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I have to compute this integral:

$$ \int d\Omega\;\sin\theta\cos\phi|Y_{\ell m}(\theta,\phi)|^2, $$

in the physicists notation ($\theta$ is the azimuthal angle, from $0$ to $\pi$). I say that $d\Omega$ is even, $\cos\phi$ is even, $|Y_{\ell m}(\theta,\phi)|^2$ is even, and $\sin\theta$ is odd, therefore the integral must be 0. However I am not sure about this result, because I am solving a problem with rotational symmetry about the $z$ axis, meaning that $x/r = \sin\theta\cos\phi$ is the function I'm integrating. If I choose to integrate $y/r = \sin\theta\sin\phi$ I would get that the integrand is even, and therefore the integral wouldn't be 0! Are my results correct?

Any thoughts on this are helpful for me. Cheers.

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As noted in the OP, the range of integration of the azimuthal angle $\theta$ is 0 to $\pi$. Hence the relevant symmetry operation is not $\theta\mapsto -\theta$ but $\theta\mapsto \pi-\theta$, and $\sin \theta$ is symmetric (not antisymmetric!) under this operation. The integral, therefore, need not vanish.

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