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

Section 15.6 Triple Integrals 1033
SOLUTION The solid E is shown in Figure 9. If we regard it as a type 1 region, then we
need to consider its projection D 1 onto the xy-plane, which is the parabolic region in
Figure 10. (The trace of y − x 2 1 z 2 in the plane z − 0 is the parabola y − x 2 .)
z
y
y=≈+z@
y=4
TEC Visual 15.6 illustrates how solid
regions (including the one in Figure 9)
project onto coordinate planes.
x
0
4
E
y
D¡
0
y=≈
x
FIGURE 9
R e gion of int e gratio n
FIGURE 10
Projection onto xy-plane
From y − x 2 1 z 2 we obtain z − 6sy 2 x 2 , so the lower boundary surface of E is
z − 2sy 2 x 2 and the upper surface is z − sy 2 x 2 . Therefore the description of E
as a type 1 region is
E − hsx, y, zd | 22 < x < 2, x 2 < y < 4, 2sy 2 x 2 < z < sy 2 x 2 j
and so we obtain
D£
FIGURE 11
Projection onto xz-plane
z
≈+z@=4
0
_2 2
The most difficult step in evaluating
a triple integral is setting up an expression
for the region of integration (such
as Equation 9 in Example 2). Remem -
ber that the limits of integra tion in
the inner integral contain at most two
variables, the limits of integration in
the middle integral contain at most one
variable, and the limits of integration in
the outer integral must be constants.
x
y y sx 2 1 z 2 dV − y 2 y sy2x2
22 y4 sx 2 1 z 2 dz dy dx
x 2 2sy2x 2
E
Although this expression is correct, it is extremely difficult to evaluate. So let’s
instead consider E as a type 3 region. As such, its projection D 3 onto the xz-plane is the
disk x 2 1 z 2 < 4 shown in Figure 11.
Then the left boundary of E is the paraboloid y − x 2 1 z 2 and the right boundary is
the plane y − 4, so taking u 1 sx, zd − x 2 1 z 2 and u 2 sx, zd − 4 in Equation 11, we have
yyy sx 2 1 z 2 dV − y Fy 4
E
D3
Although this integral could be written as
sx 2 1 z
x 2 1z
dyG 2 dA − y s4 2 x 2 2 z 2 dsx 2 1 z 2 dA
2
y 2 22 ys42x2 2s42x 2 s4 2 x 2 2 z 2 d sx 2 1 z 2 dz dx
it’s easier to convert to polar coordinates in the xz-plane: x − r cos , z − r sin . This
gives
y y sx 2 1 z 2 dV − y
E
D3
D3
y s4 2 x 2 2 z 2 dsx 2 1 z 2 dA
− y 2
0
y 2
0
− 2F 4r 3
3 2 r 5
s4 2 r 2 dr r dr d − y 2
d y 2
s4r 2 2 r 4 d dr
2
5G0
− 128
15
0
0
■
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