Apparent counter example to Stoke's theorem?

Question

I think I found an apparent contradiction to Stoke's theorem with this 2-differential form

$M= \overline{B^{2}}- \{ 0 \}$,

$\partial M = S^1$,

$$\omega = \frac{x~dy-y~dx}{x^2+y^2}$$ defined in $\mathbb{R}^2 - \{0\}$ and then pullbacked to $ \overline{B^{2}}- \{ 0 \}$

$d \omega = 0$ So that by Stoke's Theorem

$$ \int\limits_{\partial M} \omega = \int\limits_{M} d \omega = 0 $$

But direct calculation shows that $\int\limits_{S^1} \omega \neq 0 $

Answer 1

Your $M$ is not compact, and $\omega$ is not compactly supported. So you have not contradicted Stokes's theorem.

Answer 2

Congratulation! You found something which leads to idea of closed but not exact forms and cohomology groups.

To see the problem take $M= \overline{B^2(1)} \setminus B^2(\epsilon)$. Than $\partial M = S^1(1) - S^1(\epsilon)$. Now the Stokes theorems says: $$ \int_{S^1(1)} \omega - \int_{S^1(\epsilon)} \omega = \int_M d\omega $$ You might think that integral $\int_{S^1(\epsilon)} \omega$ goes to zero as $\epsilon$ goes to zero, but that is not true. In fact that integral does not depend on $\epsilon$, so you cannot omit it if $\epsilon$ is small.