= Standard thin-disk luminosity
For the steady zero-inner-torque disk, the <standard thin-disk dissipation flux> already includes both faces. Integrate it over projected annuli, rather than multiplying by a second factor of two:
$$
L_{\rm thin}=\int_{R_{\rm in}}^\infty 2\pi R F_{\rm diss}(R)\,dR
=\frac{GM\dot M}{2R_{\rm in}}
=\frac12\dot M R_{\rm in}^2\Omega_{K,\rm in}^2.
$$
This is the decrease in <orbital energy> per unit <time> as matter arrives from arbitrarily large <radius> on a circular orbit. Half the decrease in gravitational <potential energy> remains as orbital <kinetic energy> at the inner edge. Accretion onto a slowly rotating stellar surface can release additional <energy> in a <boundary layer>; a black-hole inner boundary requires the appropriate relativistic treatment.<Dynamics of Astrophysical Discs, Lecture 5: Boundary Conditions and Steady Accretion by Gordon I. Ogilvie>[Section 5.2, printed p. 4, luminosity integral and stellar boundary-layer interpretation]{ref}
For a finite outer <radius> $R_{\rm out}$, integration of this same specified flux gives instead
$$
L(R_{\rm out})=\frac{GM\dot M}{2R_{\rm in}}\left[1-3\frac{R_{\rm in}}{R_{\rm out}}+2\left(\frac{R_{\rm in}}{R_{\rm out}}\right)^{3/2}\right].
$$
It tends to zero as $R_{\rm out}\downarrow R_{\rm in}$ and to $L_{\rm thin}$ as $R_{\rm out}\to\infty$. This finite-annulus result inherits the same steady <torque> profile; it does not impose an additional zero-torque outer boundary.
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