= Solution
Write $y=\nu\Sigma$. A steady inward <accretion rate> has $F_M=-\dot M$, so the integrated angular-momentum equation is
$$
\partial_r\mathcal G-r\mathcal T=\dot M\frac{dh}{dr}.
$$
For $\beta=0$, $\mathcal T=0$ and therefore
$$
\mathcal G=\dot M(h-h_{\rm in})
$$
after imposing the <zero-torque inner boundary condition> at $r_{\rm in}$. In a <Keplerian accretion disk>, $h\propto r^{1/2}$ and $\mathcal G=3\pi hy$, so
$$
\boxed{\nu\Sigma
=\frac{\dot M}{3\pi}
\left[1-\left(\frac{r_{\rm in}}r\right)^{1/2}\right]}.
$$
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