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.
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.