Let $\Phi: [0,1] \times L \to (M, \omega), \: (t,x) \mapsto \phi^t (x)$ be a Hamiltonian isotopy generated by a 1-periodic Hamiltonian, where $L$ is a compact Lagrangian submanifold in $(M, \omega$).
I want to show that the area swept by such an Hamiltonian isotopy is zero.
If $\omega = d \theta$ is exact and $L$ is exact i.e. $\theta\vert_L$ exact, this is just Stokes theorem:
$$ \int_{L \times [0,1]} \Phi^\ast(w) = \int_{L \times [0,1]} d \Phi^\ast(\theta) = \int_{L \times \{0\}} \Phi^\ast \theta - \int_{L \times \{1\}} \Phi^\ast \theta = \int_L \theta - \int_{L}(\phi^1)^\ast\theta = 0, $$
Q. Does this also hold in the non-exact case? A. Yes. Here is my attempt:
So, in general, $d\Phi_{(t,x)} = \frac{d}{dt} \phi^t(x) + \phi^t_\ast = X_H(\phi^t(x))+ \phi_\ast^t$, thus, $$ \int_{L \times [0,1]} \Phi^\ast\omega = \int_{L \times [0,1]} \omega \circ \frac{d}{dt} \phi^t + (\phi^t)^\ast \omega $$ Now, $(\phi^t)^\ast \omega = \omega$, hence the second expression vanishes for $L$ is Lagrangian. Moreover, let $\phi$ be the time-1-map. The, $$ \int_{L \times [0,1]} \omega \circ \frac{d}{dt} \phi^t = \int_{L \times [0,1]} \omega(X_H, \cdot) \circ \Phi = \int_{L \times [0,1]}dH \circ \Phi = \int_L H \circ \phi - \int_L H. $$ But this is zero since $H$ is invariant under its flow. Indeed, $$\frac{d}{dt}(H \circ \phi^t) = dH(X_H) = \omega(X_H, X_H) = 0.$$ Is this argument right?
Edit: In the exact case, I do not use the assumption that $\Phi$ is a Hamiltonian isotopy. I realize that this means that this argument works also for a family of symplectomorphisms $\{\phi_t\}$. However, note that $M = T^\ast S^1$ and $L = S^1$ is not a counterexample since $\theta = p dq$ is not exact on $S^1$.
Edit 2: This is not surprising, of course. In an exact symplectic manifold, symplectomorphisms and Hamiltonian diffeomorphisms coincide.