Page 239 - Calculus Demystified

Applications of the Integral
CHAPTER 8
226
2
rotation, it will generate a disk of radius 2−x, and hence area A(x) = π(2−x) .
Thus the volume generated over the segment 1 ≤ x ≤ 2is
2

2
V 2 = π(2 − x) dx.
1
In summary, the total volume of our solid of revolution is
V = V 1 + V 2
! "
3 1 3 2

x −(2 − x)
= π +
3 3
0 1

1 1
= π − 0 + −0 − −
3 3
2π
= .
3
EXAMPLE 8.6
2
The portion of the curve y = x between x = 1 and x = 4 isrotated about
the x-axis (Fig. 8.16). What volume does the resulting surface enclose?

Fig. 8.16

SOLUTION
2
2
At position x, the curve is x units above the x-axis. The point (x, x ), under
2
rotation, therefore generates a circle of radius x .The disk that the circle bounds