 0 0.4x2 42x dx 840 dx 8400 2133.33 Producer surplus ...
View Full Document

This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online