Compute the contraction of $i_\mathbb{X}\beta$ Question: Let $\beta=zdx\wedge dy+xzdx\wedge dz$, and let $\mathbb{X}$ be the vector field on $\mathbb{R}^3$ given by $\mathbb{X}=(0,-x,-1)$. Compute $i_\mathbb{X}\beta$, combining terms where possible.
Answer so far: $$i_\mathbb{X}\beta=i_\mathbb{X}(zdx\wedge dy+xzdx\wedge dz)=i_\mathbb{X}(zdx\wedge dy)+i_\mathbb{X}(xzdx\wedge dz)$$
I know that $i_\mathbb{X}(a\wedge b)=(i_\mathbb{X}a)\wedge b+(-1)^ka\wedge(i_\mathbb{X}b)$ for $a$ a $k$-form. In this case we are only dealing with the wedge product of $1$-forms so this formula will actually be: 
$$i_\mathbb{X}(a\wedge b)=(i_\mathbb{X}a)\wedge b-a\wedge(i_\mathbb{X}b).$$
Problem: I am stuck on the first part of the contraction:
$$i_\mathbb{X}(zdx\wedge dy)=i_\mathbb{X}(zdx)\wedge dy-zdx\wedge i_\mathbb{X}(dy)=z\cdot 0\wedge dy-zdx\wedge(-x).$$
I'm pretty sure this first part disappears $(=0)$, what happens to this second part?
Any hints would be greatly appreciated, but if I could get a worked answer I would be very greatful
 A: You can further simplify by using the fact that $i_{\mathbb{X}}$ is linear over smooth functions, i.e. $i_{\mathbb{X}}(f\omega) = fi_{\mathbb{X}}\omega$. To see this, note that $f$ is a $0$-form, so by the Leibniz property 
$$i_{\mathbb{X}}(f\omega) = i_{\mathbb{X}}f\wedge\omega + (-1)^0fi_{\mathbb{X}}\omega = fi_{\mathbb{X}}\omega$$ 
where we have used the fact that $i_{\mathbb{X}}f$ is a $-1$-form and therefore must be zero.
Therefore, $$i_{\mathbb{X}}(z\,dx\wedge dy) =  z\,i_{\mathbb{X}}(dx\wedge dy) = z(i_{\mathbb{X}}dx\wedge dy - dx\wedge i_{\mathbb{X}}dy).$$
As $dx$ is a one-form, $i_{\mathbb{X}}dx = dx(\mathbb{X}) = dx((0, -x, -1)) = 0$ - $dx$ is the one-form which returns the first component of the vector field. Likewise, as $dy$ is a one-form, $i_{\mathbb{X}}dy = dy((0, -x, -1)) = -x$ - $dx$ is the one-form which returns the second component of the vector field. Therefore
$$i_{\mathbb{X}}(z\,dx\wedge dy) =  z(i_{\mathbb{X}}dx\wedge dy - dx\wedge i_{\mathbb{X}}dy) = z(0\wedge dy - dx\wedge (-x)) = z(xdx) = xz\, dx.$$
Note, the last step of your working reduces to $xz\, dx$ once you realise that wedging with a $0$-form is just multiplication.

In general, if the $k$-form $\omega$ is written as the wedge product of $1$-forms, $\omega = \alpha^1\wedge\dots\wedge\alpha^k$, then 
$$i_{\mathbb{X}}\omega = \sum_{i=0}^k(-1)^{i-1}\alpha^i(\mathbb{X})\alpha^1\wedge\dots\wedge\alpha^{i-1}\wedge\alpha^{i+1}\wedge\dots\wedge\alpha^k.$$
This can be shown by repeated application of the Leibniz rule.
