If $S$ is a compact surface oriented by a 2-form $\eta$. Then $\eta$ is not an exact form. In fact, if $\exists \omega$ a 1-form s.t. $\eta = d \omega$, we'd have by Stoke's Theorem and $\partial S = \emptyset$:
$$ \int_S \eta = \int_S d\omega = \int_{\partial S} \omega = \int_{\emptyset} \omega = 0$$
Then $\eta_p = 0$ for some $p \in S$ what contradicts the fact of $S$ beeing oriented by $\eta$.
I'm looking for an example showing that the result is false if $S$ isn't compact.
Thank you
Take for instance $\mathbb{R}^{2}$ with volume form $dx\wedge dy=d(xdy)$.
In fact, the second de Rham cohomology group of $\mathbb{R}^{2}$ is trivial, which shows that any differential form of top degree (which is automatically a closed form) is exact.