= Solution
Integrate the profile over the thickness:
$$
\boxed{Q(\theta)=Uh-\frac{g\cos\theta}{3\nu}h^3.}
$$
This is a volume flux per unit axial length. In a steady layer with no evaporation, sources or axial variation, mass conservation gives $\partial_\theta Q=0$ to leading lubrication order. Therefore $Q$ is constant around the cylinder even though the local speed and thickness vary.
Assume $U,Q>0$ for the wall-entrained branch, and substitute $h=(Q/U)H$ into the flux formula. After division by $Q$,
$$
1=H-\frac{gQ^2}{3\nu U^3}\cos\theta\,H^3.
$$
Consequently
$$
\boxed{\frac1{H^2}-\frac1{H^3}=\alpha\cos\theta,\qquad
\alpha=\frac{gQ^2}{3\nu U^3}.}
$$
The cubic correction measures gravitational drainage relative to transport by the rotating wall. The nondimensionalization would need modification for zero or oppositely directed mean flux.
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