Interpreting the closed condition of a symplectic form

Question

In McDuff's What is Symplectic Geometry?, she writes: "the fact that $\omega$ is closed means that the symplectic area of a surface $S$ with boundary does not change as $S$ moves, provided that the boundary is fixed."

I don't see why we need that the boundary be fixed. I tried to come up with some argument, but it must be flawed. Let me reproduce it here, anyways, and perhaps someone could explain its faults.

Let $(M,\omega)$ be a symplectic manifold with compact surface $S$ with boundary. Let $I=[0,1]$. Suppose we have a family of diffeomorphisms $F:M \times I \to M$ such that $F(x,0)=Id$. We'll "move" $S$ with $F$. Then $$0 = \int_{S \times I} F|_S^* \; d\omega = \int_{\partial S \times I} F^*|_{\partial S} \; \omega + \int_S F_1|_S^* \; \omega - \int_S \omega.$$

If we could express $\partial S \times I$ as the boundary of something, then by Stokes, the first integral of the three vanishes. But it wouldn't matter if $F$ didn't fix the boundary. In any event, I don't think I have the correct view of what $\partial (S \times I)$ even is.

Answer 1

If you consider a surface $S$ with boundary; its boundary is an union of curves ($c_1,...,c_n$) $H^2(S)=0$ implies that $\int_S\Omega =\sum_{i=1}^{i=n}\int_{c_i}\alpha$ where $d\alpha=\omega$. Therefore if $S,T$ have the same boundary, $\int_S\omega=\int_T\omega$.

Top homology of a manifold with boundary

Answer 2

The manifold with boundary $\partial S \times I$ satisfies $$\partial(\partial S \times I) = \partial S \times \partial I \neq \varnothing ,$$ so $\partial S \times I$ is not the boundary of any manifold with boundary.

Answer 3

This is a late addition, but here is a simple counterexample. Take a disc $D$ in the standard plane and compare to a disc $D'$ of different radius. They have different areas. One can be moved into the other by a translation and stretching a little.