 Solution

A rotation by $\pi$ about the vertical symmetry axis maps a horizontal angular velocity ${\mathbf{\Omega }}^{\prime }$ to $-{\mathbf{\Omega }}^{\prime }$ but leaves a possible vertical force unchanged. Linearity and uniqueness of Stokes flow therefore require that vertical force to equal its own negative, so it vanishes. A horizontal force and a horizontal couple are allowed by the same symmetry.

Write $\lambda =1-ϵ$ and examine the lower pole, where $\theta \ll 1$. Since $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$,Thus $h=O(ϵ\Delta )$ for $\theta =O({ϵ}^{1/2})$, a circular patch of radius

In the broad region, the surface speed is $O({\Omega }^{\prime }a)$, the shear is $O(\mu {\Omega }^{\prime }a/\Delta )$, the area is $O(a^2)$, and the moment arm is $O(a)$. HenceIn the narrow patch, the shear rises to $O(\mu {\Omega }^{\prime }a/(ϵ\Delta ))$, while its area falls to $O(ϵa^2)$; the moment arm remains $O(a)$. Therefore

Dropping the moment arm gives the shear-force scale. Pressure produces the same horizontal order after multiplication by the small surface slope. Both regions consequently contribute