 Solution

For $u=(-y,x,\alpha )$, one has $\nabla \cdot u=0$, and therefore

Close the upper half-ellipsoid with its unit-disk base $D$ in the plane $z=0$. By symmetry,On the base, the outward normal is $n=(0,0,-1)$, soThe first two components integrate to zero over the disk, while the third integrates to $-\pi {\alpha }^2$. The closed-surface tensor identity therefore givesMultiplying by ${\beta }^2$,