## An Example of an Iterated Integral

My description of how to evaluate a multiple integral over some region other than an -dimensional interval by using iterated integrals might not have been the clearest, so I’m hoping an example will help illustrate what I mean. Let’s calculate the Jordan content of a three-dimensional sphere of radius , centered at the origin. This will also help cement the fact that Jordan content is closely related to what we mean by “volume”.

So, we have our sphere

We can put it inside the interval , so we know that the Jordan content is

Now we want to peel off an integral from the inside. Let’s first integrate over the variable . Projecting the sphere onto the plane we’re left with the circle

and we need to write as the region between the graphs of two functions on this projection. And indeed, we can write

Thus we can write

Next we’ll integrate with respect to . Projecting onto the line we find . And then we find

which lets us peel off another integral from the inside

Finally, we can start evaluating these integrals. The innermost integral goes immediately, using the fundamental theorem of calculus. An antiderivative of with respect to is , and so we find

So we put this into our integral, already in progress:

Here’s where things start to get difficult. I’ll just tell you (and you can verify) that an antiderivative of with respect to is

So the fundamental theorem of calculus tells us

Technically, this is an improper integral (which we haven’t really discussed), so we need to take some limits. The denominators are limits as some dummy variable approaches zero from above. Thus we continue evaluating our integral

This one is easy: an antiderivative of is , and so

And this is exactly the well-known formula for the volume of a sphere of radius . There are easier ways to get at this formula, of course, but this one manages to illustrate the technique of iterated integrals with variable limits of integration on the inner integrals.

It looks like you skipped a step, between “The denominators 0+ are limits as some dummy variable approaches zero from above.” and “Thus we continue evaluating our integral”.

It is likely that we are supposed to know that arctan(∞) is π/2, but since arctan was introduced as a “I’ll just tell you..”, a statement to that effect would have been nice.

Comment by Blaise Pascal | January 4, 2010 |

I didn’t mean “I’ll just tell you” to introduce the notion of arctan at all. It was meant to skip the derivation of the antiderivative rather than go through all the mess of trigonometric substitutions. This is perfectly valid, since all we need for the FToC is

anantiderivative, wherever it comes from.Comment by John Armstrong | January 4, 2010 |

[…] there are other places that so-called “improper integrals” come up, like the example I worked through the other day. In this case, the fundamental theorem of calculus runs into trouble at the endpoints […]

Pingback by Improper Integrals II « The Unapologetic Mathematician | January 15, 2010 |