= Solution
Use <cylindrical coordinates> $(R,\phi,z)$, put $s=(R^2+z^2)^{1/2}$, and write the purely azimuthal velocity as $\mathbf u=R\Omega\mathbf e_\phi$. The steady radial, azimuthal, and vertical momentum equations are
$$
0=-\frac1\rho\frac{\partial p}{\partial R}
-\frac{GM_{\rm BH}R}{s^3}+R\Omega^2,
\qquad
0=-\frac1{\rho R}\frac{\partial p}{\partial\phi},
$$
$$
0=-\frac1\rho\frac{\partial p}{\partial z}
-\frac{GM_{\rm BH}z}{s^3}.
$$
Thus $\nabla p=\rho\mathbf g_{\rm eff}$, where the <effective gravity in a rotating frame> is
$$
\boxed{\mathbf g_{\rm eff}
=-\frac{GM_{\rm BH}}{s^3}(R\mathbf e_R+z\mathbf e_z)
+R\Omega^2\mathbf e_R}.
$$
When <radiation pressure> supplies the pressure support, force balance gives
$$
\boxed{\frac\kappa c\mathbf F+\mathbf g_{\rm eff}=0,
\qquad
\mathbf F=-\frac c\kappa\mathbf g_{\rm eff}}.
$$
On the upper conical surface $z=R\tan\alpha$, the tangent and outward normal are
$$
\mathbf t=\cos\alpha\,\mathbf e_R+sin\alpha\,\mathbf e_z,
\qquad
\mathbf n=-\sin\alpha\,\mathbf e_R+cos\alpha\,\mathbf e_z.
$$
Because a pressure surface is an <equipotential surface>, $\mathbf g_{\rm eff}\cdot\mathbf t=0$. Since $s=R/\cos\alpha$, this fixes
$$
R\Omega^2=\frac{GM_{\rm BH}\cos\alpha}{R^2},
\qquad
\boxed{\mathbf g_{\rm eff}
=-\frac{GM_{\rm BH}}{R^2}sin\alpha\cos\alpha\,\mathbf n}.
$$
The lower face gives the reflected result. One face has area element $dA=2\pi R,dR/\cos\alpha$. Adding both faces and using $F=c|\mathbf g_{\rm eff}|/\kappa$ yields
$$
dL=2F,dA
=\frac{4\pi GM_{\rm BH}c}{\kappa}
\sin\alpha\frac{dR}{R}.
$$
Consequently
$$
\boxed{L_{\max}=L_{\rm Edd}\sin\alpha
\log\frac{R_2}{R_1}}.
$$
For a broad radial range, each logarithmic interval can radiate an Eddington-scale contribution because rotation reduces the surface-normal gravity. The factor $\sin\alpha$ makes a thin wedge faint and lets a geometrically thick funnel expose a larger effective gravity and emitting area.
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