= Solution
Put $q=r_0^2$ and $\eta=d\psi+\tfrac12\cos\theta\,d\phi$. Expanding the <metric tensor> in the stationary direction gives the useful exact identities
$$
g_{tt}=-1+\frac q{r^2},\qquad g_{t\psi}=-\frac{\alpha q}{r^2},\qquad
k=\left(-1+\frac q{r^2}\right)dt-\frac{\alpha q}{r^2}\eta.
$$
Taking the <exterior derivative> gives
$$
dk=\frac{2q}{r^3}dt\wedge dr
+\frac{2\alpha q}{r^3}dr\wedge\eta
+\frac{\alpha q\sin\theta}{2r^2}d\theta\wedge d\phi.
$$
The fibered form of the <metric tensor> has <determinant> $\det g=-r^6\sin^2\theta/16$. Its inverse components needed for the flux are
$$
g^{tt}=-\frac h f,\qquad g^{t\psi}=-\frac{h\Omega}f,\qquad g^{rr}=f.
$$
Therefore
$$
(dk)^{tr}=f\left[g^{tt}(dk)_{tr}+g^{t\psi}(dk)_{\psi r}\right]
=-\frac{2q}{r^3}h(1-\alpha\Omega)=-\frac{2q}{r^3},
$$
since $h(1-\alpha\Omega)=1$. This calculation includes the rotational mixed term; dropping it prematurely would miss the exact cancellation.
With the stipulated negative coordinate <orientation>, $\epsilon_{tr\psi\theta\phi}=-r^3\sin\theta/4$. The pullback of the <Hodge star operator> to the constant-$t,r$ three-dimensional <spacelike submanifold> is consequently
$$
(\star dk)_{\psi\theta\phi}=\epsilon_{\psi\theta\phi tr}(dk)^{tr}
=\frac q2\sin\theta.
$$
The coordinate ranges and standard sphere identifications give
$$
\int\star dk=\frac q2(2\pi)^2\int_0^\pi\sin\theta\,d\theta=4\pi^2q.
$$
Thus the five-dimensional <Komar mass>, in the normalization given and $G_5=1$, is
$$
\boxed{M=\frac{3\pi r_0^2}{8}.}
$$
The 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.
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