Integral of Differential 1-form Let $\omega$ be the closed $1$-form 
$\omega = \frac{xdy - ydx}{x^2 + y^2}$ and let $S$ be the unit circle and let $C = \{(x,y) : (x-3)^2 +y^2 = 1\}$.
I'm trying to find $\int _S \omega$ and $\int _C \omega$.  I believe I should use Stokes' Theorem to solve these integrals, especially the second one, but am not exactly sure how to go about this.
I parametrize $S$ by, $\phi (x,y) = (cos(\theta), sin(\theta))$, and I know that $Area(S) = \pi$
 A: For the second integral, $\int_C \omega$, note that the integrand $\omega=\frac{x dy - y dx}{x^2+y^2}$ is defined on the interior of the circle. This means that we can apply the Poincaré lemma, which says that all forms on a contractible manifold ar exact. Hence there is some 2-form $\omega'$ such that $d\omega'=\omega$. Let's call the closure of the interior of the circle for $D$, so that $\partial D = C$. Then by Stokes's theorem
$$
\int_C \omega = \int_{\partial D} \omega = \int_D d \omega = \int_D d(d \omega')=?
$$
For the second one, you are right that you need to parametrize. If we use the parametrization $\theta \mapsto (\cos \theta, \sin \theta)$, then the integral takes the form
$$
\int_C \omega = \int_0^{2\pi} \frac{\cos \theta d (\sin \theta) - sin \theta d(\cos \theta)}{\cos^2 \theta + \sin^2 \theta} = \int_0^{2\pi} d \theta = 2 \pi.
$$
It is interesting what is happening here: In the first case you get zero, somehow because calculating the integral is homotopy invariant. In the second case, the presence of a singularity in the integrand forbids you from applying the Poincaré lemma, but at the same time makes things more interesting. This proves that $\omega$ is not exact, meaning that $H^1_{dR}(\mathbb R^2 \backslash \{ 0\})\neq 0$. This also hints at much of complex analysis, where computing residues (which is what we've just done) is a big deal.
(after having read this answer, you should close your browser window, and go through the calculations yourself! Reading other people's answers help a lot, but doing the calculations yourself helps even more.)
