Iterated Integrals V
Our iterated integrals worked out nicely over -dimensional intervals because these regions are simple products of one-dimensional intervals. When working over more general sets, though, it’s not so nice. Still, we can often do almost as well.
Unlike our earlier approach, though, we won’t peel off an integral from the outside first, but from an inside. That is, instead of writing
Underlying all of these equations is the assumption that each of the integrals we write down exists, or at least the inner ones don’t fail to exist often enough to cause the outer ones to also fail to exist.
Okay, so let’s say that we’ve got some bounded region contained in an -dimensional interval . Just like before, we project into the plane with to get the -dimensional region . This will be contained in the projection of . Again, we write for the remaining variables.
Here’s where our restriction comes in: assume that is contained between the graphs of two integrable functions on . That is, we want
where and are two integrable functions defined on .
Now we can try to evaluate the integral
We first replace by because is identically zero wherever is outside of , and thus so will the whole inner integral. Then we replace the limits of integration for the inner integral by and because for any fixed the function is zero for below the former and above the latter.
If we’re lucky, itself can be written in a similar form, as the region between the graphs of two integrable functions in another coordinate direction, and so on. Each inner integral’s integrand depends on the remaining variables of later integrals, as before. But now the bounds of integration can also depend on later variables, which adds new complication. However, once a given variable has been integrated away, nothing further out ever depends on it again, and so this whole procedure is still well-defined.