 Solution

Put $q=r_0^2$ and $\eta =d\psi +\frac{1}{2}\cos \theta d\varphi$. Expanding the metric tensor in the stationary direction gives the useful exact identitiesTaking the exterior derivative givesThe fibered form of the metric tensor has determinant $\det g=-r^6{\sin }^2\theta /16$. Its inverse components needed for the flux areThereforesince $h(1-\alpha \Omega )=1$. This calculation includes the rotational mixed term; dropping it prematurely would miss the exact cancellation.

With the stipulated negative coordinate orientation, ${ϵ}_{tr\psi \theta \varphi }=-r^3\sin \theta /4$. The pullback of the Hodge star operator to the constant-$t,r$ three-dimensional spacelike submanifold is consequentlyThe coordinate ranges and standard sphere identifications giveThus the five-dimensional Komar mass, in the normalization given and $G_5=1$, isThe flux is already independent of $r$ in the vacuum region, so its limit at infinity has the same value. The negative orientation in the question is crucial for the positive sign; reversing that orientation reverses the flux.