Un-curl operator?

I would like to use Stokes' Theorem to find the area of a surface over a given region. This is given by:

$A = \oint\vec{F}\centerdot d\vec{r}$

but only if the following condition holds:

$(\vec{\triangledown}\times\vec{F})\centerdot \vec{n} = 1$

where $\vec{n}$ is the normal to the surface. How do I come up with a vector field, $\vec{F}$, that satisfies this condition? I found a paper that discusses an inverse-curl operator here, but this is only useful if I know what $(\vec{\triangledown}\times\vec{F})$ is and need to find $\vec{F}$. Any ideas?

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Is your surface nonplanar? If so, this is going to be difficult. Imagine a sphere with a small hole in it; what does $\nabla\times F$ have to look like? – Rahul Apr 15 '11 at 21:43
@Rahul Narain: Yes, the surface is non-planar. In fact, the specific surface I'm interested in is $z=\sin(x)\sin(y)$. If I use $\vec{F}=\langle0,x,0\rangle$ it satisfies the condition, but it doesn't encode any information about z, so it would give the same answer regardless of z. – okj Apr 15 '11 at 21:51
Essentially this is a vector boundary problem in $\mathbf{G}=\nabla\times\mathbf{F}$ defined by $$\nabla\cdot\mathbf{G}=0,$$ $$\mathbf{G}|_{\partial\Omega}\cdot\mathbf{n}=1.$$ The vector field $\mathbf{F}$ can be recovered from $\mathbf{G}$ via the Helmholtz theorem. – anon Aug 15 '11 at 2:00
@anon You should have posted this as answer, this question still shows up as unanswered – Tobias Kienzler Aug 2 '13 at 10:03

Essentially this is a vector boundary problem in $\mathbf G = \nabla \times \mathbf F$ defined by \begin{align} \nabla \cdot \mathbf G &=0\\ \mathbf G \vert_{\partial \Omega}\cdot \mathbf n &= 1 \end{align} The vector field $\mathbf F$ can be recovered from $\bf G$ via the Helmholtz theorem.