Differential Forms and Area I think I'm just misunderstanding something here, but in $\mathbb{R}^2,$ there exists a $1$-form (in fact infinitely many) $\omega$ such that for any region $S,$ we have $\int_{\partial S} \omega = \textrm{area}(S).$ Can we say the same on the sphere in $\mathbb{R}^3$? From what I can see, the answer is no, as the area form on the unit sphere is given by $x dy \land dz + y dz\land dx + zdx\land dy,$ which is not exact (or even closed). However, thinking about the punctured sphere (i.e. the sphere minus the north pole), it seems that such a form should exist since it is diffeomorphic to $\mathbb{R}^2.$ Is this line of reasoning correct?

By Stokes' theorem we have $\int_{\partial S} \omega = \int_S d \omega$, so the problem is whether a given volume form is exact, or equivalently whether it's zero in top de Rham cohomology. On a compact oriented Riemannian manifold, a volume form is never exact (in fact it generates top de Rham cohomology), but in the noncompact case this need not be true. 
$S^2$ is compact but $S^2 \setminus \{ 0 \}$, like $\mathbb{R}^2$, is noncompact and in fact contractible, so its second de Rham cohomology is trivial. 
(What you've written by itself is not enough because the diffeomorphism between the punctured sphere and $\mathbb{R}^2$ you have in mind doesn't respect the volume forms you have in mind.) 
