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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 thereforeTaking 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 yieldsThe 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 $H=ϵr$, the gravitational potential on this surface is ${\Phi }_s=-GM/[r\sqrt{1+{ϵ}^2}]$. The first integral evaluated at the surface says $C={\Phi }_s(r)-\int r{\Omega }^{2}dr$. Differentiating along it givesand therefore $\boxed{{\Omega }^2=(1+{ϵ}^2{)}^{-1/2}GM/r^3}$. 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 profileIt is nonnegative for $K>0$. Integrating over height, with $y=z/r$, givesThus 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.