Solve in 75 seconds: $\displaystyle{\iiint_{\Omega} {\dfrac{(3x + y − 4)^2}{9x^2 + y^2 + z^2 − 18x − 2y + 10}} \,dz\,dy\,dx}$ (how in the world do I solve this in 75 seconds)
$${\iiint\limits_{\Omega} {\dfrac{(3x + y − 4)^2}{9x^2 + y^2 + z^2 − 18x − 2y + 10}} \ \mathrm dz\ \mathrm dy\ \mathrm dx}$$
where $\displaystyle{\Omega = \{ (x, y, z) \in \mathbb{R}^3 \mid 9 (x − 1)^2 + (y − 1)^2 + z^2 \le 1 \}}.$
It's $\dfrac{8\pi}{27}$. I'm not asking for solutions. This is a competition problem. I'm just amazed how people could solve this that fast. Perhaps, exposure to a lot of problems like this during training enables them. Maybe this is easy for you too.
 A: I have no real experience with this level of competition, but this would be my guess regarding how to work with this integral quickly.
First off, based on the region of integration being the interior of an ellipsoid we might get the idea to transform it into a spherical integral, so it will help to rewrite our integrand in terms of $3(x-1), y-1, $ and $z,$ since they're our squared terms in the equation for the ellipsoid. It's rather easy to see that our integrand ends up as
$$\iiint_\Omega \frac{\left(3(x - 1) + (y-1)\right)^2}{(3(x - 1))^2 + (y-1)^2 + z^2} dV$$
and by a quick substitution $x' = 3(x - 1), y' = y - 1, z' = z$ we get a new integral:
$$\iiint_E \frac{(x' + y')^2}{x'^2 + y'^2 + z'^2} \frac{dV'}3$$
where $E$ is the unit sphere. From here we can compute directly with spherical coordinates, but I think the fastest way is to notice that we can discard the odd $\frac{2x'y'}{\rho^2}$ term over the even region, so our integrand reduces to $\left(\frac{r}{\rho}\right)^2 = (\sin \phi)^2$ and we get
$$\begin{align}\frac13 \iiint_E \frac{x'^2 + y'^2}{x'^2 + y'^2 + z'^2} dV' & = \frac83 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \int_0^1 \sin^2{\phi} \cdot \rho^2 \sin\phi \ d\rho \ d\phi\  d\theta \\ & = \frac83 \cdot \int_0^1 \rho^2 d\rho \cdot \int_0^\frac\pi2 \sin^3 \phi \cdot \int_0^\frac\pi2 d\theta \\ & = \frac83 \cdot \frac13 \cdot \frac23 \cdot \frac\pi2 = \frac{8\pi}{27}\end{align}$$
and for someone in a competitive mindset and with sufficient practice and mechanical acumen I can believe this being done sufficiently quickly.
