I'm having trouble understanding how exactly the Jacobian is transforming volumes.
To understand this in more detail... if I have a two-dimensional integral and I'm trying to change the integration from (x,y) coordinates to (u,v), I want to re-express dx*dy as |J|dudv.
Let $J(x_0) = [a b; c d]$ be the Jacobian from (x,y) space to (u,v) space at $x_0$ and say we're mapping an orthotope in (x,y) space with volume dx*dy whose columns are [dx 0; 0 dy] to (u,v) space. So we have:
C = [a b; c d] * [dx 0; 0 dy] = [adx bdy; cdx ddy], and the determinant of this is (ad-bc)dxdy = |J|dxdy.
So that's fine, but how do I show that dudv = |J|dxdy? I feel like there's some connection here I'm missing. I understand that the columns of C represent the axes of the parallelipiped in (u,v) space, but I don't understand how exactly to connect this to du*dv.
Any thoughts?
Thanks.
If we have vector-valued variables $\mathbf{u},\,\mathbf{v}$ of the same dimension $n$ (in your case $n=2$), the chain rule relating their infinitesimals is $\text{d}u_i=\sum_{j}J_{ij}\text{d}v_j$ with $J_{ij}:=\frac{\partial u_i}{\partial v_j}$, the entries of the Jacobian matrix $J$. In vector notation, we can just write $\text{d}\mathbf{u}=J\text{d}\mathbf{v}$. Some function $f(J)>0$ should exist with $\text{d}^n\mathbf{u}=f(J)\text{d}^n\mathbf{v}$. If we transform twice, say $\text{d}\mathbf{u}=J_1\text{d}\mathbf{v},\,\text{d}\mathbf{v}=J_2\text{d}\mathbf{w}$, it becomes clear we need $f(J_1 J_2)=f(J_1) f(J_2)$. This, combined with the multilinearity of $f$ (because if the $u_i$ are multiplied by constants we'd expect to absorb these factors into $f$), proves $f=|\det J|$. Indeed, if $f$ were allowed a sign in such a way it changes sign when swapping two variables, we'd have enough properties to prove $f=\det J$.