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

$\iint f(x,y)dxdy \overset{?}{=} \iint f(\mathbf{r}(u,v))\left\|\begin{bmatrix}x_u & x_v\\y_u & y_v\end{bmatrix} \right\|dudv$

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    $\begingroup$ Here's a relevant thread. math.stackexchange.com/questions/267267/… $\endgroup$ – littleO Aug 10 '16 at 6:21
  • $\begingroup$ The change of variables formula is a hard theorem, and your derivation is nowhere even close to a proof. I also can't tell what you're really asking. But maybe this is what you want: say $dx=x_udu+x_vdv$, $dy=y_udu+y_vdv$. Then $$dxdy=(x_udu+x_vdv)(y_udu+y_vdv)=x_uy_udu^2+x_uy_vdudv+x_vy_udvdu+x_vy_vdv^2$$ But in the algebra of differential forms, $du^2=dv^2=0$ and $dvdu=-dudv$. Therefore $dxdy=(x_uy_v-y_ux_v)dudv$. $\endgroup$ – symplectomorphic Aug 10 '16 at 6:26
  • $\begingroup$ Yes proof is not the right word. I meant more along the lines of an intuitive derivation or train of thought for how to integrate the original integral if I change coordinate systems $\endgroup$ – ApplePi Aug 10 '16 at 6:33
  • $\begingroup$ @ApplePi: then what exactly is your question? You say "The area differentials for the two coordinate systems being equal does not seem immediately obvious to me." But you just "showed" that $dA=\det J\,dudv$. What else could $dA$ be but $dxdy$? $\endgroup$ – symplectomorphic Aug 10 '16 at 6:36
  • $\begingroup$ dA=det J dudv in the uv plane. The whole problem with changing coordinate systems is that there's a stretching of space.I found an expression for dA in the uv plane for a small change in u and v, but I don't see why that area is necessarily equal to dxdy in the xy plane as the change of coordinate systems has stretched the space $\endgroup$ – ApplePi Aug 10 '16 at 6:54

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