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:
∭x2+2y2x2+4y2+z2dv=∭x2+4y2+z2−2y2−z2x2+4y2+z2dv=∭1−z2+2y2x2+4y2+z2dv
But by symmetry x↔z, we have:
∭x2+2y2x2+4y2+z2dv=∭z2+2y2x2+4y2+z2dv
So:
∭x2+2y2x2+4y2+z2dv=12∭1dv=1243π=23π
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