Page 240 - Calculus Demystified

Applications of the Integral
CHAPTER 8
2 2
has area A(x) = π · (x ) . Thus the described volume is 227
4 5 4

4 x 1023π
V = π · x dx = π · = .
1 5 1 5
Math Note: The reasoning we have used in the last two examples shows this:
If the curve y = f(x), a ≤ x ≤ b, is rotated about the x-axis then the volume
enclosed by the resulting surface is
b
2
V = π ·[f(x)] dx.
a
You Try It: Calculate the volume enclosed by the surface obtained by rotating
√
the curve y = x + 1, 4 ≤ x ≤ 9, about the x-axis.
EXAMPLE 8.7
3
The curve y = x ,0 ≤ x ≤ 3, isrotated about the y-axis. What volume
does the resulting surface enclose?
SOLUTION
It is convenient in this problem to treat y as the independent variable and x as
the dependent variable. So we write the curve as x = y 1/3 . Then, at position y,
the curve is distance y 1/3 from the axis so the disk generated under rotation will
] .
have radius y 1/3 (Fig. 8.17). Thus the disk will have area A(y) = π ·[y 1/3 2
Also, since x ranges from 0 to 3 we see that y ranges from 0 to 27. As a result,

Fig. 8.17