1. Suppose we consider some function on the domain . For example:
We revolve this about the axis by 360 degrees. So this looks like
Question: what is the area occupied by this surface?
2. The trick is to write the area as the sum of the area of smaller strips. So for example, one such strip might be:
What is its area? We approximate this as a cylinder with radius
and height is the segment’s arclength
Thus the cylinder’s surface area is
We have many cylinders, and we add them all up. This sum is a very natural integral:
This can be somewhat simplified as
Example 1. Consider the curve on the domain . What is the area for its surface of revolution?
Solution. We see the arc-length differential for this function is
The differential area for the surface is
Integrating gives us the area
But now we have to perform this integral!
We let and . Thus our integral (7) becomes
This can be solved directly as
Now we unroll our substitution, transforming (9) into
That concludes our example.
Example 2 [Variation of Math.SE]. Consider the curve
What is the area for the surface of revolution about the axis?
Solution. We have the surface’s area be described by the integral
We need to express the integrand as an explicit function of . So we begin with
Great, so now we can consider its square
Now we can consider the arc-length portion of the integrand
Great, what now?
We can set up the integral in Equation (12) as
which we will consider as the sum of two integrals:
Now we have to evaluate each integral in turn.
We use substitution for (17a), take
(18) and .
Then we have
This is a trivial integral to evaluate
We unroll our substitution, letting and we get
Great this gives us , but what about our other integral in Equation (17b)?
So we want to evaluate the integral
Lets use the substitution
(22) and .
Note this means that . So our integral is
Lets break this up into more manageable problems.
We see that we have one term be
Using integration by parts we have, letting and , that
Thus integration by parts yields
But these are trivial integrals, we find
This is one term for Equation (23). The other is
Combining Equations (26) and (27) gives us
Now we must unroll our substitution, again. So we plug in , giving us
We then evaluate this to find
This is the other half of our problem.
Now Equation (21) gave us and Equation (30) gave us . We can now combine these together to get the area
That concludes our problem.
Exercise 1. Let . Consider the surface of revolution about the axis. Calculate the area for the surface of revolution restricted along the interval . [from Math.SE].