 Solution

For cylindrical rotation $u_{\varphi }=R\Omega (R)$, the equilibrium Euler momentum equation is $-R{\Omega }^2{\mathbf{e}}_R=-\nabla p/\rho -\nabla \Phi$. HenceThe integral is the centrifugal potential of cylindrical rotation. On a regular barotropic fluid branch define the specific enthalpy $h(\rho )$ by $dh=dp/\rho$. ThenFor a regular equation of state with $dp/d\rho >0$, $h$ is invertible, so $\rho$ and $p=p(\rho )$ depend only on $\Psi$ within a connected equilibrium region. In particular $dp/d\Psi =-\rho$. This local conclusion uses an invertible barotropic branch; distinct disconnected fluid regions can have different integration constants.