Apologies for perhaps a very trivial question, but I'm slightly doubting my understanding of Jacobians after explaining the concept of coordinate transformations to a colleague.
Basically, as I understand it, the Jacobian (intuitively) describes how surface (or volume) elements change under a particular coordinate transformation. In essence, they characterise the difference in how area (or volume) is measured in the two different coordinate systems (I think this is right?!)
When one (informally) derives the Jacobian (aka Jacobian determinant) corresponding to a coordinate transformation the usual approach is to consider a mapping between two coordinate systems (sticking to $2D$ for simplicity), $(u,v)$ and $(x,y)$ respectively, defined by $$x=x(u,v)\, , \;\; y=y(u,v)$$ and then consider a square in the $uv$ plane whose area is given by $\Delta u\Delta v$. One then maps this area into the $xy$ plane, and to a linear approximation this corresponds to a parallelogram with sides $\left(\frac{\partial x}{\partial u}\Delta u\, ,\, \frac{\partial y}{\partial u}\Delta u\, ,\,0\right)$ and $\left(\frac{\partial x}{\partial v}\Delta v\, ,\, \frac{\partial y}{\partial v}\Delta v\, ,\,0\right)$. The corresponding area in the $xy$ plane can then be approximated as $$\Delta A\approx \bigg\lvert\left(\frac{\partial x}{\partial u}\Delta u\, ,\, \frac{\partial y}{\partial u}\Delta u\, ,\,0\right)\times\left(\frac{\partial x}{\partial v}\Delta v\, ,\, \frac{\partial y}{\partial v}\Delta v\, ,\,0\right)\bigg\rvert =\bigg\lvert\left(\frac{\partial x}{\partial u}\, ,\, \frac{\partial y}{\partial u}\, ,\,0\right)\times\left(\frac{\partial x}{\partial v}\, ,\, \frac{\partial y}{\partial v}\, ,\,0\right)\bigg\rvert\Delta u\Delta v$$ We then define the Jacobian as $J(u,v)=\left(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}\right)$ such that $$\bigg\lvert\left(\frac{\partial x}{\partial u}\, ,\, \frac{\partial y}{\partial u}\, ,\,0\right)\times\left(\frac{\partial x}{\partial v}\, ,\, \frac{\partial y}{\partial v}\, ,\,0\right)\bigg\rvert\Delta u\Delta v =\bigg\lvert\left(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}\right)\bigg\rvert\Delta u\Delta v =\lvert J(u,v)\rvert\Delta u\Delta v$$ Taking the infinitesimal limit the approximation becomes exact and we have that $$dA=\lvert J(u,v)\rvert dudv$$ This procedure can then be extended to higher dimensional cases.
Now, assuming that I've explained the above correctly, my issue is that often in textbooks this relation is written as $$dxdy=\lvert J(u,v)\rvert dudv$$ Of course, in the derivation above where we considered finite changes in coordinates, the area in the $xy$ plane is not simply $\Delta x\Delta y$ (since the map transforms the square in the $uv$ plane to a parallelogram in the $xy$ plane to linear approximation), but when we take the infinitesimal limit we then express the infinitesimal area element as $dxdy$. Is this simply because, in the $xy$ plane we are integrating over the independent variables $x$ and $y$ and so we simply denote the area element as $dxdy$, and as such the infinitesimal area elements in the $uv$ and $xy$ planes are related as $$dxdy=\lvert J(u,v)\rvert dudv$$ I maybe getting myself into a fuss over nothing, but I just want to clarify things before I give out false/incorrect information.
There is no requirement that $dx \, dy$ corresponds to the geometrical area element. Suppose the geometrical area element $dA = f(x,y) \, dx \, dy$, then let $g(u,v) = (x(u,v), y(u,v))$, and $dA$ can be written in the $u,v$ coordinates as
$$dA = (f \circ g)(u,v) J(u,v) \, du \, dv$$
I have purposefully not used the absolute value here, as the map $g$ could change the orientation of the surface; using absolute value would lose information. Granted, in many cases this information is not used or needed.
The relation between $dx \, dy$ and $J(u,v) \, du \, dv$ remains true regardless, but again, there is still the arbitrary factor $f$ that relates $dx \, dy$ to the true geometrical area element.
From another point of view, the factor $f$ in the area element could be considered part of the integrand instead, so I expect this is why that factor is often ignored: what is the "true" area element is somewhat arbitrary, or perhaps best dictated by the problem.