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If I want to compute the area element for a coordinate trasformation, let's say cartesian to polar \begin{align*} x &= r \cos \vartheta \\ y &= r \sin \vartheta \end{align*} to find $d A$ in the new coordinates I have to do \begin{equation*} d A = d x dy = |J|d r d\vartheta = r d r d\vartheta \end{equation*} But I asked myself, suppose I didn't know better, what would I do? Of course, instead of looking at the area element in the new coordinates, I would blindly compute \begin{align*} d x &= \cos \vartheta d r - r \sin \vartheta d\vartheta \\ d y &= \sin \vartheta d r + r \cos \vartheta d\vartheta \end{align*} and I would conclude (neglecting quadratic terms in $dr$ and $d\vartheta$) that \begin{equation*} d x d y = r \cos^2\vartheta d r d\vartheta - r \sin^2\vartheta d\vartheta d r \end{equation*} which is wrong. Considering $d x d y$ as an exterior product among forms I could still save the day because $dr d\vartheta = -d\vartheta d r$ but remaining in the "realm" of calculus 1 I am a bit lost. Why is this method not working?

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Actually, it involves cross product instead:

\begin{align*} \mathbf{r} &= \mathbf{r}(u,v) \\ dA &= \left| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right| \, du \, dv \end{align*}

which holds for surface too.

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    Are those 2d-vectors? I get the meaning though, the cross product gives me the area of the small parallelogram and that is the way to go. – 2017-02-20
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    You can always append an extra coordinate. It's a usual trick to find an area of a parallelogram spanned by vectors $\mathbf{a}$ and $\mathbf{b}$: $$A=|\mathbf{a} \times \mathbf{b}|$$ By the way, we can also denote angular velocity by using curl: $$\boldsymbol{\omega}=\frac{1}{2} \nabla \times \mathbf{r}$$ – 2017-02-20
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In general, for 3-dimensional curvilinear co-ordinates $u_i=\mathrm{constant}$, $i=1,2,3$ ; (it can be extended to n-dimensions), we can express any vector $\vec C$ as $$\vec C=h_1 \hat e_1+h_2 \hat e_2+h_3 \hat e_3$$ where $\hat e_i$ is the $\mathrm{i}^{\mathrm{th}}$ unit vector along the co-ordinate curve $u_i=\mathrm{constant}$ and $h_i$ is the scale factor of the $\mathrm{i}^{\mathrm{th}}$ unit vector $\hat e_i$.

Now, for such co-ordinates, a line element $\vec {dl}$ of $\vec C$ $= h_1 du_1+h_2 du_2+h_3 du_3$

And an area element $\vec {dA}$ is defined as $\vec {dA}= h_1 h_2 h_3\cdot (du_1)( du_2)(du_3)$

For planar polar co-ordinates, $u_1=r$ and $u_2=\theta$; and $h_1=1$ and $h_2=r$.

Nothing as such for $u_3$ as here the dimension is $2$.

So, in this case, $\vec {dA}= 1 \cdot r\cdot (dr)( d\theta)=\mathrm{r \,dr \,d\theta}$

Your procedure is wrong, since the area element is not defined in that way. This is the actual definition.