Volume Element in the Integral for Variational Method I'm learning how to use variational method to calculate the energy of ground state helium. I understand how we can set up a trial wavefunction that is a product of two hydrogen-like wavefunctions; however, when finding the integral $\int_{a}^{b} \phi H \phi dr_1dr_2$ (for $\phi=\phi_1(r_1)\phi_2(r_2)$), what would the volume element be? Would it be $16\pi^2 r_1^2 r_2^2$? I'm assuming the denominator is one for normalized wavefunction of hydrogen-like atom.
If you are working in spherical coordinates, and if the integrand has no dependence on the angular coordinates, then that volume element would be correct ($4\pi$ from each integration over $\theta_i \phi_i$ and $r_i^2$ for each spherical shell); however, the interelectronic potential $\frac{1}{r_{12}}$ depends not only on the radial coordinates, but also on the angular coordinates of each electron, so you can't just integrate those out.