Question: Let, $F$=$r\over{||r||}^2$ = ${{xi + yj}}\over{x^2+y^2}$
Lets $\alpha$ be a simple closed curve in a plane that doesn't pass through origin. Show ${\int F\cdot nds }$ is equal to zero or 2pi according as the origin inside or outside $\alpha$
My work so far: I did the computation and found the flux using ${\int F\cdot nds }$ to be 2$\pi$. I also used green's therom and found that div(F) is equal to zero. I think these numbers are right, if you could double check though that would be nice. I understand that those two terms are different because div(F) is not defined at origin because you would be dividing by zero. At origin div(F) blows up and everywhere else its zero. But I dont know how to approach the question that is asked. I dont know what my teachers means by "Show ${\int F\cdot nds }$ is equal to zero or 2pi....". Is there more computation needed to answer this question? If so, what should I do next?
Here's how I would prove the hard case:
Step 1: Show that $\int_\alpha F\cdot n\, ds=\int_{C_r} F\cdot n\, ds$, where $C_r$ is a closed circle of radius $r$, chosen small enough so that the corresponding disk fits inside the closed region created by $\alpha$. You can show this fact that splitting up the region enclosed by $\alpha$ into two regions that avoid the origin (where a clockwise circle is created around). Divergence theorem applies.
Step 2: Solve with parameterization on the new integral.