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I tried to solve this triple integral but couldn't integrate the result. $$\int\int\int\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$$ 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?

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    How about _spherical coordinates_? You're integrating over the unit sphere (centered at the origin). – 2017-01-12
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    @StackTD I tried to do so as well but I couldn't transform it. – 2017-01-12
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    Wolfram gives the result in terms of elliptic integrals so I don't think its going to work nicely by hand. (Assuming I haven't typoed it into Wolfram.) – 2017-01-12
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    Transforming shouldn't be the problem, but it may still be an ugly integral to do by hand. The numerical result I got from [WolframAlpha](http://www.wolframalpha.com/input/?i=int_0%5E1+(+int_0%5E%7B2*pi%7D+(int_0%5Epi+(r%5E2*sin(f)*(2+(-3+%2B+cos(2+t))+sin%5E2(f))%2F(-7+%2B+3+cos(2+f)+%2B+6+cos(2+t)+sin%5E2(f)))+df)+dt+)+dr) seems to suggest the exact answer might be $2\pi/3$ though. – 2017-01-12
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    And you can get there elegantly... – 2017-01-12

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

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    I'm missing something obvious, why does symmetry make the integrand constant? – 2017-01-12
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    @GFauxPas: because $(x^2+2y^2)+(z^2+2y^2)=(x^2+4y^2+z^2)$ ! – 2017-01-12
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    @GFauxPas it makes those integrals with numerators $x^2+2y^2$ and $z^2+2y^2$ _equal_ and adding them (giving double the value we're looking for) yields the integral of the simple constant $1$. – 2017-01-12
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Trivial by symmetry: the unit ball $B\subset\mathbb{R}^3$ is invariant under the transformation $(x,y,z)\mapsto(z,y,x)$, hence:

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