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

Section 6.2 Volumes 445
that its height is s3 y − s3s1 2 x 2 . The cross-sectional area is therefore
Asxd − 1 2 ? 2s1 2 x 2 ? s3 s1 2 x 2 − s3 s1 2 x 2 d
and the volume of the solid is
V − y 1 21
Asxd dx − y 1 21 s3 s1 2 x 2 d dx
− 2 y 1
0 s3 s1 2 x 2 d dx − 2s3Fx 2 x 3
1
3G0
− 4s3
3
n
Example 8 Find the volume of a pyramid whose base is a square with side L and
whose height is h.
SOLUtion We place the origin O at the vertex of the pyramid and the x-axis along its
central axis as in Figure 14. Any plane P x that passes through x and is perpendicular to
the x-axis intersects the pyramid in a square with side of length s, say. We can express s
in terms of x by observing from the similar triangles in Figure 15 that
x
h − sy2
Ly2 − s L
and so s − Lxyh. [Another method is to observe that the line OP has slope Lys2hd and
so its equation is y − Lxys2hd.] Therefore the cross-sectional area is
Asxd − s 2 − L2
h 2 x 2
y
y
P
O
x h s L
x
O x
h
x
FIGURE 14
FIGURE 15
y
h
The pyramid lies between x − 0 and x − h, so its volume is
V − y h
Asxd dx − y h
0
0
L 2
h 2 x 2 dx − L2
h 2 x 3
h
3G0
− L2 h
3
n
y
0
FIGURE 16
x
Note We didn’t need to place the vertex of the pyramid at the origin in Example 8.
We did so merely to make the equations simple. If, instead, we had placed the center of
the base at the origin and the vertex on the positive y-axis, as in Figure 16, you can verify
that we would have obtained the integral
V − y h
0
L 2
h sh 2 2 yd2 dy − L2 h
3
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