= Solution
For $u=(-y,x,\alpha)$, one has $\nabla\cdot u=0$, and therefore
$$
a_i=\partial_j(u_iu_j)
=(u\cdot\nabla)u_i
=(-x,-y,0)_i.
$$
Close the upper half-ellipsoid with its unit-disk base $D$ in the plane $z=0$. By symmetry,
$$
\int_V(-x,-y,0)\,dV=0.
$$
On the base, the outward normal is $n=(0,0,-1)$, so
$$
u_iu_jn_j=-\alpha u_i
=(\alpha y,-\alpha x,-\alpha^2)_i.
$$
The first two components integrate to zero over the disk, while the third integrates to $-\pi\alpha^2$. The closed-surface tensor identity therefore gives
$$
\int_Su_iu_jn_j\,dS
=(0,0,\pi\alpha^2)_i.
$$
Multiplying by $\beta^2$,
$$
\boxed{F=(0,0,\pi\alpha^2\beta^2)}.
$$
Back to article page
