Prove or disprove: All radial vector fields are conservative This was a question a calculus student had asked me, and unfortunately I believe I gave an incorrect proof.

DEFINITION: A radial vector field is defined by ${\bf F}(x,y) = g(x,y){\bf r},$ where $g$ is a scalar function and ${\bf r}= \langle x,y\rangle$ is a vector.

At the time, I did not know the complete definition of a radial vector field. I proved it like so: 
PROOF: Let $a,b\in\mathbb R$. Take ${\bf F}(x,y) = ax{\mathrm i} + by\mathrm j$. Consider the function $f(x,y) = {1\over2}ax^2 + {1\over2}by^2$. Then clearly $\nabla f(x,y) = ax\mathrm i + by\mathrm j = {\bf F}(x,y)$, hence all radial vector fields are conservative.
However, this is not correct because it only works for fixed constants $a,b$. For a general function $g(x,y)$, what might we be able to do? The more I think about it, the less I think the conjecture is true but I can't think of a counter example.
 A: Consider the radial vector field
$$
F(x, y) = (x^2, xy) = x (x, y).
$$
Note that the $x$-component of this field is everywhere nonnegative, and mostly positive. Integrate it over the sides of the square with vertices $(0,0), (1, 0), (1,1), (0, 1)$ and the $x$-coordinate of the result will be nonzero. But for a conservative field, the integral over any cycle is always zero. 
Hence not every radial field is conservative. 
Alternatively: A field $F$ on a simply connected domain is conservative only if $\partial F_y/\partial x =\partial F_x / \partial y$ everywhere. In the case of this field, that would require that 
$$y = 0$$
which is not everywhere true. 
A: As written, you're right, because the $x$ component is (taking the non-unit $\mathbf{r}$ for convenience) $g(x,y) x$ and the $y$ component is $g(x,y) y$. There is no reason to expect $\frac{\partial}{\partial y}(g(x,y) x)=\frac{\partial g}{\partial y} x$ to be the same as $\frac{\partial g}{\partial x} y$.
A case that would work would be $f(x) e_1 + g(y) e_2$. In this case you can just find antiderivatives of $f$ and $g$ and add them together.
A: I think your question is already answered above. However there is a subclass of radial vector fields defined as  $\mathbf{v} = f(r) \mathbf{r}$ where $r = \lvert \mathbf{r} \rvert$, $\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}$ which are always conservative. Such vector fields occur naturally when considering gravitational fields for example. Thus sometimes they are taken as definition of a radial vector field.
Proof. $\mathbf{v} = f(r) \mathbf{r}$ is conservative if we can find $F$ such that
$$
\nabla F(r) = F'(r) \frac{\mathbf{r}}{r} = \mathbf{v}
$$
which implies that $F'(r) = r f(r)$. Since any continuous function has an anti derivative, we can always find $F$ as long as $f$ is continuous.
Example. Take $f(r) = \cos r$. Then $F(r) = \sin r - r \cos r$.
