The solution to Schrodinger's equation are wave functions $\Psi (r,\theta ,\phi )$ of the form, $\Psi (r,\theta ,\phi )= R(r)\Theta(\theta)\Phi(\phi)$ Where, the probability of finding an electron in a region, V:$0
$N^2\int_{0}^{\infty }r^2R^2(r)dr\int_{0}^{2\pi }\Phi^2(\phi)d\phi\int_{0}^{\pi }\Theta^2(\theta)\sin \theta d\theta =1$
Now, consider the $2p_{y}$ orbital; $\Psi _{2p_{y}}=\frac{1}{4\sqrt{2\pi a_{0}^{5}}}re^{-\frac{r}{2a_{0}}}sin \theta sin \phi$
And now the question itself; Evaluate the three integrals,
One. $\int_{0}^{\infty }r^2R^2(r)dr$
Two. $\int_{0}^{2\pi }\Phi^2(\phi)d\phi$
Three. $\int_{0}^{\pi }\Theta^2(\theta)\sin \theta d\theta$
and show $\int_{V}^{}\Psi ^2dV=1$
I understand this may be alot but any help would be greatly appreciated. Integration is something I really need to work on. Thanks guys.