= Solution
Steady <mass conservation> gives $\dot m=-2\pi R\Sigma u_R$. With specific angular momentum $l=R^2\Omega$, multiply the angular-momentum equation by $2\pi$ and define the signed <viscous torque in an accretion disk>
$$
\mathcal G(R)=2\pi\nu\Sigma R^3\frac{d\Omega}{dR}.
$$
Then
$$
\frac{d\mathcal G}{dR}
=-\dot m\frac{dl}{dR}.
$$
Integration from $R_1$ to $R_2$ gives
$$
\boxed{
\mathcal G_2-\mathcal G_1=-\dot m(l_2-l_1)
}.
$$
The viscous power generated in an annulus is $2\pi R F_{\rm diss}\,dR=\mathcal G\,d\Omega$. Taking the inner torque to vanish and writing $l_{\rm in}=l(R_{\rm in})$ gives $\mathcal G=-\dot m(l-l_{\rm in})$, hence
$$
L_{\rm gen}
=-\dot m\int_{R_{\rm in}}^{R_{\rm out}}
(l-l_{\rm in})\frac{d\Omega}{dR}\,dR.
$$
<Integration by parts> yields
$$
L_{\rm gen}
=\dot m\left[
\int_{l_{\rm in}}^{l_{\rm out}}\Omega\,dl
-\Omega_{\rm out}(l_{\rm out}-l_{\rm in})
\right].
$$
For steady circular force balance, $de=\Omega\,dl$; equivalently, the stated equality of gravitational- and rotational-potential differences makes the integral $e_{\rm out}-e_{\rm in}$. Therefore
$$
\boxed{
L_{\rm gen}(R_{\rm in},R_{\rm out})
\simeq\dot m
\left[e_{\rm out}-e_{\rm in}
-\Omega_{\rm out}(l_{\rm out}-l_{\rm in})\right]
}.
$$
When $R_{\rm out}\gg R_{\rm in}$, the outer energy and boundary term vanish. For an approximately Keplerian inner orbit, $e_{\rm in}=-R_{\rm in}^2\Omega_{\rm in}^2/2$, so
$$
L_{\rm gen}\simeq
\frac12\dot mR_{\rm in}^2\Omega_{\rm in}^2.
$$
Comparison with part (c) gives
$$
\boxed{
\frac{L_{\rm gen}}{L_{\rm thin}}
\simeq
\left(\frac{\Omega_{\rm in}}{\Omega_{K,\rm in}}\right)^2
}.
$$
A Keplerian inner flow generates the standard thin-disk power. A pressure-supported sub-Keplerian slim disk generates less through shear, and its emergent luminosity can be smaller still because <photon trapping in an accretion flow> carries part of that generated energy through the inner edge.
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