Proving no vector potential for gravitation field defined on all of $\mathbb{R}^3 -$ origin Let:
$$F=\frac{x,y,z}{(x^2+y^2+z^2)^{3/2}}$$ Show that there is no vector potential for F which is defined on all of $\mathbb{R}^3 - \text{origin}$
I can find a vector potential which is not well-behaved on the z-axis quite easily, but I’m not sure how to show that it’s impossible to find one for all $\mathbb{R}^3 -$ origin. 
My professor suggested I assume a vector potential exists then calculate the following in two different ways:
$$\iint_SF\bullet\mathbf{n} dS$$ 
where S is the unit sphere.
I’m not sure how to calculate this integral though, and I don’t see how it would help.
 A: Suggestion: Using ideas that Helmholtz proved, the gravitational vector field can be decomposed into the sum of an irrotational vector field (call it $\phi$) and divergence-less vector field (call it $A$).  This means a vector field $F$ can be written as $F = -\nabla \phi + \nabla \times A.$  Obviously the gravitational field is curl free (conservative), so $A = 0$ and you should try to find a scalar field $\phi$ such that the gradient equals the gravitational field.  This is a standard procedure that can be found in any vector calc book or wikipedia.
As you will see, you cannot find a potential for $F$ on all $\mathbb{R^3}$.
A: Hint: If there were a vector field $G$ defined in the complement of the origin such that $F = \nabla \times G$, then Stokes' theorem would imply
$$
\iint_{S} F \cdot n\, dS = 0
$$
because the sphere has empty boundary. On the other hand, you can calculate
$$
\iint_{S} F \cdot n\, dS
$$
explicitly (without calculus, even!), since on the unit sphere you have $n = (x, y, z)$ and $x^{2} + y^{2} + z^{2} = 1$.
