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.
Best Answer
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.