I have a vector field F = $\frac{1}{(x^2+y^2+\frac{\beta^2}{4})^{\textrm{1/2}}}(-xi-yj -\frac{\beta}{2}k)$, where $\beta = \frac{4\pi}{\lambda}A\sin(\frac{2\pi}{\lambda}z)\left[A\cos(\frac{2\pi}{\lambda}z)+b\right]$. F represents the unit (inward) normals to the surface $x^2 +y^2 = A^2cos^2(\frac{2\pi}{\lambda}z) + 2Abcos(\frac{2\pi}{\lambda}z) + b^2$. In other words, $x^2 +y^2 = R^2(z)$ : it is like the surface of revolution of a cosine wave with amplitude A and offset b in the z-axis.
I wish to find a vector G such that $\nabla$ $\times$ $ \bf{G} = \bf{F} $. The purpose of this is to use Stokes theorem to change an integral over my surface into a line integral. However, F is not divergence free, and so finding a G seems impossible. Is there a divergence free vector, let's say $\bf{\Phi}$, which is equal to F on the surface?
As a simpler example of what I am trying to do, Consider an open-ended cylinder of radius R. The field of unit normals to the cylinder is $\textbf{n} = \frac{1}{R}(-xi -yj +0k)$. This is not divergence free and so finding a vector U satisfying $\nabla \times \bf{ U = n}$ does not work. However, the vector $\textbf{w} = \frac{R}{x^2+y^2}(-xi -yj +0k)$ is the same as $\textbf{n}$ on the surface of the cylinder, is divergence-free and has a $\textbf{U}$ that can be easily found: $\textbf{U} = \frac{Rz}{x^2+y^2}(-yi + xj +0k)$.
The question describes a very interesting approach of solving surface integrals.
Some ideas:
The normal to a surface defined by $f(x,y,z)=c$ is the gradient $\nabla f$ which you denote by $\mathbf{F}\,.$ Saying that $\mathbf{F}$ is divergence free is nothing else than $\Delta f=0\,.$ Therefore, $\nabla f$ is divergence free if and only if $f$ is harmonic.
If $f$ is not harmonic you are then looking for a harmonic function $g$ whose gradient equals $\nabla f$ on a boundary (the surface). In other words, we have a Laplace equation $\Delta g=0$ with boundary conditions who generalize the famous ones named after Carl Neumann.