Application of Green's Theorem when undefined at origin Problem:
Let $P={-y \over x^2+y^2}$ and $Q={x \over x^2+y^2}$ for $(x,y)\ne(0,0)$.
Show that $\oint_{\partial \Omega}(Pdx + Qdy)=2\pi$ if $\Omega$ is any open set containing $(0,0)$ and with a sufficiently regular boundary. 
Working:
Clearly, we cannot immediately apply Green's Theorem, because $P$ and $Q$ are not continuous at $(0,0)$. So, we can create a new region $\Omega_\epsilon$ which is $\Omega$ with a disc of radius $\epsilon$ centered at the origin excised from it. 
We then note ${\partial Q \over \partial x} - {\partial P \over \partial y} = 0$ and apply Green's Theorem over $\Omega_\epsilon$. Furthermore, $\oint_C(Pdx + Qdy)=2\pi$ if $C$ is any positively oriented circle centered at the origin.
I get the general scheme of how to approach this problem, however I am unsure of how to argue it in a rigorous manner. 

If your boundary is smooth, then you may parametrize it by $x(t) = r(t) \cos t$ and $y(t) = r(t) \sin t$, with $t \in [0,2\pi]$. Then your differential form $P \Bbb d x + Q \Bbb d y$ becomes
$$\frac {-r(t) \sin t \Big (r'(t) \cos t - r(t) \sin t \Big) + r(t) \cos t \Big( r'(t) \sin t + r(t) \cos t \Big)} {r^2 (t)} = 1$$
which, integrated from $0$ to $2\pi$, will give $2 \pi$.
Please note that the application of Green's theorem on $\Omega _\varepsilon$ will tell you that the integral on $C$ is equal to the integral on the small circle of radius $\varepsilon$ surrounding the origin, but won't calculate things for you. You'll still have to parametrize the circle and do an explicit integral.
