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

Section 15.2 Double Integrals over General Regions 1001
For single integrals, the region over which we integrate is always an interval. But for
double integrals, we want to be able to integrate a function f not just over rectangles but
also over regions D of more general shape, such as the one illustrated in Figure 1. We suppose
that D is a bounded region, which means that D can be enclosed in a rectangular
region R as in Figure 2. Then we define a new function F with domain R by
1 Fsx, yd −H 0
f sx, yd if
if
sx, yd is in D
sx, yd is in R but not in D
y
y
D
R
D
0
x
0 x
FIGURE 1
FIGURE 2
If F is integrable over R, then we define the double integral of f over D by
2 y
D
y f sx, yd dA − y
R
y Fsx, yd dA where F is given by Equation 1
Definition 2 makes sense because R is a rectangle and so yy R
Fsx, yd dA has been previously
defined in Section 15.1. The procedure that we have used is reasonable because the
values of Fsx, yd are 0 when sx, yd lies outside D and so they contribute nothing to
the integral. This means that it doesn’t matter what rectangle R we use as long as it contains
D.
In the case where f sx, yd > 0, we can still interpret yy D
f sx, yd dA as the volume of the
solid that lies above D and under the surface z − f sx, yd (the graph of f ). You can see that
this is reasonable by comparing the graphs of f and F in Figures 3 and 4 and remembering
that yy R
Fsx, yd dA is the volume under the graph of F.
z
z
graph of f
graph of F
0
0
x
D
y
x
D
y
FIGURE 3
FIGURE 4
Figure 4 also shows that F is likely to have discontinuities at the boundary points
of D. Nonetheless, if f is continuous on D and the boundary curve of D is “well behaved”
(in a sense outside the scope of this book), then it can be shown that yy R
Fsx, yd dA exists
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