Every compact orientable surface with $S^1\times\{0\}$ as its boundary intersects the $z$-axis 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!
 A: Consider $M_{s}^{g}$ to be the manifold in question with genus $g$. Consider $M^{g}$ to be $M_{s}^{g}$ filled the circle into a region homeomorphic to $\mathbb{D}^{2}$ such that it has no intersection with $M^{g}_{s}$. This can always be done. By the classification theorem we have $M_{g}$ to be a closed oriented surface with $g$ handles. 
I claim the following:
1) $M_{g}$ must contain at least one point in $z$-axis in its interior. 
2) Any line passing through this point must intersect with $M_{g}$ at some point. 
The second one is clear since $M_{g}$ are all compact. 
The first one need a contradiction type argument. If $M_{g}$ has an intersection point with $z$-axis, then we are done. If $z$-axis is disjoint from $M_{g}$, then we can put a box such that $M_{g}$ is in the box, and $z$-axis is outside of the box. Now up to homeomorphism we can deform $M_{g}$ to the 'standard' type such that $\mathbb{S}^{1}$ remain a circle on one of the $g$ handles or on its surface area(hence contractible). In the former case the center of the circle is in $M_{g}$. In the later case it is clear the line must intersect $M_{g}$ at some point.  
