Page 256 - Calculus Demystified

Applications of the Integral
CHAPTER 8
EXAMPLE 8.22 243
2
Calculate the length of that portion of the graph of the curve 16x = 9y 3
between the points (0, 0) and (6, 4).
SOLUTION
We express the curve as
3 3/2
x = y , 0 ≤ y ≤ 4.
4
9 1/2
Then dx/dy = y . Now, reversing the roles of x and y in (.), we find that
8
the requested length is

4 # 4
1 +[(9/8)y 1/2 2 1 + (81/64)y dy.
] dy =
0 0
This integral is easily evaluated and we see that it has value [2·(97) 3/2 −128]/
243.

Notice that the last example would have been considerably more difficult (the
integral would have been harder to evaluate) had we expressed the curve in the
form y = f(x).

You Try It: Write the integral that represents the length of a semi-circle and
evaluate it.

8.5.2 SURFACE AREA

Let f(x) be a non-negative function on the interval [a, b]. Imagine rotating the
graph of f about the x-axis. This procedure will generate a surface of revolution,
as shown in Fig. 8.31. We will develop a procedure for determining the area of such
a surface.

y = f (x)

a b

Fig. 8.31