I know that what follows is wrong and I know one needs to use the jacobian to calculate a new volume element $dV$ after a change of variables. However, I don’t understand why the following goes wrong:
Say we have the following change of variables (i.e. we could be doing a multivariable integral): $$ x_i = x_i(u_0,...,u_n), $$ where as usual $x_i$ is the $i$th component of some vector $x$ (same for $u$). One learns that the differential of $x_i$ is given by $$ dx_i= \frac{\partial x_i}{\partial u_j}du_j, $$ where summation is implied. Then, if $dV$ in $x$ is given by $$ dV=\Pi_i dx_i $$ why cant we say $$ dV = \Pi_i dx_i = \Pi_i (\frac{\partial x_i}{\partial u_j}du_j). $$
My instinct tells me the problem might lie in the definition of $dV=\Pi_i dx_i$, if incorrect please explain why.
Thanks!
If you are familiar with differential forms the answer is simply that the volume element is a decomposable top form, i.e. the wedge product of $n$ 1-forms. In Euclidean space with the usual coordinates $x^i$ $d V= dx^1\wedge \cdots \wedge dx^n$ and if you plug in the transformation law $dx^i \mapsto \sum _j \partial x^i /\partial u ^j du^j$ you get out a factor $\det (\partial x^i /\partial u^j)$. Duh...
More intuitively and working in 2 dimensions for ease, suppose you have a small surface area of sides $dx$, $dy$ centred at $(x,y)$. Its "volume" (well, area here) is $dx \times dy$, where $\times$ is just good old standard multiplication. Now suppose you work in polar coordinates and the surface element is centred at $(r,\theta)$, and it "extends" by $d\theta$ on the angular part and by $dr$ in the radial one. By drawing a figure you should be able to convince yourself that you have a small portion of an annulus whose "volume" is (approximatively) $rd\theta \times dr$.
Reasoning in terms of volume it should be possible to get the general expression without making use of the language of differential forms, but I am too lazy to try right now. Differential forms are a convenient language as the determinant comes out automatically so you don't have to think hard about the geometry described by the coordinates you are using.
Finally, it may be helpful to point out why Cartesian coordinates behave so nicely: it is because the $dx^i$ are orthonormal stuff - where stuff and hence orthonormal can be made precise but the intuitive meaning should be clear, after all the Euclidean metric is $dx^2+dy^2 +\cdots$. Any other orthonormal stuff (coframe) $e^i$ would lead to a similarly nice volume element $e^1\wedge\cdots\wedge e^n$. Since in 2D $dx^2+dy^2=dr^2+r^2 \sin^2\theta d\theta^2$, an orthonormal coframe is $e^1=dr$, $e^2 = r \sin\theta d\theta$.
If a manifold admits an orthonormal frame which is also coordinate, $e^i=dx^i$, near every point then that manifold is flat. In other words coordinate and orthonormal are both nice properties but you usually cannot have both!