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

440 Chapter 6 Applications of Integration
y
y − sr 2 2 x 2 . (See Figure 4.) So the cross-sectional area is
r
y
Asxd − y 2 − sr 2 2 x 2 d
Using the definition of volume with a − 2r and b − r, we have
_r
0
r
x
x
V − y r Asxd dx − y r sr 2 2 x 2 d dx
2r
2r
FIGURE 4
− 2 y r
sr 2 2 x 2 d dx
0
− 2Fr 2 x 2 x 3
r
3G0
− 2Sr 3 2
3D
r 3
(The integrand is even.)
− 4 3 r 3 n
TEC Visual 6.2A shows an animation
of Figure 5.
Figure 5 illustrates the definition of volume when the solid is a sphere with radius
r − 1. From the result of Example 1, we know that the volume of the sphere is 4 3 ,
which is approximately 4.18879. Here the slabs are circular cylinders, or disks, and the
three parts of Figure 5 show the geometric interpretations of the Riemann sums
o n
Asx i d Dx − o n
s1 2 2 x i2 d Dx
i−1
i−1
when n − 5, 10, and 20 if we choose the sample points x i * to be the midpoints x i . Notice
that as we increase the number of approximating cylinders, the corresponding Riemann
sums become closer to the true volume.
(a) Using 5 disks, VÅ4.2726 (b) Using 10 disks, VÅ4.2097 (c) Using 20 disks, VÅ4.1940
FIGURE 5
Approximating the volume
of a sphere with radius 1
Example 2 Find the volume of the solid obtained by rotating about the x-axis the
region under the curve y − sx from 0 to 1. Illustrate the definition of volume by
sketching a typical approximating cylinder.
SOLUtion The region is shown in Figure 6(a). If we rotate about the x-axis, we get
the solid shown in Figure 6(b). When we slice through the point x, we get a disk with
radius sx . The area of this cross-section is
Asxd − (sx ) 2 − x
and the volume of the approximating cylinder (a disk with thickness Dx) is
Asxd Dx − x Dx
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