= Solution
Take the positive constant ratio $\lambda=p_r/p_g$. The <equation of state> implies
$$
\rho=\frac{4\sigma}{3c\lambda\mathcal R}T^3,\qquad p=(1+\lambda)\mathcal R\rho T\propto T^4.
$$
This is a <polytropic equation of state> of index three. Differentiating the <pressure> and using vertical <hydrostatic equilibrium> gives $4(1+\lambda)\mathcal R\rho\,T_z=-\rho\Omega^2z$. Hence
$$
T=T_0-\frac{\Omega^2z^2}{8(1+\lambda)\mathcal R}.
$$
Define the semi-thickness $H$ by $T(H)=0$ and put $s=1-z^2/H^2$. Then the <constant gas-radiation ratio vertical disc structure> is
$$
\boxed{T=T_0s,\qquad\rho=\rho_0s^3,\qquad p=p_0s^4,\qquad |z|<H,}
$$
with
$$
\boxed{T_0=\frac{\Omega^2H^2}{8(1+\lambda)\mathcal R},\quad\rho_0=\frac{\sigma\Omega^6H^6}{384c\lambda(1+\lambda)^3\mathcal R^4},\quad p_0=\frac{\rho_0\Omega^2H^2}{8}=\frac{\sigma\Omega^8H^8}{3072c\lambda(1+\lambda)^3\mathcal R^4}.}
$$
The <surface density> is the integral across both faces. The supplied polynomial integral gives $\int_0^1(1-x^2)^3dx=16/35$, so $\Sigma=(32/35)\rho_0H$. Substitution yields
$$
\boxed{420\lambda(1+\lambda)^3\Sigma=\left(\frac{\mu_m m_p}{k}\right)^4\frac\sigma cH^7\Omega^6.}
$$
All coefficients here are local in radial position; the derivation fixes the vertical profile rather than a global accretion rate. The zero-temperature surface is a formal boundary of this bulk model; optically thick diffusion ultimately fails in a thin photospheric layer.
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