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.

• "Perhaps, exposure to a lot of problems like this during training enables them." -- That's how it tends to go; not just in competition math, but math in general. A lot of math proofs have this or that technique that makes you wonder "how in the hell did you think of that", and then you start looking and see it over and over, or realize it's just a staple of the field. Exposure is invaluable when it comes to problem-solving. Jun 28, 2022 at 3:02
• I agree! Especially with the proofs. I've just revisited some papers and saw this and shared it. Jun 28, 2022 at 3:03
• What competition was this? Jun 28, 2022 at 3:07
• UP Nationwide Search for the Math Wizard (Philippines) Jun 28, 2022 at 3:09
• The fastest way to do this problem is probably to recognize it as $1-\cdots$ where $\cdots$ is an integral that can be done very quickly by symmetry in spherical coordinates. Jun 28, 2022 at 3:11

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.

• Why not $\frac13 \iiint_E \frac{x'^2 + y'^2}{x'^2 + y'^2 + z'^2} dV' = \frac13 \cdot \frac23 \iiint_E 1 dV' = \frac13 \cdot \frac23 \cdot \frac{4}{3}\pi$? Jun 28, 2022 at 4:03
• @RiverLi Hadn't thought of that, but sure that also works out, since by symmetry $\iiint_E \frac{x'^2}{\rho^2} dV' = \iiint_E \frac{y'^2}{\rho^2} dV' = \iiint_E \frac{z'^2}{\rho^2} dV'.$ Certainly faster once you get it, but somehow I doubt it would come to mind quickly enough to be worth it unless you had specifically memorized it beforehand, so I'm not entirely convinced it's practical. But again, I'm no expert, that's just my two cents. Jun 28, 2022 at 4:27
• Yes, here computing using spherical coordinates is not complicated, your answer is nice. Jun 28, 2022 at 5:46