= Solution
Take $\gamma>1$ for the <ideal gas>. At fixed <cylindrical radius>, put $\mathcal R=k/(\mu m_p)$, where $k$ is the <Boltzmann constant>, $\mu$ the <mean molecular weight> and $m_p$ the <proton> mass. Also write $a=9\alpha\Omega/4$. Vertical <hydrostatic equilibrium>, <thermal equilibrium> and the <ideal gas> law require
$$
\frac{dP}{dz}=-\Omega^2z\rho,\qquad aP=AT^{1+\beta},\qquad P=\mathcal R\rho T.
$$
Consequently, with $K=A/a$,
$$
P=KT^{1+\beta},\qquad\rho=\frac K{\mathcal R}T^\beta,\qquad
(1+\beta)\mathcal R\frac{dT}{dz}=-\Omega^2z.
$$
Integrating from the midplane yields
$$
T=T_0-\frac{\Omega^2z^2}{2(1+\beta)\mathcal R},\qquad
H^2=\frac{2(1+\beta)\mathcal R T_0}{\Omega^2}.
$$
The <polytropic vertical structure in stellar gravity> therefore has, for $|z|\leq H$,
$$
\boxed{T=T_0\left(1-\frac{z^2}{H^2}\right),\quad
\rho=\rho_0\left(1-\frac{z^2}{H^2}\right)^\beta,\quad
P=P_0\left(1-\frac{z^2}{H^2}\right)^{1+\beta},}
$$
where
$$
P_0=\frac{4A}{9\alpha\Omega}T_0^{1+\beta},\qquad
\rho_0=\frac{P_0}{\mathcal R T_0}.
$$
This is a <polytropic equation of state> $P\propto\rho^{1+1/\beta}$ with <polytropic index> $\beta$. For $|z|>H$, set $\rho=P=0$; <temperature> has no physical gas value in that vacuum. Here $H$ is the finite semi-thickness, rather than an exponential <disk scale height>.
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