We compute surface area with double integrals.

*arc length over an interval*is

*surface area over a region.*Consider the surface over some region in the -plane. To compute the surface area, first consider the surface area of a ‘‘patch’’ of the surface, determined by and below:

hence

Summing these ‘‘patches’’ together leads to a double integral.

*defining*the concept of the ‘‘area of a surface.’’ While we already have a notion of the area of a region in the

*plane*, we did not yet have a solid grasp of what ‘‘the area of a surface in

*space*’’ means. Double integrals make this notion of surface area precise.

Let’s train ourselves to use our new tools by computing the surface areas of known surfaces. We start with a triangle.

It’s common knowledge that the surface area of a sphere of radius is . While there are several ways to confirm this formula, we will use a double integral. Our computation will involves using our formula for surface area, polar coordinates, and improper integrals!

Since our function only defines the top upper hemisphere of the sphere, we double our surface area result to get the total area:

Since the region that we are integrating over is the circle, we are likely to have greater success with our integration by converting to polar coordinates. Using the substitutions

we write

Simplifying, we now write: We evaluate this integral by making the substitution

and paying close attention to the limits of integration, we now have However, since the integrand of the inner integral is not defined at , the inner integral is in fact an improper integral. Let’s carefully evaluate the inner integral. Write with me:

So we may now write

Thus confirming our previous formula.

Let’s find the surface area of a general region now.

Thus the surface area is described by the double integral As with integrals describing arc length, double integrals describing surface area are in general hard to evaluate directly because of the square-root. This particular integral can be easily evaluated, though, with judicious choice of our order of integration. Integrating with order requires us to evaluate This is not too easy. The computation involves integration by parts and .

However, integrating in the order has as its first integral and this is easy to evaluate, So we proceed with the order . In this case, the limits of integration are already given in the statement of the problem. Write with me:

Setting, you can use substitution to find:

Note that the surface has a much greater area than the region in the -plane. This is because the -values of the surface change dramatically over .

In practice, technology helps greatly in the evaluation of such integrals. High
powered computer algebra systems can compute integrals that are difficult, or at
least time consuming, by hand, and can at the least produce very accurate
approximations with numerical methods. In general, just knowing *how* to set up the
proper integrals brings one very close to being able to compute the needed value.
Most of the work is actually done in just describing the region in terms of polar or
rectangular coordinates. Once this is done, technology can usually provide a good
answer.