Awesome Solution to a MIT Multivar Problem
by Wolstenholme, Sep 3, 2015, 4:23 AM
OK, so in the last few days BOGTRO and I tested out of MIT's multivariable calculus class (class 18.02). One of the harder problems on the test was the following:
Consider the ball of radius
centered at the point
. What is the average distance from the origin to a point in this ball?
So, the natural inclination would be to do a simple triple integral after translating to spherical coordinates. However, despite the nice bounds
and
, the bounds for
require the Law of Sines/Cosines which is just dumb. So, BOGTRO and I (with Chris Shao also aiding in the conception of this beautiful idea) present the following "clearly-intended" solution - prepare your bodies!
get rekt mit
Unfortunately, BOGTRO and I kept up messing up our calculations so this took us an hour lol.
Consider the ball of radius


So, the natural inclination would be to do a simple triple integral after translating to spherical coordinates. However, despite the nice bounds



get rekt mit
Invert about the sphere centered at the origin with radius
(we could have easily used
instead of
, having
instead functions as a sanity check). The ball (which we shall henceforth denote by
is sent to a half-space whose boundary lies
above the
-plane. Thinking about this inversion in terms of substitution, we have that point
is sent to the point
where
![\[ u = \frac{r^2x}{x^2 + y^2 + z^2} \: \: v = \frac{r^2y}{x^2 + y^2 + z^2} \: \: w = \frac{r^2z}{x^2 + y^2 + z^2} \]](//latex.artofproblemsolving.com/7/1/d/71d826c5c024b80f3986498a7c8587f703f3a50b.png)
Now some easy derivatives yield the Jacobian
![\[ \frac{d(u, v, w)}{d(x, y, z)} = \frac{r^6}{(x^2 + y^2 + z^2)^6}\begin{vmatrix}y^2 + z^2 - x^2 && -2xy && -2xz\\-2yx && z^2 + x^2 - y^2 && -2yz\\-2xz && -2yz && x^2 + y^2 - z^2\end{vmatrix} = -\frac{r^6}{(x^2 + y^2 + z^2)^3} \]](//latex.artofproblemsolving.com/9/1/2/912accce53ee6d3d6a3ffac6905973f2fda584f0.png)
Also note that
so
and also
. Therefore

Since we no longer have to do the dumb Law of Cosines/Sines, we can NOW convert to spherical coordinates without major trouble!!! Our integral becomes

which is the correct answer. GG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!









![\[ u = \frac{r^2x}{x^2 + y^2 + z^2} \: \: v = \frac{r^2y}{x^2 + y^2 + z^2} \: \: w = \frac{r^2z}{x^2 + y^2 + z^2} \]](http://latex.artofproblemsolving.com/7/1/d/71d826c5c024b80f3986498a7c8587f703f3a50b.png)
Now some easy derivatives yield the Jacobian
![\[ \frac{d(u, v, w)}{d(x, y, z)} = \frac{r^6}{(x^2 + y^2 + z^2)^6}\begin{vmatrix}y^2 + z^2 - x^2 && -2xy && -2xz\\-2yx && z^2 + x^2 - y^2 && -2yz\\-2xz && -2yz && x^2 + y^2 - z^2\end{vmatrix} = -\frac{r^6}{(x^2 + y^2 + z^2)^3} \]](http://latex.artofproblemsolving.com/9/1/2/912accce53ee6d3d6a3ffac6905973f2fda584f0.png)
Also note that




Since we no longer have to do the dumb Law of Cosines/Sines, we can NOW convert to spherical coordinates without major trouble!!! Our integral becomes

which is the correct answer. GG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Unfortunately, BOGTRO and I kept up messing up our calculations so this took us an hour lol.