16 MULTIPLE INTEGRALS

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CHAPTER 16 MULTIPLE INTEGRALS (b) ∫ 1 ∫ 1 ∫ |xy| 0 −1 0 1 dzdx dy = 2 ∫ 1 ∫ 1 ∫ xy 0 0 0 dzdx dy = 2 ∫ [ ] 1 1 12 0 x 2 y dy = 2 ∫ 1 0 0 2 1 ydy = 1 2 y2] 1 = 1 0 2 . This is the total volume between z = 0andz = xy. Because we take the absolute value, the volumes do not cancel. 9. Since the surface area is 4π,weneedtofindφ so that the area lit is π. π = ∫ φ ∫ 2π 0 0 sin φ dθ dφ = 2π ∫ φ 0 sin φ dφ = 2π (− cos φ + cos 0),so 1 2 = 1 − cos φ ⇒ cos φ = 1 2 ⇒ φ = π 3 . 10. (a) ∫ 1 0 ∫ x − √1−(x−1) 2 (x + y) dydx + ∫ 2 1 ∫ 2−x − √1−(x−1) (b) ∫ √ 0 ∫ 1+ 1−y 2 −1 √ (x + y) dx dy + ∫ 1 ∫ 2y 1− 1−y 2 0 y (x + y) dx dy. 2 (x + y) dydx Note that the circular part of the curve is y =− √ 1 − (x − 1) 2 or x = 1 ± √ 1 − y 2 . 11. (a) ∫ π/2 ∫ 1/(sin θ+cos θ) 0 0 r 3 sin θ cos θ dr dθ + ∫ π ∫ 1/(sin θ−cos θ) π/2 0 r 3 sin θ cos θ dr dθ (b) 0 ∫∫ 1 12. R 9 − ( x 2 + y 2) 3/2 dA (a) 1 2 (4π − π) = 3π 2 (b) Since the semicircles satisfy x 2 + y 2 = 1andx 2 + y 2 = 4, we have on x 2 + y 2 = 1, 1 9 ( x 2 + y 2) 3/2 = 1 8 and on x2 + y 2 1 = 4, 9 ( x 2 + y 2) 3/2 = 1. (c) A lower bound is the minimum value times the area, that is, 1 8 · 3π 2 = 3π 16 . An upper bound is the maximum value times the area, that is, 1 · 3π 2 = 3π 2 . 13. (a) { (r, θ) | 0 ≤ r ≤ 1, π 4 ≤ θ ≤ 7π } 4 (b) ∫∫ Pac-Man xdA= ∫ 1 ∫ 7π/4 0 π/4 r 2 cos θ dθ dr = ∫ 1 [ 0 r 2 sin θ ] 7π/4 π/4 dr =−√ 2 ∫ √ 1 0 r2 dr =− 2 3 ∫∫Pac-Man ydA= ∫ 1 ∫ 7π/4 0 π/4 r 2 sin θ dθ dr = ∫ 1 [ 0 −r 2 cos θ ] 7π/4 π/4 dr = 0 14. (a) (b) ∫ 1 ∫ √ y ∫ xy 0 y 3 0 dzdx dy = ∫ 1 ∫ √ 3 x ∫ xy 0 x 2 0 dz dydx 15. x = 2u + v, y = u + 2v 938
CHAPTER 16 SAMPLE EXAM SOLUTIONS (a) (b) The Jacobian is ∂x/∂u ∂x/∂v ∣ ∣∣ ∂y/∂u ∂y/∂v ∣ ∫∫S (3x + 2y) dA = ∫ 1 0 = 3 ∫ 1 0 ∣∣ ∣∣∣ ∣∣∣ ∣ = 2 1 1 2 ∣∣ = 3, so ∫ 1 0 [3 (2u + v) + 2 (u + 2v)] 3 du dv [ 1 3u 2 + 3uv + u 2 + 4uv] dv 0 = 3 ∫ 1 0 (3 + 3v + 1 + 4v) dv = 3 [ 4v + 7 2 v2] 1 0 = 45 2 16. ∫ 9 1 17. (a) ∫ π/2 ∫ π/4 0 π/6 ρ2 sin φ dφ dθ dρ = π 2 = ∫ 9 [ 1 −ρ 2 cos φ ] π/4 [ 13 ρ 3] 9 √ 3 − √ 2 2 π/6 dρ = π 2 = 182 1 3 ∫ √ √ 9 3 − 2 1 ( √3 − √ 2 ) π 2 ρ 2 dρ T maps the unit square in the uv-plane to the unit circle in the xy-plane. (b) The area of S is π. ∫ 1 ∫ 1 ∫ 1 −1 −1 (ax + by) dydx −1 2ax dx 18. f ave = ∫ 1 ∫ 1 −1 −1 1 dydx = = 0 4 939
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16 MULTIPLE INTEGRALS 16.1 DOUBLE I
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SECTION 16.1 DOUBLE INTEGRALS OVER
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GROUP WORK 2, SECTION 16.1 An Odd F
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GROUP WORK 2, SECTION 16.1 An Odd F
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16.2 ITERATED INTEGRALS SUGGESTED T
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GROUP WORK 1, SECTION 16.2 Regional
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GROUP WORK 3, SECTION 16.2 The Shap
Page 15 and 16: SECTION 16.3 DOUBLE INTEGRALS OVER
Page 17 and 18: GROUP WORK 1, SECTION 16.3 TypeIorT
Page 19 and 20: GROUP WORK 3, SECTION 16.3 Bounding
Page 21 and 22: SECTION 16.4 DOUBLE INTEGRALS IN PO
Page 23 and 24: SECTION 16.4 DOUBLE INTEGRALS IN PO
Page 25 and 26: GROUP WORK 2, SECTION 16.4 Fun with
Page 27 and 28: 16.5 APPLICATIONS OF DOUBLE INTEGRA
Page 29 and 30: SECTION 16.5 APPLICATIONS OF DOUBLE
Page 31 and 32: GROUP WORK 3, SECTION 16.5 A Slick
Page 33 and 34: SECTION 16.6 TRIPLE INTEGRALS • S
Page 35 and 36: ( ∫ 1 ∫ √ 4−x 2 0 1 SECTION
Page 37 and 38: GROUP WORK 2, SECTION 16.6 Setting
Page 39 and 40: DISCOVERY PROJECT Volumes of Hypers
Page 41 and 42: SECTION 16.7 TRIPLE INTEGRALS IN CY
Page 43 and 44: GROUP WORK, SECTION 16.7 A Partiall
Page 45 and 46: 16.8 TRIPLE INTEGRALS IN SPHERICAL
Page 47 and 48: SECTION 16.8 TRIPLE INTEGRALS IN SP
Page 49 and 50: GROUP WORK 1, SECTION 16.8 Describe
Page 51 and 52: GROUP WORK 2, SECTION 16.8 Setting
Page 53 and 54: APPLIED PROJECT Roller Derby This p
Page 55 and 56: SECTION 16.9 CHANGE OF VARIABLES IN
Page 57 and 58: SECTION 16.9 CHANGE OF VARIABLES IN
Page 59 and 60: GROUP WORK 2, SECTION 16.9 Transfor
Page 61 and 62: Compute ∫∫∫ R 10 dV and ∫
Page 63 and 64: (b) Evaluate ∫∫ Pac-Man xdAand
Page 65: (b) 3. (a) CHAPTER 16 SAMPLE EXAM S
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16 MULTIPLE INTEGRALS

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