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

1006 Chapter 15 Multiple Integrals
y
D
y=1
y=x
ExamplE 5 Evaluate the iterated integral y 1 0 y1 x sinsy2 d dy dx.
SOLUTION If we try to evaluate the integral as it stands, we are faced with the task
of first evaluating y sinsy 2 d dy. But it’s impossible to do so in finite terms since
y sinsy 2 d dy is not an elementary function. (See the end of Section 7.5.) So we must
change the order of integration. This is accomplished by first expressing the given iterated
integral as a double integral. Using (3) backward, we have
0
FIGURE 15
D as a type I region
y
1
x=0
0
D
x=y
FIGURE 16
D as a type II region
1 x
x
where
y 1
0 y1 x sinsy2 d dy dx − y sinsy 2 d dA
D − hsx, yd | 0 < x < 1, x < y < 1j
We sketch this region D in Figure 15. Then from Figure 16 we see that an alternative
description of D is
D − hsx, yd | 0 < y < 1, 0 < x < yj
This enables us to use (5) to express the double integral as an iterated integral in the
reverse order:
y 1
0 y1 x sinsy2 d dy dx − y sinsy 2 d dA
D
− y 1
− y 1
0 yy 0
0
D
sinsy 2 d dx dy − y 1
0 fx sinsy2 dg x−0
x−y
dy
y sinsy 2 d dy − 2 1 1
2 dg cossy2 0 −
1
2s1 2 cos 1d ■
Properties of Double Integrals
We assume that all of the following integrals exist. For rectangular regions D the first
three properties can be proved in the same manner as in Section 5.2. And then for general
regions the properties follow from Definition 2.
6 y
D
7 y
D
y f f sx, yd 1 tsx, ydg dA − y f sx, yd dA 1 y
y c f sx, yd dA − c y
D
D
y f sx, yd dA
If f sx, yd > tsx, yd for all sx, yd in D, then
D
y tsx, yd dA
where c is a constant
y
D
D¡ D
8 y f sx, yd dA > y tsx, yd dA
D
D
The next property of double integrals is similar to the property of single integrals
given by the equation y b f sxd dx − a yc f sxd dx 1 a yb c
f sxd dx.
If D − D 1 ø D 2 , where D 1 and D 2 don’t overlap except perhaps on their boundaries
(see Figure 17), then
0
FIGURE 17
x
9 yy
D
f sx, yd dA − yy
D 1
f sx, yd dA 1 yy
D 2
f sx, yd dA
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