Evaluate the following triple integral

Question

I am unsure how to get started on this integral. When I try to evaluate with the traditional methods I know, I get lost halfway through. Is this type of integral only possible to evaluate with polar coordinates? \begin{equation} \int_{-2}^{-1}\int_{-2}^{-1}\int_{-2}^{-1} \frac{x^2}{x^2+y^2+z^2} dx dy dz\end{equation}

Answer 1

The best way to solve this would be to exploit symmetries using the divergence theorem, and without polar coordinates. There are two vector fields whose divergences are this integrand that we will utilize

$$F_x = \left(x-\sqrt{y^2+z^2}\tan^{-1}\left(\frac{x}{\sqrt{y^2+z^2}}\right),0,0\right)$$

$$F_z = \left(0,0,\frac{x^2}{\sqrt{x^2+y^2}}\tan^{-1}\left(\frac{z}{\sqrt{x^2+y^2}}\right)\right)$$

Then the divergence theorem tells us

$$I = \iint\limits_{[-2,-1]^2 \\x=-1}F_x\cdot\hat{x}\:dydz+\iint\limits_{[-2,-1]^2\\ x=-2}F_x\cdot-\hat{x}\:dydz = \iint\limits_{[-2,-1]^2 \\z=-1}F_z\cdot\hat{z}\:dxdy+\iint\limits_{[-2,-1]^2\\ z=-2}F_z\cdot-\hat{z}\:dxdy $$

We can denote this as $I = I_{-1} - I_{-2}$ where

$$I_k = \iint\limits_{[-2,-1]^2}k-\sqrt{s^2+t^2}\tan^{-1}\left(\frac{k}{\sqrt{s^2+t^2}}\right)\:dsdt $$

$$ = \iint\limits_{[-2,-1]^2}\frac{s^2}{\sqrt{s^2+t^2}}\tan^{-1}\left(\frac{k}{\sqrt{s^2+t^2}}\right)\:dsdt = \iint\limits_{[-2,-1]^2}\frac{t^2}{\sqrt{s^2+t^2}}\tan^{-1}\left(\frac{k}{\sqrt{s^2+t^2}}\right)\:dsdt$$

by symmetry (since the integration variables are all $2$D dummies, we can eliminate the geometric distinction between $x$, $y$, and $z$ at this stage). Thus we can add the integral to itself three times

$$3I_k = \iint\limits_{[-2,-1]^2}k-\left(\sqrt{s^2+t^2}-\frac{s^2+t^2}{\sqrt{s^2+t^2}}\right)\tan^{-1}\left(\frac{k}{\sqrt{s^2+t^2}}\right)\:dsdt$$

$$= \iint\limits_{[-2,-1]^2}k\:dsdt = k \implies I_k = \frac{k}{3}$$

therefore the final answer is

$$I = -\frac{1}{3} + \frac{2}{3} = \frac{1}{3}$$