If $\omega=x dy - y dx$ is a $1$-form, how do I compute $d\omega$?

Question

If $\omega=x dy - y dx$ is a $1$-form, how do I compute $d\omega$?

I am having trouble understanding the definition of $1$-form and computing $d\omega$.

I know a $1$-form is an expression of the form $F(x,y)dx+G(x,y)dy$.

Answer 1

I know a $1$-form is an expression of the form $F(x,y)dx+G(x,y)dy$.

Good. You may also know that the differential of a function (zero-form) is a one-form: $$ dF = \frac{\partial F}{\partial x} \,dx + \frac{\partial F}{\partial y}\,dy $$ and that you can differentiate a basic one-form like so: $$ d(F\,dx) = dF \wedge dx $$ If you combine the previous two equations, and the two similar ones for $d(G\,dy)$, and the fact that the wedge product is skew-symmetric, you have all you need.

Answer 2

If $\omega=P dx+Q dy$, use the identity $\displaystyle d\omega= \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \ dxdy$.

So, if $\omega=x dy - y dx$, then we have $$d\omega=2 \ dxdy.$$