I have tried to compute the "classical" surface element in polar coordinates for volume integration (i.e. $dx\ dy=r \ dr\ d\theta$) through this method: $$ \left\{ \begin{array}{c} x=r \cos \theta\\ y=r \sin\theta \\ \end{array} \right. = \begin{cases} dx=-r \sin\theta \ d\theta +dr \cos\theta \\[2ex] dy=r \cos\theta \ d\theta +dr \sin\theta \end{cases}$$
And then multiply the two differential elements (and simply the $(dr)^{2}$ and $(d\theta)^{2}$ to $0$). But I always seem to end up with $dx\ dy=r(\cos(\theta)^{2}-\sin(\theta)^{2})\ dr\ d\theta$ instead of$+$ as needed to get $r\ dr\ d\theta$. Is this a problem of assumptions or algebra mistakes?
When the variables are changed between $x,y$ and $u,v$, your $dx dy$ comes from the area of a rectangle, but it is mapped to a region that is not a rectangle. So instead of doing $du dv$, you will have to do $|d\vec{u}\times d\vec{v}|$, the area of a parallelogram, which gives you the jacobian. In fact, here is the formula,
$$dx dy =|\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}| du dv.$$