The problem of finding the surface area of a solid of revolution is closely related to that of finding the length of a graph. First consider the following picture of the frustum of a cone in which it is desired to find the lateral surface area. In this picture, the frustum of the cone is the left part which has an l next to it and the lateral surface area is this part of the area of the cone.
To do this, imagine painting the sides and rolling the shape on the floor for exactly one revolution. The wet paint would make the following shape.
What would be the area of this wet paint? Its area would be the difference between the areas of the two sectors shown, one having radius l_{1} and the other having radius l + l_{1}. Both of these have the same central angle equal to

Therefore, Theorem 2.3.13, this area is

The view from the side is
and so by similar triangles, l_{1} = lr∕

Now consider a function f, defined on an interval
Let A

where ≈ denotes that these are close to being equal and the approximation gets increasingly good as h → 0. Therefore, rewriting this a little yields

Therefore, taking the limit as h → 0, and using A

What would happen if you revolved about the y axis? I will leave it to you to verify this would lead to the initial value problem

As before, this results in the following simple procedure for finding the surface area of a surface of revolution.
Procedure 8.1.4 To find the surface area of a surface obtained by revolving the graph of y = f

Similarly, to get the area of the graph rotated about the y axis, compute

Example 8.1.5 Find the surface area of the surface obtained by revolving the function y = r for x ∈
Using the above initial value problem, solve

The solution is A
Example 8.1.6 Find the surface area of a sphere of radius r.
Here the function involved is f

and so, by the procedure described above, the surface area is
