James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

1010 Chapter 15 Multiple Integrals
65–69 Use geometry or symmetry, or both, to evaluate the
double integral.
65. y y sx 1 2d dA,
D
D − hsx, yd | 0 < y < s9 2 x 2 j
66. y y sR 2 2 x 2 2 y 2 dA,
D
D is the disk with center the origin and radius R
67. y y s2x 1 3yd dA,
D
D is the rectangle 0 < x < a, 0 < y < b
CAS
68. y y s2 1 x 2 y 3 2 y 2 sin xd dA,
D
D − hsx, yd | | x | 1 | y | < 1j
69. y y sax 3 1 by 3 1 sa 2 2 x 2 d dA,
D
D − f2a, ag 3 f2b, bg
70. Graph the solid bounded by the plane x 1 y 1 z − 1 and
the paraboloid z − 4 2 x 2 2 y 2 and find its exact volume.
(Use your CAS to do the graphing, to find the equations of
the boundary curves of the region of integration, and to
evaluate the double integral.)
Suppose that we want to evaluate a double integral yy R
f sx, yd dA, where R is one of the
regions shown in Figure 1. In either case the description of R in terms of rectangular
coordinates is rather complicated, but R is easily described using polar coordinates.
y
y
≈+¥=1
≈+¥=4
R
0
x
R
0
≈+¥=1
x
FIGURE 1
(a) R=s(r, ¨) | 0¯r¯1, 0¯¨¯2πd
(b) R=s(r, ¨) | 1¯r¯2, 0¯¨¯πd
y
P(r, ¨)=P(x, y)
Recall from Figure 2 that the polar coordinates sr, d of a point are related to the rectangular
coordinates sx, yd by the equations
r
y
r 2 − x 2 1 y 2 x − r cos y − r sin
¨
O
FIGURE 2
x
x
(See Section 10.3.)
The regions in Figure 1 are special cases of a polar rectangle
R − hsr, d |
a < r < b, < < j
which is shown in Figure 3. In order to compute the double integral yy R
f sx, yd dA, where
R is a polar rectangle, we divide the interval fa, bg into m subintervals fr i21 , r i g of equal
width Dr − sb 2 adym and we divide the interval f, g into n subintervals f j21 , j g
of equal width D − s 2 dyn. Then the circles r − r i and the rays − j divide the
polar rectangle R into the small polar rectangles R ij shown in Figure 4.
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