Does the potential exist? 
Verify Green's theorem for $X(x,y)=(xy^2,-yx^2)$ in the circle of radius $R$ with center $(0,0)$.

I think there is a mistakein the field. I suppose the first thing I should do is to find a function $F$ such that $\nabla F= X$. I don't think it is possible: $\displaystyle{\int xy^2 = \frac{1}{2}x^2y^2+g(y)}$ and then it should be $\displaystyle{\frac{\partial}{\partial y}\left(\frac{1}{2}x^2y^2+g(y)\right)}=-yx^2$ but $\displaystyle{\frac{\partial}{\partial y}\left(\frac{1}{2}x^2y^2+g(y)\right)=yx^2+g'(y)}$ since $g$ only has the $y$ variable is impossible to get $-yx^2$
 A: If the potential function exists (but it doesn't), then the integral would be zero.  Green's theorem works just fine here.  But when you "verify" it, you'll have to parametrize the circle and grind out the line integral the hard way.  You don't need a potential here.
A: The Green theorem allows us to write
$$
\oint_C \vec{X}\cdot d\vec{r} = \iint_S (\frac{\partial X_2}{\partial x}-\frac{\partial X_1}{\partial y})\; dS
$$
where $S$ is the disc of radius $R$ centered at $(0,0)$, $X_2=-yx^2$ and $X_1=xy^2$. Therefore, the integral equals
$$
\int_0^{2\pi}\int_0^R (-2r^2\cos \theta\sin\theta-2r^2\cos \theta\sin\theta)rdrd\theta = 0
$$
Alternatively, 
$$
\oint_C \vec{X}\cdot d\vec{r} = \int_0^{2\pi} \vec{X}(t) ||\vec{v}(t) ||\; dt = \int_0^{2\pi} \pmatrix{R^3\cos t \sin^2 t\\-R^3\cos^2 t \sin t}\cdot \pmatrix{R\cos t \\ R\sin t}dt = \int_0^{2\pi} 0\; dt = 0
$$
I think what is important to note here is that although the integral equals $0$ on a closed curve, the field $\vec{X}$ does not have a potential. It works the other way around: if $\vec{X}$ did have a potential, then we could have written $\oint_C \vec{X}\cdot d\vec{r}=0$ without any computations.
