INTEGRATION in several variables, POLAR COORDINATES


The Astrodome in Houston might be modelled mathematically as the region below a sphere
$x^2+y^2+z^2  =  R^2$
and above a circular disk $D$. It would be natural for several reasons, both practical (air-conditioning) and financial (seating arrangements, cost etc.), to ask for its volume
$\displaystyle{\int\int_D\, \sqrt{R^2 - x^2 - y^2}\, dx dy.}$
And mathematically, we should be able to provide the answer by evaluating the double integral. Right now we could do that with quite a bit of effort, but the fact that everything in sight is rotationally symmetric suggests that the effort would be considerably less if only we could change to polar coordinates. For functions of one variable simplification of an integral was what the method of substitution was all about. 

    Let's see why and how we'd change variables $(x,\, y) \, \to \, (r,\,\theta)$  from Cartesian to polar coordinates:

   Suppose $D$ is the semicircle shown to the right and that
$ I  =  \displaystyle{\int}\displaystyle{ \int_D}\, f(x, y)\,dxdy$
for some unspecified $f$.

In Cartesian coordinates,
$ I  =  \displaystyle{\int_{-2}^2}\left(\displaystyle{\int_0^{\sqrt{4-x^2}}}\, f(x, y)\, dy\right) dx,$
which may or may not be relatively easy to evaluate. But in polar coordinates,
$ I  =  \displaystyle{\int_0^2 }\left(\displaystyle{\int_0^{\pi}}\,f(r\cos \theta, r \sin \theta)\, r d\theta\right)dr.$
There are two big differences now:
     (i) the integral is taken over a rectangle in $r,\, \theta$ coordinates, and we know that a rectangular domain of integration is usually a lot simpler to deal with than a non-rectangular one, while
     (ii) a factor $r$ has been added to the integral.

Only later will we see why this crucial factor $r$ has crept in.


     In Cartesian coordinates
$D  =  \{\,(x, y)\,: \, 0 \le y \le \sqrt{4-x^2},\,\, -2 \le x \le 2 \,\},$

     while in polar coordinates

$D  =  \{\,(r, \theta)\, : \, 0 \le r \le 2,\,\, 0 \le \theta \le \pi\,\}.$


     A good specific case where both $D$ and the function $f$ are well-suited to polar coordinates is the following: use a cylindrical drill of radius 4 inches to drill a hole through the center of a sphere of radius 5 inches as shown to the right below. In mathematical terms, the drill removes a cylindrical plug of radius 4 inches. But in reality, the sphere could be an apple from which we've removed a core, or we could say that the Astrodome is the top half of such a cylindrical plug - can you see that? Would it even be true to say that there's no real mathematical difference between the Astrodome and the top half of an apple core?

    Problem: find the volume remaining after the 4 inch drill has removed a cylindrical plug though the center of a sphere of radius 5 inches.

When the sphere is regarded as the graph of
$ x^2 + y^2 +z^2  =  25,$
then from overhead what's left of the sphere after drilling looks like the annulus


Thus in polar coordinates what remains has volume
$\displaystyle{2\int_4^5}\left(\displaystyle{\int_0^{2\pi}}\, \sqrt{25 - r^2}\, r d\theta \right) dr  =  4\pi \displaystyle{\int_4^5}\, \sqrt{25 - r^2}\,rdr;$
notice that we need the factor 2 because without it we'd only get the part of the solid in the upper hemisphere. Now we can start to appreciate the factor $r$ because if the substitution $u = 25 - r^2$ is used the last integral reduces to
$\displaystyle{2\pi \int_0^9}\, \sqrt{u}\,du  =   36 \pi \,$cu ins.

   
     Question 1: can you see how to use volumes of revolution from single variable calculus to calculate the volume of the cylindrical plug or the volume of what's left after drilling?

   Question 2: if a square plug of side length 4 inches is removed through the center of the sphere, how would you calculate the volume of what's left? Would you use the polar coordinate method to the left or the volumes of revolution method of Question 1?