Let $\omega=z d x \wedge d y,$ a 2-form in $\mathbb{R}^{3}$. Let $M = \{(x,y,z) \in \mathbb{R}^3 : z = 1 + x^2 + y^2\}$. Determine whether the restriction of $\omega$ to M is exact. If so, construct a 1-form $\eta$ on M such that $\omega = d\eta$.
This is an old exam question, and I am not sure how I should approach it. If it is a one form I would try finding a closed curve and integrate the form over it, but with a 2-form I cannot think of any efficient tools to solve this.
On
Idea: brute force it. Suppose $\omega=d\tau$ for some $1$-form $\tau=fdx+gdy+hdz.$ Then,
$d\tau=(g_x-f_y)dx\wedge dy+(h_z-f_z)dx\wedge dz+(h_y-g_z)dy\wedge dz=zdx\wedge dy.$
Restricting to $M$, we get
$(g_x-f_y+2y(h_z-f_z)+2x(g_z-h_y))dx\wedge dy=(1+x^2+y^2)dx\wedge dy.$
Take $h=0,\ g(x,y,z)=x+\frac{1}{3}x^3,\ f(x,y,z)=-\frac{1}{3}y^3$
Parametrising $M$ by coordinates $x$ and $y$, $\omega$ restricted to $M$ is $\omega = (1+x^2+y^2)\,dx\wedge dy$. This equals $d\eta$, where $$ \eta = \frac{1}{2}\left(1+\frac{x^2+y^2}{2}\right)(x\,dy - y\,dx). $$ So $\omega$ restricted to $M$ is exact.
To see how one can find $\eta$: integrating $\omega$ over a region $x^2+y^2\le R$ gives $2\pi \left(\frac{R^2}{2}+\frac{R^4}{4}\right)$. If $\omega = d\eta$, by Stoke's theorem this must equal $\int_{r=R}\eta$, which suggests $\eta = \left(\frac{r^2}{2}+\frac{r^4}{4}\right)d\theta$. Converting back to Cartesian coordinates gives the result.