Take a two-dimensional symplectic submanifold $M$ in $\mathbb{R}^3$. Now, I want to show that the symplectic form $\omega$ is invariant under the Hamiltonian flow $\phi^{t}: M \rightarrow M.$
Is it now sufficient to show that for any two-dimensional submanifold $N$ of $M$ we have $\int_{N} \omega = \int_{\phi^{t}(N)} \omega$ for any $t$ and $N$? So does this imply that $(\phi^t)^* \omega = \omega$? Since $ \int_{\phi^{t}(N)} \omega = \int_N (\phi^t)^* \omega$ this reduces to the question whether $\int_{N} \omega = \int_N (\phi^t)^* \omega$ implies $(\phi^t)^* \omega = \omega$ ?
Edit: Since I received answers trying to propose different ways of solving this problem, I should mention that I want to prove it exactly the way I proposed.
Let $X_t$ denote the time dependent vector field generating $\phi_t$, and let $H_t$ be a smooth family of corresponding Hamiltonian functions. Then$$\mathcal{L}_{X_t}\omega=di_{X_t}\omega+i_{X_t}d\omega=ddH_t=0,$$and hence $\omega$ is preserved by $\phi_t$. Note that this holds for any Hamiltonian flow in any symplectic manifold, not only surfaces in $\mathbb{R}^3$.