## Surface Area

This module deals with the surface area of solids of revolution. Consider the portion of a curve for revolved about a horizontal axis to create a solid. In earlier modules the goal was to find the volume of such a solid, but now the focus is on finding the surface area. As always, the method will be to find the surface area element and integrate it. The surface area element which works well is the thin band shown here:

## Surface area of a cone

The first step towards finding the surface area element is to find the lateral surface area of a more simple solid: the cone. Consider a cone whose base has radius and lateral height (the lateral height is the distance from the tip of the cone to a point on the circumference of the base; see the left diagram below).

To find the area, consider cutting the cone along the straight dotted line from base circumference to tip and unrolling the cone. The result is a portion of a circle whose radius is , as shown on the right in the diagram above. Note that the circumference of the base of the cone, , becomes the length of arc of the unrolled cone. This means that the unrolled cone is a fraction of the full circle of radius , and that fraction is (the ratio of the circumference of the partial circle to the circumference of the whole circle). Thus the surface area of the cone is .

The surface area of a cone can be used to find the area of a frustum of a cone whose top radius is , bottom radius is , and lateral height (as in the below diagram). The area of this frustum is . Expressed another way, the area is , where is the average of the two radii of the frustum.

## Surface area element

Now, the surface area element can be found. When the curve is partitioned into sufficiently small pieces, the surface area element is just the area of the frustum formed by rotating the arclength element about the axis (see the diagram):

Thus, the surface area element is , where is the distance from the curve to the axis of rotation, and is the arclength element (i.e. ). In the (common) case where the axis of rotation is the -axis, one finds that .

Thus, the surface area resulting from revolving the curve for about the -axis is given by

**Example**

Consider the sphere of radius . If the sphere is cut into slices of equal width, which slice has the most surface area?

If we center the sphere at the origin, we can think of the sphere as the surface of revolution obtained by revolving the curve

about the -axis. First, we compute the arclength element:

Plugging this into the surface area element, we find

Note that this is independent of ! This means that every slice of the sphere has equal surface area.

For example, if we were to slice the sphere into four slices of equal thickness, then a middle slice goes from to , and its surface area

The end-cap slice, on the other hand, goes from to , so its surface area is

So we see that the pieces have equal surface area.

**Example**

Consider the surface generated by revolving the curve for about the -axis.

Find the values of for which the surface has finite surface area. Then find the values of for which the solid of revolution has finite volume.

The surface area, in terms of , is

Unfortunately, this integral is not computable using standard methods, but we can use a binomial expansion to determine the leading order term of the integrand, which will tell us whether the integral converges or not. We see that

Therefore, the leading order term in this integral is , which we know converges for and diverges for (from our study of p-integrals). So this surface of revolution has finite area if and only if .

Turning to the volume of this solid, it is best to use slices perpendicular to the -axis, which leads to discs whose radius is :

The volume element is therefore

Thus, the volume is

We know this is convergent if , i.e. . So the volume of the solid is finite if .

This leads to the surprising fact that for

the volume of the solid is finite, but the surface area is infinite.

## Rotations about the y-axis

Suppose we want to know the surface area which results from revolving the curve

about the -axis. There are two main ways one can go about finding this surface area:

- Express everything as a function of (including range of inputs), and then use the above formula but with the roles of and switched.
- Leave things in terms of , but adjust the formula slightly.

The first method expresses the curve as

where and . Then express the surface area element as

Putting it together, the surface area can be expressed as

Again, this is really just a reuse of the original formula, with the roles of and flipped.

The second method is sometimes simpler to apply because it involves less algebra. The main observation to make is that the radius in the surface area element is simply when the curve is revolved around the -axis:

So the surface area element can be written

This integral is with respect to , and so it should be integrated over the original range of :

**Example**

Compute the surface area of the surface resulting from revolving the curve

about the -axis:

Using the first method requires some algebra. The curve becomes

So the area element is

So the surface area is

Using the second method, we have

So the surface area is

So we get the answer with (perhaps) slightly less algebra involved.

## EXERCISES

- Compute the surface area resulting from revolving the curve , about the -axis.