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Let $$ I = \int_{S^2} x^iy^jz^k dx dy dz \int_{S^1} u^l v^m du dv $$ an integral to be computed ($S^n$ being the n-sphere).

Can someone confirm me that as soon as one of the exponent $i,j,k,l,m$ is odd then $I$ is zero?

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3 Answers 3

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This is trivial: an odd power of a coordinate is an odd function and by symmetry the integral vanishes.

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Let $\boldsymbol{x}=(x_1,x_2,\cdots,x_n)$ be a $n$-component unit vector (i.e. $\boldsymbol{x}^2=1$) on the unit sphere $S^{n-1}$. The integral of monomials of $x_i$ over the sphere is given by $$\int_{S^{n-1}}\prod_{i=1}^{n}x_i^{m_i}=\frac{\prod_{i=1}^n(m_i-1)!!(n-2)!!}{(\sum_{i=1}^n m_i+n-2)!!},$$ if all $m_i$ are even. The integral is 0 by symmetry if any of the exponent $m_i$ is odd. Here $n!!$ denotes the double factorial of $n$.

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I answer positively to my answer :)

and in the case where $i,j,k,l,m)$ are even, it suffices to integrate on the domain $D=\{(\theta,\phi) \in [0,\pi/2]^2\}$ ($\theta$ is the longitude and $\phi$ the colatitude) and to multiply by 8. The integral is exact and can be compute with the $\Gamma$'s function.

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