differential form and cylindrical coordinate Problem. 

If $r, \theta, z$ are the cylindrical coordinate functions on $\mathbb
> R^3$ , then $x = r\cos\theta, y = r\sin\theta, z = z$. Compute the
  volume element dx dy dz of $\mathbb R^3$ in cylindrical coordinates.
  (That is, express dx dy dz in terms of the functions $r, \theta, z$,
  and their differentials.)

My solution.
$dx = \cos\theta dr - r\sin\theta d\theta$
$dy = \sin\theta dr + r\cos\theta d\theta$
$dz = dz$
$$ dx\wedge dy\wedge dz = ... = rdrd\theta dz$$
Actually, these steps are from internet or answer page. I have no idea of how to connect $dxdydz$ and $drd\theta dz$. I don't know why $wedge$ operator is done. 
Additionally, I want to know how to convert spherical coordinate using 1-form.
 A: Begin with coordinate transformation:
$ ds^2 = g_{ab}dx^adx^b $
$ = g_{ab}\frac{\partial x^a}{\partial \zeta^{\alpha}}\frac{\partial x^b}{\partial \zeta^{\beta}} d\zeta^{\alpha}d\zeta^{\beta}$
$ = \gamma_{\alpha\beta} d\zeta^{\alpha}d\zeta^{\beta}$
where 
$ \gamma_{\alpha\beta}=g_{ab}\frac{\partial x^a}{\partial \zeta^{\alpha}}\frac{\partial x^b}{\partial \zeta^{\beta}} $
This relates the Cartesian metric to the cylindrical metric:
$ g_{ab}=
\begin{pmatrix}
    1       & 0 & 0 \\
    0       & 1 & 0 \\
    0       & 0 & 1
\end{pmatrix}$
$ \gamma_{\alpha\beta}=
\begin{pmatrix}
    1       & 0 & 0 \\
    0       & r^2 & 0 \\
    0       & 0 & 1
\end{pmatrix}$
The volume form is given as:
$dV = \sqrt{|det(\gamma_{\alpha\beta})|}d\zeta^{(0)} \land d\zeta^{(1)} \land d\zeta^{(2)} $
$= rd\zeta^{(0)} \land d\zeta^{(1)} \land d\zeta^{(2)} $
$= rdr \land d\theta \land dz $


*

*note that: $ Adx \land dy \land dz = Adxdydz $ in most shorthand.


Wedge products are used because they allow for one to construct area and volume elements from infinitesimal length elements $ds$. So $dx$ would be an infinitesimal length element, $dx \land dy$ would be an infinitesimal area, and so on. 
To convert to spherical coordinates, you would just redefine the second metric as:
$ \gamma_{\alpha\beta}=
\begin{pmatrix}
    1       & 0 & 0 \\
    0       & r^2 & 0 \\
    0       & 0 & r^2sin^2(\theta)
\end{pmatrix}$
with $ d\zeta^{(2)} = d\phi$
