Volume element in different coordinate system

[teddd]

Very simple question:

Let $x^0,x^1,...,x^n$ be some fixed coordinate system, so that the infinitesimal volume element is $dV=dx^0dx^1...dx^n$.
Then any change to a new (primed) coordinate system $x^{0'},x^{1'},...,x^{n'}$ transforms the volume to $$dV=\frac{\partial (x^0,x^1,...,x^n)}{\partial (x^{0'},x^{1'},...,x^{n'})}dx^{0'}dx^{1'}...dx^{n'}$$ where $\frac{\partial (x^0,x^1,...,x^n)}{\partial (x^{0'},x^{1'},...,x^{n'})}$ is the determinat of the jacobian of the transformation.

So let's try to do this in a concrete example: the transformation from cartesian $x,y$ to polar $r,\theta$ coordinates.
The Jacobian is simply $r$ and so i get to $dV=dxdy=rdrd\theta$.

Doing the math i get $$dV^{pol.}=r(\cos\theta dx+\sin\theta dy)(-\frac{1}{r}\sin\theta dx+\frac{1}{r}\cos\theta dy)$$since $dr=\cos\theta dx+\sin\theta dy$ and $d\theta=-\frac{1}{r}\sin\theta dx+\frac{1}{r}\cos\theta dy$.
But then:$$dV^{pol.}=-\cos\theta\sin\theta dx^2+\cos^2\theta dxdy-\sin^2\theta dxdy + \cos\theta\sin\theta dy^2$$ and is clear that $dV^{cart.}\neq dV^{pol.}$ !
Where is my mistake?? Thanks for your help guys!

 

[Petr Mugver]

You can't just multiply differentials.

Try using the formal rules

dx dx = dy dy = 0

and

dx dy = -dy dx

 

[HallsofIvy]

$$x= r cos(\theta)$$ $$y= r sin(\theta)$$

$$dx= cos(\theta)dr- r sin(\theta)d\theta$$

$$dy= sin(\theta)dr+ r cos(\theta)d\theta$$

But when you multiply differentials, you have to use "skew product" Petr Mugver mentions.

A good way of thinking about it is as vectors
$$dx= cos(\theta)dr\vec{i}- r sin(\theta)d\theta\vec{j}$$
$$dy= sin(\theta)dr\vec{i}+ r sin(\theta)d\theta\vec{j}$$

Now the product is the cross product of the two "vectors".

$$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ cos(\theta) & - rsin(\theta) & 0 \\ sin(\theta) & rcos(\theta) & 0 \end{array}\right|= (rcos^2(\theta)+ rsin^2(\theta))drd\theta\vec{k}= r drd\theta \vec{k}$$