TITLE: Compute the wedge product
QUESTION [4 upvotes]: This is my first time computing the wedge product, I am not sure if I have done it correctly as I do not have solutions. If I have gotten the answer right or am doing the right method please say. Also, am I applying the wedge product and using addition in the correct order? How could I have used brackets better?

Let $\alpha$ be the $1$-form and $\beta$ the $2$-form on $\mathbb{R}^3$ given by
  \begin{align*}
\alpha &= (x+y)dy+(x^2-y^2)dz\\
\beta &=zdx\wedge dy+xzdx\wedge dz.
\end{align*}
  Compute $\alpha\wedge\beta$.

My answer:
\begin{align*}
\alpha\wedge\beta(x,y,z) =&\ [(x+y)dy+(x^2-y^2)dz]\wedge[zdx\wedge dy+xzdx\wedge dz]\\
=&\ [(x+y)zdy\wedge dx\wedge dy]+[(x^2-y^2)xzdz\wedge dx\wedge dz]+[(x+y)xzdy\wedge dx\wedge dz]\\
& +[(x^2-y^2)zdz\wedge dx\wedge dy].
\end{align*}
We have $dx\wedge dx=dy\wedge dy=dz\wedge dz=0$, therefore we can cancel some terms and get
$$\alpha\wedge\beta(x,y,z)=(x+y)xzdy\wedge dx\wedge dz+(x^2-y^2)zdz\wedge dx\wedge dy.$$
Since $dy\wedge dx\wedge dz=-dz\wedge dx\wedge dy$, we get
$$\alpha\wedge\beta(x,y,z)=((x+y)xz+(y^2-x^2)z)dy\wedge dx\wedge dz =(xyz+y^2z)dy\wedge dx\wedge dz.$$
Any feedback would be greatly appreciated

REPLY [1 votes]: Your answer is correct. A couple of comments though.
One usually doesn't write $\alpha\wedge\beta(x, y, z)$, just $\alpha\wedge\beta$. 
As pointed out in the comments, one usually arranges the differential one-forms $dx, dy, dz$ in alphabetical order, though this is not necessary.
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