Let $M$ be a compact orientable surface (manifold in $\mathbb R^3$) with boundary $S^1\times\{0\}$.Show that $M$ intersects the $z$-axis.
Some ideas:
$1)$Since $M$ is a compact orientable manifold with boundary, one way to solve this problem would be using Stokes.Assuming that $M$ doesn't intersect the $z$-axis, we should be able to integrate some differential form $\omega$ on $M$ or on $\partial{M}$ and get a contradiction.I just don't know what form I could integrate. Any suggestions?
$2)$When I first encountered this problem I tried this: by contradiction, if $M$ doesn't intersect the $z$-axis then ,if we take the projection $\pi:\mathbb R^3\rightarrow \mathbb R^2$, $\pi(x,y,z)=(x,y)$, the image of $M$ by $\pi$ is closed since $M$ is compact. Then if it doesn't intersect the $z$-axis, $(0,0)\notin \pi(M)$.Now,since the complement of $\pi(M)$ is open then there is a cylinder $B_{\epsilon}(0,0)\times \mathbb R$ that doesnt intersect $M$.I don't know what to do from here, it seems to be possible to finish the problem with the compacity of the surface, I don't think that we need the its orientability.I'm guessing this by intuition.Is there any way to finish this without the orientability, if not, is there any counterexample?
Thanks!