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

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:
$$\iiint \frac{x^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v = \frac{1}{2}
 \iiint 1 \mbox{d}v = \frac{1}{2}\frac{4}{3}\pi = \frac{2}{3}\pi$$
