I'm trying to derive the formula (?) for multivariate change of variables using vector calculus but am getting stuck on the final steps. I am hoping that someone can help me finish the derivation from where I'm stuck rather than provide an alternative derivation.
Show: $\iint f(x,y)dxdy = \iint f(u,v)\left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv$ where x=x(u,v) and y=y(u,v)
suppose we have $\iint f(x,y)dxdy$ and we want to integrate using (u,v) coordinates.
Let $\mathbf{r}(u,v)=\begin{bmatrix}x(u,v)\\y(u,v)\end{bmatrix}$, then the integral can be written $\iint f(\mathbf{r}(u,v))dxdy$
if we change u by some small amount du, we can approximate the change in $\mathbf{r}(u,v)$ by $\mathbf{r}_udu = \begin{bmatrix}x_u\\y_u\end{bmatrix}du$.
Likewise we can approximate the change in $\mathbf{r}(u,v)$ as $\mathbf{r}_vdv = \begin{bmatrix}x_v\\y_v\end{bmatrix}dv$ for some small change in v.
The area spanned by these two vectors can be found from the absolute value of the determinant, giving the area differential:
$dA = \left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv $
So we can approximate the volume under f of this chunk as:
$f(\mathbf{r}(u,v))\left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv$
From the above, I can see the jacobian and how it relates to change of variables. I like the way I arrived there. The problem is that I don't really understand why I would equate this with the corresponding volume in the (x,y) plane. The area differentials for the two coordinate systems being equal does not seem immediately obvious to me:
$f(x,y)dxdy \overset{?}{=} f(\mathbf{r}(u,v))\left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv $
Then to get the full area under f we would just need to integrate those volumes over their boundaries
$\iint f(x,y)dxdy \overset{?}{=} \iint f(\mathbf{r}(u,v))\left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv$