show that if $\mathbf{F}(x,y)=\frac{-y\mathbf{i}+x\mathbf{j}}{x^2+y^2}$, then $\oint\mathbf{F}\dot{}d\mathbf{r}=a\pi$ for every simple closed path that encloses the origin. Find the constant $a$.
I first calculated the curl of the vector field and it was $\mathbf{0}$. Which means that there exists a scalar field $f$ such that $\mathbf{F}=\nabla f$ So the integral becomes $\oint\mathbf{\nabla }f\dot{}d\mathbf{r}=f(\mathbf{r(a)}-f(\mathbf{r(a)})=0)$ and Hence $a=0$.. But apparently the mark scheme says it should be $a=2$. Any idea where I am going wrong?