Problem with Stokes' theorem: how can I apply it?

Question

Let $\gamma$ be a smooth curve in $M:= \mathbb{R}^2 \setminus \{(0,0)\}$ very close to the origin that goes once around the origin and assume that the image of $\gamma$, which we denote by $N$, is a submanifold of $M$. Endow $N$ with the counterclockwise orientation. Compute $\int_N i^* \alpha$ where $\alpha = \frac{xdy-ydx}{x^2 + y^2}$ and $i: N \to M$ is the inclusion.

Attempt: Not entirely sure how to begin. I'm sure I must apply Stokes' somehow, but I'm not sure how. Maybe I will need to add some extra cirkles or something like that.

Answer 1

Assuming that $\gamma$ is a smooth, closed non-self-intersecting curve, given the standard orientation, then $\int_N i^* \alpha$ is just integration over the image of $\gamma$. And since $0\notin \text{im}\ \gamma$, compactness of $\text{im}\ \gamma$ gives us an $r>0$ such that $C\subseteq \text{im}\ \gamma$ where $C=\{(x,y)=(r\cos t,r\sin t):0\le t<2\pi \}.$ And now Green's theorem says $\int_N i^* \alpha=\int_Ci^*\alpha=2\pi.$

Answer 2

Since $N$ is just the image of $\gamma$, we can compute the integral over $N$ via a parameterization using $\gamma$. For our parameterization, use $\gamma(t) = (x(t),y(t))= (\cos(t),\sin(t))$ for $t\in[0,2\pi)$. Noting that $x(t)^2+y(t)^2 = 1$, we then compute the pullback: $$ i^*\alpha = i^*\left(\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right) = i^*\left(\frac{-y}{x^2+y^2}\right)i^*(dx)+i^*\left(\frac{x}{x^2+y^2}\right)i^*(dy) \\= -\sin t\cdot d(\cos(t))+\cos t\cdot d(\sin(t)) = \sin^2t+\cos^2t=1. $$ Therefore, $$\int_Ni^*\alpha = \int_0^{2\pi}1dt = 2\pi.$$