16 MULTIPLE INTEGRALS

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
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