Monday Math 4

Coordinate transformations and double integration

Suppose we have the double integral:

Note that we have a pole on the edge of our region, at .

I will demonstrate two methods for doing this, the first more direct, the second much easier to do, but requiring a bit more insight.

1. Direct method:

If we try to do the integral as ordered (x first), we see our x integral is an elliptic integral. So we reverse the order of integration (Fubini’s Theorem):

Now consider the inner integral: this is an integral of the form
, where .
Using the change of variables , our integral becomes:

One can use successive integration by parts to reduce the power of the numerator
(see #11 here [or just use #3 in the same table]), and obtain



So our original integral becomes:

Using the fact that , we can see that

so:

Using the substitution , we find

Note that while is technically undefined at w=0, we see the limit as w goes to 0 exists (the singularity is removable) and, in fact,
.
Thus

2. “Polar transformation” method:


Instead we use the substitution . Then our integral becomes:

Treating r and θ as the standard polar coordinates, our region of integration is the upper half of the unit disk. Using the fact that along with the conversion factors between polar and cartesian coordinates (and Fubini’s Theorem):

The inner integral is simple:


and so:

This gives us the same answer, but with much less work.

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