Change of Variables via Line Element Differential vs. Jacobian If I had a double integral
$\int \int_R f(x,y) dxdy $
I would change variables to polar coordinates by expressing the position vector
$\vec{r} \ = \ \vec{r}(x,y) \ = \ x \vec{e}_x \ + \ y\vec{e}_y$
in terms of polar coordinates
$\vec{r} \ = \ \vec{r}(r,\theta) \ = \ r\cos(\theta) \vec{e}_x \ + \ r\sin(\theta)\vec{e}_y$
& use the fact that the line element
$d\vec{r} \ = \ \frac{\partial \vec{r}}{dr}dr \ + \ \frac{\partial \vec{r}}{d\theta}{d\theta} \ = \ ||\frac{\partial \vec{r}}{dr}||dr \ \vec{e}_r \ + \ ||\frac{\partial \vec{r}}{d\theta}|| {d\theta} \ \vec{e}_\theta \ = \ dr \ \vec{e}_r \ + \ r {d\theta} \ \vec{e}_\theta$
shows that 
$||\frac{\partial \vec{r}}{dr}||||\frac{\partial \vec{r}}{d\theta}|| dr d\theta \ = \ rdrd\theta$ 
is the area element & modify my integral accordingly. 
Similarly for any other change of variables between orthogonal coordinate systems. What is the relationship between this way of thinking and the Jacobian way of thinking when you are changing variables? I keep forgetting things when thinking of Jacobians yet I can re-derive on the spot any area/volume element I want doing the above, can somebody give me intuition on how & why the Jacobian is superior to the above & how the above is a special case of the Jacobian? If you think you can make it more interesting I'd love to see how you'd relate this to wedge products, though that's not needed.
 A: First lets consider the change in area when we use a coordinate independent transformation. We will consider a transformation from coordinates (x,y) to coordinates (u,v) via the following,
$$u = a x + b y$$
$$v = c x + d y$$
note that $a$,$b$,$c$, and $d$ are constants independent of $x$ or $y$.
Now consider a rectangle in the (x,y) system with edges defined by the vectors $\vec{l}$ and $\vec{w}$ with $ \vec{l} \cdot \vec{w} $=0 (orthogonal). 
For concreteness let the vectors be, $$ \vec{l} = l \vec{x}$$ $$\vec{w} = w \vec{y}$$.
The area of the $xy$ rectangle is given by the magnitude of the cross product between $\vec{l}$ and $\vec{w}$. We will denote this area $A_{xy}$.
$$ A_{xy} = \vert \vec{l} \times\vec{w} \vert = \vert \ lw \vec{z} \ \vert = lw$$
Now since $\vec{l}$ and $\vec{w}$ are vectors we know their components will change under coordinate transformation by the same rule that $x$ and $y$ follow. Therefore we can conclude that the transformed vectors will be,
$$ l'_u = a (l) + b (0) = al$$
$$ l'_v = c (1) + d (0) = cl$$
$$ w'_u = a (0) + b (w) = bw$$
$$ w'_v = c (0) + d (w) = dw$$
$$ \Rightarrow \vec{l'} = al \vec{u} + cl \vec{v} \qquad \vec{w'} = bw \vec{u} + dw \vec{v}$$ 
Notice that these are no longer orthogonal and in fact now define the edges of a parallelogram. We will now compute the area of this parallelogram in the $uv$ system.
$$ A_{uv} = \vert \vec{l'} \times \vec{w'} \vert = \vert \left( (al)(bw)-(cl)(dw) \right) \vec{\zeta} \vert = \vert ab-cd\vert lw  = \vert ab-cd\vert A_{xy}$$
$$ \Rightarrow \quad A_{uv} = J A_{xy} $$
Where $J= \vert ab-cd \vert $ is the conversion factor which converts $xy$ area into $uv$ area. 
Now this isn't a jacobian yet because our transformation was independent of the coordinates. However if our transformation is differentiable we can apply similar reasoning locally.
Let $f(x,y)$ and $g(x,y)$ be continuously differentiable functions of $x$ and $y$. Furthermore require that the change of coordinates defined below be invertible,
$$ u = f(x,y)$$
$$ v = g(x,y)$$
Let $(u_0,v_0)$ be the image of $(x_0,y_0)$ under this transformation and consider a neighborhood of $(x_0,y_0)$ small enough for the tangent plane approximations of $f$ and $g$ to be good. Then consider the following,
$$ u-u_0 = f(x,y) - f(x_0,y_0) \approx \frac{\partial f}{\partial x} (x-x_0) + \frac{\partial f}{\partial y} (y-y_0)$$
$$ v-v_0 = g(x,y) - g(x_0,y_0) \approx \frac{\partial g}{\partial x} (x-x_0) + \frac{\partial g}{\partial y} (y-y_0)$$
Now we have the rule for the transformation of displacements near $(x_0,y_0)$ which is the same as rule for transformations of vectors at that point. From this point on we apply the same reasoning above to arrive at the conversion factor for area.
The distinction between this approach and yours is that your transformations are orthogonal. This means that they locally preserve the angles between displacement vectors, so a small rectangle will be sent to a small rectangle instead of a parallelogram.
