I've come across this general integral $$S^{ij} = \oint_{\text{loop on the surface}} x^i dx^j - x^j dx^i \approx \text{Area Enclosed By The Loop.}$$
which, in $(x,y)$ coordinates is
$$S^{xy} = \oint_{\text{loop on the surface}} x dy - y dx \approx \text{Area Enclosed By The Loop.}$$
and this is easily seen to be $2\pi r^2$ when $x$ and $y$ are parameterized and integrated around a circle.
My question is this.
Should I also be able to calculate that as
$$S^{r\theta} = \oint^{\theta=2\pi} _{\theta=0} r d\theta - \theta dr = r \theta-0=2\pi r$$
Why doesn't this also work?
(I have transformed the rank 2 tensor to polar coordinates and, unless I messed up, it does give this answer...which isn't an "area." So the meaning of the tensor seems to have changed and that doesn't seem right or acceptable.)
You cannot replace one system of coordinates by another without invovling some "correction" terms: here you cannot just use the substitution $(x,y) \to (r,\theta)$ by just saying $x = r$ and $y=\theta$ in the integral.
Here is a proper way: say $x = r \cos \theta$ and $y = r \sin \theta$. Hence, \begin{align} \mathrm{d}x &= \cos \theta \mathrm{d}r - r \sin \theta \mathrm{d}\theta,\\ \mathrm{d}y &= \sin \theta \mathrm{d}r + r \cos\theta \mathrm{d}\theta. \end{align} From this, we deduce \begin{align} x \mathrm{d}y - y \mathrm{d}x &= r\cos\theta\left(\sin \theta \mathrm{d}r + r \cos\theta \mathrm{d}\theta \right) - r\sin\theta\left(\cos \theta \mathrm{d}r - r \sin \theta \mathrm{d}\theta \right) \\ &= r^2 \mathrm{d}\theta. \end{align} And from now, you should be able to recover the right answer.
Comment. A quick way to show the result is by defining $\alpha = x\mathrm{d}y - y\mathrm{d}x$, then $\mathrm{d}\alpha = 2\mathrm{d}x\wedge \mathrm{d}y$, which is twice the standard volume form in coordinates. By Stokes formula, it follows that the integral is twice the area enclosed, that is, $2 \times \pi r^2$ if $r$ is constant.