VOLUME D 179
Fig. 22-18
Fig. 22-19
22.30
Let 9? be the region bounded by y = x
3 , x = l, x = 2, and y-x-\. Find the volume of the solid
generated when SI is revolved about the x-axis.
y = x3 lies above y = x-l for l<;t<2 (see Fig. 22-19). So, we can use the circular ring formula:
37477/21.
Fig. 22-20
Solving y = 2* - 4 and y
2 = 4x simultaneously, we obtain y
2 — 2y — 8 = 0, y = 4 or y = —2.
Thus, the curves intersect at (4,4) and (1, -2), as shown in Fig. 22-18. We integrate along the y-axis, using
the circular ring formula:
22.31
Same as Problem 22.30, but revolving about the y-axis.
We use the difference of cylindrical shells: V = 2ir J
2 x[x
3 - (x - 1)] dx = 2ir J\ (x
4 - x +x)dx =
2TT(\X5 - \X* + kx2) ]2 = 277[(f - 1 + 2) - (I - I + I)] = 1617T/15.
22.32
Let 5? be the region bounded by the curves y = x
2 - 4x + 6 and y = x + 2. Find the volume of the solid
generated when 3ft is rotated about the ^-axis.
Solving y = x
2 -4x + 6 and y = x + 2, we obtain x
2 -5x + 4 = Q, x = 4 or x=l. So, the
curves meet at (4,6) and (1,3). Note that y = x
2 - 4x + 6 = (x -2)
2 + 2. Hence, the latter curve is a
parabola with vertex (2, 2) (see Fig. 22-20). We use the circular ring formula: V= TT J7 {(x + 2)
2 - [(x - 2)
2 +
2]
2 }dx= 7rJ 1
4 [(^ + 2)
2 -(^-2)
4 -4(x-2)
2 -4]rfA: = 7r(K^ + 2)
3 ~H^-2)
5 -!^-2)
3 -4^)]: = 7r[(72f - f - 16) - (9 + L + | - 4)1 = 1627T/5.
V = TTtf((XX-l)2]dX = TTtf(x6-(X-l)2]dx = irO^-H*-!)3)]? = ^[(¥-~i)-0-0)] =