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.