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 $-yx^2$
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.