Choose local coordinates $(\theta, \phi) \in (0,2\pi) \times (0,2\pi)$ where $w = \cos\theta, x = \sin\theta, y = \cos\phi, z = \sin\phi$. Then you can just pullback the integral to $(0,2\pi) \times (0,2\pi)$ since the complement of this chart in $T^2$ is of measure zero.
We have $dw = -\sin\theta \, d\theta,\ dy = -\sin\phi \, d\phi$. So you end up with $ \int_0^{2\pi} \int_0^{2\pi} \sin^2\theta\cos\phi\sin^2\phi \, d\theta \, d\phi, $ which is a fairly straightforward integral ($u$-sub for the $\phi$ part and half angle for the $\theta$ part).