= Solution
For cylindrical rotation $u_\phi=R\Omega(R)$, the equilibrium <Euler momentum equation> is $-R\Omega^2\mathbf e_R=-\nabla p/\rho-\nabla\Phi$. Hence
$$
\frac{\nabla p}{\rho}=-\nabla\Psi,\qquad \Psi=\Phi-\int_0^R R'\Omega^2(R')\,dR'.
$$
The integral is the <centrifugal potential of cylindrical rotation>. On a regular <barotropic fluid> branch define the <specific enthalpy> $h(\rho)$ by $dh=dp/\rho$. Then
$$
\boxed{h(\rho)+\Psi=\text{constant}.}
$$
For 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.
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