$\omega = x^2dx + xydy + xzdz$ over $S^2 = \{(x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^2 = 1\}$ Verify if the following differential forms over $S^2$ are closed and/or exact:
$\omega_1 = x^2dx + xydy + xzdz$
$\omega_2 = xdy - ydx$
$\omega_3 = zdxdy - ydxdz + xdydz$.
What I have done: since  $F(x,y,z) = x^2 + y^2 + z^2 - 1 = 0$, $\partial F = 0$, is that right? Thus $xdx +ydy +zdz = 0$. $\omega_1 = x(xdx + ydy +zdz) = 0$ so $\omega_1$ is exact and closed. I couldn't do much more on the other two.
 A: Your idea for $\omega_1$ is correct. To make it rigorous, recall that at each $p \in \mathbb S^2$, we have $T_p \mathbb S^2 = \{ v\in \mathbb R^3 : v\cdot p = 0\}$. So for all $v\in T_p\mathbb S^2$ we have
$$\omega_1 (v) = x (v\cdot p) = 0$$
and so $\omega_1 |_{\mathbb S^2} = 0$. In particular it is both exact and closed. 
For $\omega_2$, using the idea from the comment, we have that 
$$d\omega_2 = 2 dx dy \Rightarrow d(\omega_2|_{\mathbb S^2}) = (d\omega_2)|_{\mathbb S^2} = dxdy|_{\mathbb S^2}.$$
At the point $p = (0,0,1)\in \mathbb S^2$, we have 
$$d(\omega_2|_{\mathbb S^2}) (v_1, v_2)=  2dxdy (v_1, v_2) = 2 \neq 0,$$
where $v_1 = (1,0,0)$ and $v_2 = (0,1, 0)$. Thus $d(\omega_2|_{\mathbb S^2})\neq 0$ and so $\omega_2|_{\mathbb S^2}$ is not closed. Thus it is also not exact.
For $\omega_3$, note that as $\mathbb S^2$ is two dimensional and $\omega_2$ is a two form, then $\omega_3|_{\mathbb S^2}$ is always closed. To show that $\omega_3|_{\mathbb S^2}$ is not exact, note that by Stokes' theorem,
$$\int_{\mathbb S^2} \omega_3 = \int_{\partial B} \omega_3 = \int_B d\omega_3 = 3\int_B dxdydz = 3\text{Vol}(B) \neq 0.$$
This implies that $\omega_3|_{\mathbb S^2}$ is not exact (try to show that using Stokes' theorem again). 
