Finding the winding number of a curve Let $\gamma(t)=(r \cos t,r \sin t)$, for some $r>0$, and let $\Gamma$ be a $C^2$-curve in $\mathbb R^2-\{\bf 0\} $, with parameter interval $[0,2 \pi]$, with $\Gamma(0)=\Gamma(2 \pi)$, such that the segments $[\gamma(t),\Gamma(t)]$ do not contain $\bf 0$ for any $t \in [0,2 \pi]$. Prove that $$ \int_\Gamma \eta =2\pi $$ where $\eta=\dfrac{y dx-x dy}{x^2+y^2}$.
I need to prove that using the fact that $\eta=d \left( \arctan \frac{y}{x} \right)$ for $x \neq 0$, and $\eta=\left( -\arctan \frac{x}{y} \right)$ for $y \neq 0$. (In less rigorous terms $\eta=d \theta$).
 A: The union of the segments $[\gamma(t),\Gamma(t)]$ $\ (0\leq t\leq 2\pi)$ is a compact set $K$ that does not contain the origin, therefore all points in $K$  have a distance $\geq \delta$ from the origin, for some fixed $\delta>0$.
It follows that one can choose an $N>0$ large enough such that with
$$t_k:={2k\pi\over N}\qquad(0\leq k\leq N)$$ and $\gamma_k$, $\Gamma_k$ being the pieces of $\gamma$ and $\Gamma$ corresponding to the interval $[t_{k-1},t_k]$, the path
$$\sigma_k:=\Gamma_k+[\Gamma(t_k),\gamma(t_k)]-\gamma_k+[\gamma(t_{k-1},\Gamma(t_{k-1})]$$
stays in a halfplane at distance ${\delta\over2}$ from the origin. In this halfplane one can choose a continuous real-valued  representative of the argument $\arg$. It follows that
$$\int\nolimits_{\sigma_k}\eta =0\qquad(1\leq k\leq N)\ .$$
Summing over $k$  we therefore obtain
$$\int\nolimits_{\Gamma}\eta -\int\nolimits_{\gamma}\eta=0\ .$$
Since the second integral obviously has value $2\pi$ the statement follows.
Remark: I purposely didn't bring $\arctan$ into the picture. The right function to work with here is $\arg:\ \dot{\mathbb R}^2\to {\mathbb R}/(2\pi{\mathbb Z})$.
