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

1034 Chapter 15 Multiple Integrals
y
1
0
z
1
y=≈
D¡
z=y
1
x
ExamplE 4 Express the iterated integral y 1 0 yx2 0 yy 0
f sx, y, zd dz dy dx as a triple integral
and then rewrite it as an iterated integral in a different order, integrating first with respect
to x, then z, and then y.
SOLUTION We can write
y 1
0 yx2 0 yy 0
f sx, y, zd dz dy dx − yyy f sx, y, zd dV
E
where E − hsx, y, zd | 0 < x < 1, 0 < y < x 2 , 0 < z < yj. This description of E
enables us to write projections onto the three coordinate planes as follows:
D
on the xy-plane:
D 1 − hsx, yd | 0 < x < 1, 0 < y < x 2 j
0
1
y
− 5sx, yd | 0 < y < 1, sy < x < 1 6
z
1
on the yz-plane:
D 2 − hsy, zd | 0 < y < 1, 0 < z < yj
z=≈
on the xz-plane:
D 3 − hsx, zd | 0 < x < 1, 0 < z < x 2 j
0
FIGURE 12
Projections of E
z=y
z
0
D£
1
x
From the resulting sketches of the projections in Figure 12 we sketch the solid E in Figure
13. We see that it is the solid enclosed by the planes z − 0, x − 1, y − z and the
parabolic cylinder y − x 2 sor x − sy d.
If we integrate first with respect to x, then z, and then y, we use an alternate description
of E:
E − 5sx, y, zd | 0 < y < 1, 0 < z < y, sy < x < 16
Thus
y y f sx, y, zd dV − y 1
0 yy 0 y1 sy
E
f sx, y, zd dx dz dy ■
1
x
x=1
FIGURE 13
The solid E
1
y=≈
y
Applications of Triple Integrals
Recall that if f sxd > 0, then the single integral y b a
f sxd dx represents the area under the
curve y − f sxd from a to b, and if f sx, yd > 0, then the double integral yy D
f sx, yd dA represents
the volume under the surface z − f sx, yd and above D. The corresponding interpretation
of a triple integral yyy E
f sx, y, zd dV, where f sx, y, zd > 0, is not very useful
because it would be the “hypervolume” of a four-dimensional object and, of course, that
is very difficult to visualize. (Remember that E is just the domain of the function f ; the
graph of f lies in four-dimensional space.) Nonetheless, the triple integral yyy E
f sx, y, zd dV
can be interpreted in different ways in different physical situations, depending on the
phys ical interpretations of x, y, z, and f sx, y, zd.
Let’s begin with the special case where f sx, y, zd − 1 for all points in E. Then the
triple integral does represent the volume of E:
12 VsEd − y y dV
E
For example, you can see this in the case of a type 1 region by putting f sx, y, zd − 1 in
Formula 6:
y y 1 dV − y Fy
E
D
u2sx, yd
u1sx, yd
dzG dA − y fu 2 sx, yd 2 u 1 sx, ydg dA
D
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