 Solution

For a twice differentiable vector potential $\mathbf{B}$, commuting partial derivatives in $\mathbf{J}=\nabla \times \mathbf{B}$ gives the necessary conditionThe potential is not unique: for any sufficiently smooth scalar $\psi$, $\mathbf{B}+\nabla \psi$ has the same curl, since the curl of a gradient is zero. This is the usual gradient freedom of a vector potential. A divergence-free condition alone need not suffice globally on every domain; here only its necessity is requested.

Stokes theorem states that for an oriented surface with compatible boundary orientation,For the integrand in this problem, a convenient globally smooth vector potential isThe boundary of the upper ellipsoid is the ellipse $z=0$. The outward normal on the upper surface induces counterclockwise orientation viewed from above, so parameterize it as $\mathbf{r}(\theta )=(a\cos \theta ,b\sin \theta ,0)$, $0\le \theta \le 2\pi$. Along this curve the first component of $\mathbf{B}$ is zero and $dz=0$, leavingSince the last angular integral is $3\pi /4$,The result is independent of the vertical semiaxis $c$. As a sign check, the field is divergence-free, so closing the cap with the downward-oriented base disc makes its flux the positive integral of $x^2$ over that disc, which has the same value.