$\omega=p(x,y)dx+q(x,y)dy\quad$ a continuously differentiable one form and $d\omega =0$
In addition, for $\alpha(t)=(r\cos t,r\sin t)$, $\int_\alpha \omega =0 $ for some $\; r \in \mathbb R$
I need to prove that
$$\int_\gamma{\omega}=0 $$ for any closed $\gamma$.
Can someone give me a hint because I don't know where to start. I know that $\; \int_\gamma{\omega}=0$ would be true if I could find a function $\; f$ so that $\;V=\text{grad}(f)=(p,q)$ but I don't think that helps.
That's a standard arguemnt using Stokes' Theorem. Let $\gamma (t)$ be a parametrization of $\gamma$. Then consider
$$F : \mathbb S^1 \times [0,1] \to \mathbb R^2, $$
where $F(t, s) = s \alpha (t) + (1-s) \gamma (t)$. Then $F(\cdot, 0) = \gamma$ and $F(\cdot, 1) = \alpha$. Using Stokes' Theorem, as $d\omega = 0$,
$$0=\int_{\mathbb S^1 \times [0,1]} F^* d\omega = \int_{\partial{(\mathbb S^1 \times [0,1])}}F^*\omega = \int_{\mathbb S^1 \times \{1\}} F^*\omega - \int_{\mathbb S^1 \times \{0\}} F^*\omega = \int_\alpha \omega - \int_\gamma \omega.$$
Thus $\int_\gamma \omega =0$ for any closed $\gamma$.