 Solution

The outward normal from the semicircular barrier is ${\mathbf{e}}_r=(-\cos \varphi ,\sin \varphi )$. The leading radial flux in the thick region isForone has $h_s=-H/\Delta$. Negligible radial flux therefore requiresThis expresses local hydrostatic blocking: the free-surface gradient opposes the downslope gravitational flux. Streamlines arriving from upslope divide at $\varphi =0$ and run around the two sides of the barrier toward $\varphi =\pm \pi /2$.

At leading order the azimuthal flux density is $q_{\varphi }\simeq h^3\sin \varphi$. Its integral across the thick region isThis must equal the upstream unit flux intercepted between the symmetry axis and that polar angle, namely $R\sin \varphi$. HenceCombining this with $H=\Delta \cos \varphi$ gives