In the Cartesian plane, the area of a rectangle formed by the vectors ${\mathbf{R}}_x\=\mathrm{dx} \mathbf{i}$ and ${\mathbf{R}}_y\=\mathrm{dy} \mathbf{j}$ is $∥{\mathbf{R}}_x\times {\mathbf{R}}_y∥\=\mathrm{dx} \mathrm{dy}$, where to compute the cross product, zeros are added as a third component in each vector.

If $\mathbf{R}\=\left[\begin{array}{c}r \cos \(\mathrm{\θ}\)\\ r \sin \(\mathrm{\θ}\)\\ 0\end{array}\right]$ is the position vector under a transformation to polar coordinates, vectors tangent to the edges of a transformed rectangle would be given by the vectors

 $\frac{\partial \mathbf{R}}{\partial r} \mathrm{dr}\=\left[\begin{array}{c}\cos \(\mathrm{\θ}\)\\ \sin \(\mathrm{\θ}\)\\ 0\end{array}\right] \mathrm{dr}$    and    $\frac{\partial \mathbf{R}}{\partial \mathrm{\θ}}d\mathrm{\θ}\=\left[\begin{array}{c}-r \sin \(\mathrm{\θ}\)\\ r \cos \(\mathrm{\θ}\)\\ 0\end{array}\right] d\mathrm{\θ}$ 

The area of the deformed rectangle, nearly a parallelogram for $\mathrm{dr}$ and $d\mathrm{\θ}$ "small," is then

$∥\frac{\partial \mathbf{R}}{\partial r} \mathrm{dr}\times \frac{\partial \mathbf{R}}{\partial \mathrm{\θ}} d\mathrm{\θ}∥$ = $∥0 \mathbf{i}\+0 \mathbf{j}\+r \mathrm{dr} d\mathrm{\θ} \mathbf{k}∥$ = $r \mathrm{dr} d\mathrm{\θ}$