How do I calculate a line integral of $ x^i dx^j - x^j dx^i $ with polar coordinates? 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.)
 A: 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.
