This question came up on a preliminary exam:
Define $$g(s,t)=(x(s,t),y(s,t))=(\cos(s)+\cos(t),\sin(s)+\sin(t)),$$ on the region $-\pi<s<\pi$, $s<t<s+\pi$. (The function $g$ is one-to-one on this region). Suppose $\omega$ is a $1$-form defined on the image of $g$ such that $\omega(g(s,t))=-\sin(t)\,dx+\cos(t)\,dy$. Show that for any region $R$ in the image of $g$ whose boundary $\partial R$ is a closed curve, we have $$\int_{\partial R}\omega = \pm\text{Area}(R).$$
My reaction to this problem is that if we knew $d\omega=1$, then we could use Stokes' theorem and be done (the $\pm$ is probably just because $\partial R$ doesn't have a defined orientation). But beyond that, I think I am making some mistake with the notation. If I simply try to substitute in the $s,t$-expressions for $x$ and $y$, to make the expression for $\omega$ depend only on $s$ and $t$, I get $\omega=dt+\cos(s-t)\,ds$.
(Related?: You can't generally integrate $\omega$ over $\partial R$, right? You need to pull back by $g$; my impression is that this is the formality behind the "substitution", but maybe I'm missing something serious.)
Eric, first of all, $\omega$ is a $1$-form on the image of $g$ and $R$ is a region contained in the image of $g$, so it certainly makes sense to integrate $\omega$ over $\partial R$.
Here's what I suggest:
(1) Compute $g^*\omega$ (you should get $\cos(s-t)\,ds + dt$).
(2) Compute $d(g^*\omega)$ and $g^*(dx\wedge dy)$. How do they compare?
(3) Using the fact that $g$ is one-to-one and $g^*$ is likewise, what can you now conclude?