= Solution
The centripetal acceleration of the purely rotating <barotropic fluid> is $-r\Omega^2\mathbf e_r$. For $p=K\rho^2$, the pressure force satisfies $\rho^{-1}\nabla p=2K\nabla\rho$, so its <barotropic enthalpy function> is $w=2K\rho$. The equilibrium equation is therefore
$$
\nabla(w+\Phi)=r\Omega^2(r,z)\mathbf e_r.
$$
Taking its curl gives $\partial_z(r\Omega^2)=0$. Thus $\Omega^2$ is independent of z; for a continuous, consistently signed rotation law, so is $\Omega$. This proves the <barotropic cylindrical rotation theorem>. Integrating the remaining radial force yields
$$
\boxed{2K\rho+\Phi_3=C,\qquad\Phi_3=\Phi-\int r\Omega^2(r)\,dr.}
$$
The constant C is common to a connected fluid region.
Interpret the semi-thickness as a free surface with $\rho(r,\pm H)=0$, taking the exterior pressure to be negligible. At $H=\epsilon r$, the <gravitational potential> on this surface is $\Phi_s=-GM/[r\sqrt{1+\epsilon^2}]$. The first integral evaluated at the surface says $C=\Phi_s(r)-\int r\Omega^2dr$. Differentiating along it gives
$$
0=\frac{GM}{r^2\sqrt{1+\epsilon^2}}-r\Omega^2,
$$
and therefore \b[$\boxed{\Omega^2=(1+\epsilon^2)^{-1/2}GM/r^3}$]. Radial pressure support makes this slightly slower than <Keplerian rotation> for nonzero thickness.
Subtracting the free-surface first integral from its value inside the disc gives the complete density profile
$$
\rho(r,z)=\frac{GM}{2K}\left[\frac1{\sqrt{r^2+z^2}}-\frac1{r\sqrt{1+\epsilon^2}}\right],\qquad |z|\leq\epsilon r.
$$
It is nonnegative for $K>0$. Integrating over height, with $y=z/r$, gives
$$
\Sigma(r)=\frac{GM}{K}\int_0^\epsilon\left[\frac1{\sqrt{1+y^2}}-\frac1{\sqrt{1+\epsilon^2}}\right]dy=\boxed{\frac{GM}{K}\left[\operatorname{arsinh}\epsilon-\frac{\epsilon}{\sqrt{1+\epsilon^2}}\right]}.
$$
Thus \b[the <surface density> is independent of r] in this <quadratic-polytrope disk with constant aspect ratio>. The factor $1/r$ in density is canceled by the scale r in the vertical extent. The free-surface assumption is essential to this determination; a prescribed nonzero boundary pressure would require its own boundary data. No radial edges are specified, so the formula describes the local equilibrium rather than a finite-mass disc extended to all radii.
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