I tried to solve this triple integral but couldn't integrate the result.
∫∫∫x2+2y2x2+4y2+z2dv and the surface to integrate in is x2+y2+z2≤1
Is there any way to transform the integral into polar coordinates?
Answer
Notice that:
∭
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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