Any solution for $\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$

Question

I tried to solve this triple integral but couldn't integrate the result. $$\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$$ and the surface to integrate in is $$x^2+y^2+z^2\le1$$ Is there any way to transform the integral into polar coordinates?

Answer 1

Notice that: $$\iiint \frac{x^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v = \iiint \frac{x^2+4y^2+z^2-2y^2-z^2}{x^2+4y^2+z^2} \mbox{d}v = \iiint 1-\frac{z^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v $$ But by symmetry $x \leftrightarrow z$, we have: $$\iiint \frac{x^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v = \iiint \frac{z^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v $$ So: $$\iiint \frac{x^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v = \frac{1}{2} \iiint 1 \mbox{d}v = \frac{1}{2}\frac{4}{3}\pi = \frac{2}{3}\pi$$

Answer 2

Trivial by symmetry: the unit ball $B\subset\mathbb{R}^3$ is invariant under the transformation $(x,y,z)\mapsto(z,y,x)$, hence:

$$\iiint_B \frac{x^2+2y^2}{x^2+4y^2+z^2}\,d\mu = \frac{1}{2}\left(\iiint_B \frac{x^2+2y^2}{x^2+4y^2+z^2}\,d\mu+\iiint_B \frac{z^2+2y^2}{x^2+4y^2+z^2}\,d\mu\right) = \frac{1}{2}\mu(B) = \color{red}{\frac{2\pi}{3}}.$$