## Change of variables, from dxdy to dudv Suppose we have a function $f(x, y)$ and we rewrite it as $f(u(x, y), v(x, y))$, i.e. perform a change of variables, then the caculation of area using double integral can be carried out with these new variables $u, v$. But we have to figure out the relation between $dudv$ and $dxdy$. Here is the thing, with the variables $x, y$, $dxdy$ is the area of the parallelogram between the vectors $(dx, 0)$ and $(0, dy)$, under the new variables, these vectors are transformed into two new vectors $(u_x dx, v_x dx)$, $(u_y dy, v_y dy)$, and $dudv$ is the area of the parallelogram between these two new vectors. $\begin{array}{cc} (dx, 0) & (u_x dx, v_x dx) \\ (0, dy) & (u_y dy, v_y dy) \end{array}$ To get the area of this parallelogram, we need the determinant of this matrix: \begin{align*} M &= \begin{pmatrix} u_x dx & u_y dy \\ v_x dx & v_y dy \end{pmatrix} \\ &= \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix} \begin{pmatrix} dx & 0 \\ 0 & dy \end{pmatrix} \\ &= J \begin{pmatrix} dx & 0 \\ 0 & dy \end{pmatrix}, \end{align*} where $J$ stands for the Jacobian matrix of $f$. So the determinant is \begin{align*} \det(M) &= \det(J) dxdy \end{align*}

Remember in $R^2$, abs of determinant is equal to area of the parallelogram between these two vectors.