Consider a change of variables in two dimensions \begin{equation} x=x(u,v) \qquad y=y(u,v) \end{equation} This is just a special case of a parametric surface!

To find the surface element of such surfaces, first foliate the surface with curves along which either $u$ or $v$ is constant. We have \begin{equation} d\rr = \Partial{\rr}{u}\,du + \Partial{\rr}{v}\,dv \end{equation} and only one of these terms is nonzero along each such curve. The surface element can now be obtained as \begin{equation} d\AA = d\rr_1 \times d\rr_2 = \left( \Partial{\rr}{u} \times \Partial{\rr}{v} \right) \, du \, dv \end{equation} We have \begin{equation} \Partial{\rr}{u} \times \Partial{\rr}{v} = \left| \begin{matrix} \xhat& \yhat&  \zhat \cr \Partial{x}{u}& \Partial{y}{u}& 0\cr \Partial{x}{v}& \Partial{y}{v}& 0\cr \end{matrix} \right| = \zhat \left| \begin{matrix} \Partial{x}{u}& \Partial{y}{u}\cr \Partial{x}{v}& \Partial{y}{v}\cr \end{matrix} \right| \end{equation}

This determinant is important enough to have its own name; we define the Jacobian of the transformation from $u,v$ to $x,y$ to be 1) \begin{equation} \Jacobian{x}{y}{u}{v} = \left| \begin{matrix} \Partial{x}{u}& \Partial{y}{u}\cr \Partial{x}{v}& \Partial{y}{v}\cr \end{matrix} \right| = \left| \begin{matrix} \Partial{x}{u}& \Partial{x}{v}\cr \Partial{y}{u}& \Partial{y}{v}\cr \end{matrix} \right| \end{equation}

Since $d\AA=dx\,dy\,\zhat$ for any (part of) the plane (oriented upward), we obtain \begin{equation} dx\,dy = \Jacobian{x}{y}{u}{v} \> du \, dv \end{equation} Unlike standard usage in multivariable calculus, we do not take the absolute value of the Jacobian determinant; we care about the relative orientation of the two sets of coordinates.

Not surprisingly, when this procedure is applied to the polar coordinate transformation \begin{equation} x=r\cos\phi \qquad y=r\sin\phi \end{equation} we get (check this yourself!) \begin{equation} \Jacobian{x}{y}{r}{\phi} = r \end{equation} so that, as we already knew, $dx\,dy=r\,dr\,d\phi$.

In practice, one is often given $u$ and $v$ in terms of $x$ and $y$, not the other way around. Rather than solving for $x$ and $y$, which can be difficult, there is a simpler way. Inverting the roles of the two sets of variables in the derivation above, we must have \begin{equation} du\,dv = \Jacobian{u}{v}{x}{y} \> dx\,dy \end{equation} from which we can conclude that \begin{equation} \Jacobian{x}{y}{u}{v} = {1\Bigg/\Jacobian{u}{v}{x}{y}} \end{equation}