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I understand that this might not be unique, but is there a (relatively) painless way to generate such a 'vector potential', so for a given field $\vec{A}$, a new field $\vec{B}$ which satifies:

$$\vec{A} = \nabla \times \vec{B}$$

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Maybe this will – sid Mar 12 '14 at 19:39
Of course, your field $\vec{B}$ is defined only up to an additive gradient. – sid Mar 12 '14 at 19:40
It looks fairly impractical then. The reason I thought of this was Stokes' theorem, I thought maybe if I'm given a vector field $\vec{F}$ to integrate over some surface, I could find a new field which curled gives $\vec{F}$, and then use the theorem to calculate a simpler line integral of the new field. But I guess that's never worth doing. – Spine Feast Mar 12 '14 at 19:44
You also need to assume $\nabla \cdot \vec A = 0$, since if $\vec A = \nabla \times \vec B$, then $\nabla \cdot \vec A = \nabla \cdot (\nabla \times \vec B) = 0$. – Robert Lewis Mar 12 '14 at 19:49
As Robert Lewis noted, you could only hope to get $\vec B$ for a zero-divergence $\vec A$. But when $\vec A$ is zero-divergence, another simplification may be more practical: find a simpler surface with the same boundary, and integrate over it. – user127096 Mar 13 '14 at 1:36

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