Let $\varphi, \psi$ and $\theta$ the following forms in $\mathbb{R}^{3}$. $$\begin{eqnarray*} \varphi & = & xdx - ydy,\\ \psi&=&zdx\wedge dy + xdy\wedge dz,\\ \theta & = & zdy. \end{eqnarray*}$$ Calculate $\varphi \wedge \psi$, $\theta\wedge\varphi\wedge\psi$, $d\varphi$, $d\psi$ and $d\theta$.
My attempt.
$\begin{eqnarray*} \varphi \wedge \psi & = & (xdx - ydy)\wedge(zdx\wedge dy + xdy\wedge dz)\\ & = & xzdx\wedge dx\wedge dy + x^2dx \wedge dy\wedge dz - yzdy\wedge dx \wedge dy - yxdy \wedge dy \wedge dz\\ & = & x^2dx \wedge dy\wedge dz - (-1)yzdy \wedge dy \wedge dx\\ & = & x^2dx \wedge dy\wedge dz \end{eqnarray*}$
$\begin{eqnarray*} \theta \wedge \varphi \wedge \psi & = & zdy \wedge (x^2dx \wedge dy\wedge dz)\\ & = & zx^2dy\wedge dx \wedge dy\wedge dz\\ & = & (-1)zx^2 dx \wedge dy \wedge dy \wedge dz\\ & = & 0. \end{eqnarray*}$
$\begin{eqnarray*} d\varphi & = & dx \wedge dx - dy\wedge dy\\ & = & 0. \end{eqnarray*}$
$\begin{eqnarray*} d\psi & = & dz \wedge dx \wedge dy + dx \wedge dy \wedge dz\\ & = & (-1)^2 dx \wedge dy \wedge dz + dx \wedge dy \wedge dz\\ & = & 2dx \wedge dy \wedge dz \end{eqnarray*}$
$\begin{eqnarray*} d\theta & = & dz\wedge dy\\ & = & (-1)dy \wedge dz\\ & = & -dy \wedge dz \end{eqnarray*}$
Is correct? Seems simples, but I'm not sure if I'm using correctly the properties.