Page 258 - Calculus Demystified

Applications of the Integral
CHAPTER 8

x 245
Fig. 8.33

We conclude that the integral
b
2 1/2

2π f(x)(1 +[f (x)] ) dx
a
represents the area of the surface of revolution.
EXAMPLE 8.23
3
Let f(x) = 2x . For 1 ≤ x ≤ 2 we rotate the graph of f about the x-axis.
Calculate the resulting surface area.
SOLUTION
According to our definition, the area is
2

2 1/2

2π f(x)(1 +[f (x)] ) dx
1

2
3 2 2 1/2
= 2π 2x (1 +[6x ] ) dx
1
2
π 3 4 1/2 3
= (1 + 36x ) (144x )dx.
54 1 2
4
This integral is easily calculated using the u-substitution u = 36x ,du =
3
144x dx. With this substitution the limits of integration become 36 and 576;
the area is thus equal to
576
π 3 1/2 π 576
(1 + u) du = (1 + u) 3/2
54 36 2 54 36
π
= [(577) 3/2 − (37) 3/2 ]
54
≈ 793.24866.