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

Section 15.2 Double Integrals over General Regions 1003
Using the same methods that were used in establishing (3), we can show that
5 y f sx, yd dA − y d syd 2
c
yh f sx, yd dx dy
h 1
syd
D
where D is a type II region given by Equation 4.
y
(_1, 2)
y=1+≈
(1, 2)
Example 1 Evaluate yy D
sx 1 2yd dA, where D is the region bounded by the
parabolas y − 2x 2 and y − 1 1 x 2 .
SOLUTIon The parabolas intersect when 2x 2 − 1 1 x 2 , that is, x 2 − 1, so x − 61. We
note that the region D, sketched in Figure 8, is a type I region but not a type II region
and we can write
D − hsx, yd | 21 < x < 1, 2x 2 < y < 1 1 x 2 j
D
y=2≈
Since the lower boundary is y − 2x 2 and the upper boundary is y − 1 1 x 2 , Equation
3 gives
_1
FIGURE 8
1
x
y sx 1 2yd dA − y 1 sx 1 2yd dy dx
21 y11x2 2x 2
D
− y 1 fxy 1 y 2 y−11x
g 2 y−2x 2 dx
21
− y 1 fxs1 1 x 2 d 1 s1 1 x 2 d 2 2 xs2x 2 d 2 s2x 2 d 2 g dx
21
− y 1 s23x 4 2 x 3 1 2x 2 1 x 1 1d dx
21
− 23 x 5
5 2 x 4
4 1 2 x 3
3 1 x 2 1
2 1 x G21
− 32
15
■
y
y=2x
(2, 4)
y=≈
Note When we set up a double integral as in Example 1, it is essential to draw a
diagram. Often it is helpful to draw a vertical arrow as in Figure 8. Then the limits of
integration for the inner integral can be read from the diagram as follows: The arrow
starts at the lower boundary y − t 1 sxd, which gives the lower limit in the integral, and
the arrow ends at the upper boundary y − t 2 sxd, which gives the upper limit of integration.
For a type II region the arrow is drawn horizontally from the left boundary to the
right boundary.
D
0 x
1 2
FIGURE 9
D as a type I region
Example 2 Find the volume of the solid that lies under the paraboloid z − x 2 1 y 2 and
above the region D in the xy-plane bounded by the line y − 2x and the parabola y − x 2 .
SOLUTION 1 From Figure 9 we see that D is a type I region and
D − hsx, yd | 0 < x < 2, x 2 < y < 2xj
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