I am studying for first time Smooth manifolds and I have some issues understunding the wedge product and the exterior differential since my teacher does not provide examples. For example, one of my excercises asks to compute $\omega \land \eta, \omega \land \alpha $ and $\omega \land \eta \land \alpha$ where
$$\omega = x \, dx - y \, dy \\ \eta = z \, dx \land dy + x \, dy \land dz \\\ \alpha = z \, dy$$
If I am not wrong, since the Wedge product is bilinear, associative and satisfies $dx \land dx = 0$ (and for any variable) then
$$\omega \land \eta = x^2 \, dx \land dy \land dz \\ \omega \land \alpha = xz \, dx \land dy \\ \omega \land \eta \land \alpha = 0$$
Is this fine? Also, I have not very clear how to calculate $d\omega, d\eta, d\alpha$ so any posible explanation would be very appreciated.
We apply the definition of the exterior derivative $d$ of $k$ forms.
Let $f= \sum f_{i_1,...,i_k}dx^{i_1}\wedge\cdots dx^{i_k}$.
Then $df=\sum df_{i_1,...,i_k}\wedge dx^{i_1}\wedge\cdots dx^{i_k} =\sum( \sum_j \frac{\partial f_{i_1,...,i_k}}{\partial x^j}dx^j)\wedge dx^{i_1}\wedge\cdots dx^{i_k}$
$d\omega= d(xdx-ydy)=dx\wedge dx-dy\wedge dy= 0$ as $dx\wedge dx=dy\wedge dy=0$
$d\eta=d(zdx\wedge dy+xdy\wedge dz)=dz\wedge dx\wedge dy + dx\wedge dy\wedge dz=2dx\wedge dy\wedge dz $
as $dz\wedge dx\wedge dy=-dx\wedge dz\wedge dy=dx\wedge dy\wedge dz$
Using these examples, try computing $d\alpha$ on your own.