I have encountered with a rather interesting vector field while studying Green's Theorem: $$\vec{F}(x,y)=-\frac{y}{x^2+y^2}\hat{i}+\frac{x}{x^2+y^2}\hat{j}$$ This field turned out to be quite special - for instance, I learned that for every closed curve such that the point $(0,0)$ is inside it, the line integral of this field on that curve is equal to $2\pi$; and if $(0,0)$ is outside the curve, then it is equal to $0$. That, as I discovered, is due to Green's Theorem, and the fact that $\vec{F}$ is not continuously differentiable at $(0,0)$.
For that reason, I was intrigued what would happen to the line integral when I take closed curves (piecewise smooth and simple) such that they go through $(0,0)$; meaning the point is neither inside or outside the curve.
Let: $$I=\oint_\gamma\vec{F}\cdot d\vec{r}$$
This is what I found:
1) For the polar rose leaf $\gamma(t)=(\sin(2t)\cos(t),\sin(2t)\sin(t)), t\in[0,\frac \pi 2]$, I got $I=\frac \pi2.$
2) For the union of the following curves: $\gamma_1(t)=(-t,1-t^2), t\in[-1,1]$ and $\gamma_2(t)=(t,0), t\in[-1,1]$ I got $I=\pi$.
3) For the triangle represented by the union of $\gamma_1(t)=(t,0), t\in[0,1]$ and $\gamma_2(t)=(1-t,t), t\in[0,1]$ and $\gamma_3(t)=(0,1-t), t\in[0,1]$, I got $I=\frac \pi2.$
4) For the circle $\gamma(t)=(\cos(t),1+\sin(t)),t\in[0,2\pi],$ I got $I=\pi.$
(All the curves are close, piecewise-smooth, positively oriented and simple)
I actually tried many more, but these are the ones I found most representing. What was interesting, is that all of the integrals I computed (if I computed correctly), were either $\frac \pi2$ or $\pi$. What I also found, is that all the curves for which I got $I=\frac \pi2$, $(0,0)$ was a singular point (meaning, the curve wasn't smooth there), and for the ones I got $I=\pi$, the point wasn't singular (meaning, the curve was smooth there).
So I came here in order to find whether my speculations are right, or wrong - for the simple reason I couldn't find a counterexample, nor couldn't I prove the matter.
Thanks for everyone!
P.S.: If my claim is partially wrong; for exmaple: The integral is $\frac \pi2$ for every curve such that $(0,0)$ is a singular point, but not necessarily $\pi$ when the point is not singular; I would happy to know at least why the right part of the claim is true.