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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.

Use a velocity potential to reconstruct an irrotational flow from one scalar field. A velocity potential $\varphi (\mathbf{x},t)$ satisfiesIt has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is unchanged as a description of the velocity field when any function of time alone is added. For twice continuously differentiable $\varphi$, the identity curl of a gradient proves zero vorticity. In incompressible flow, taking the divergence also gives ${\nabla }^2\varphi =0$, the Laplace equation. This turns a three-component velocity field into a single harmonic function with appropriate boundary conditions. ^[ref]

For uniform flow $\mathbf{u}=(U,0,0)$, take $\varphi =Ux$. With $U=2\mathrm{m}{\mathrm{s}}^{-1}$, the change from $x=0$ to $x=3\mathrm{m}$ is $\Delta \varphi =6{\mathrm{m}}^2{\mathrm{s}}^{-1}$. By the fundamental theorem for line integrals, the line integral of velocity between these points equals that change along any path in the domain; a closed curve gives zero circulation.

For the potential line vortex, choose a local angle branch in polar coordinates and set $\varphi =K\theta$. Then $u_r={\partial }_r\varphi =0$ and $u_{\theta }=r^{-1}{\partial }_{\theta }\varphi =K/r$. One counterclockwise turn changes the continued velocity potential by $2\pi K$, although the velocity field returns to its original value. With $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$, this change is $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$, exactly the circulation. A full annulus therefore admits local angle branches and a multivalued velocity potential, but no global single-valued velocity potential when $K\ne 0$. Cutting the annulus along a radial line allows a single-valued angle branch on the remaining domain. The obstruction is path independence, not a failure of the local gradient formula.