sg.symplectic geometry - Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity?

I just wanted to expand two points of Greg's answer. Both are rather trivial additions, but took me a short while to understand, so I'm putting them here for completeness's sake and my own future reference.

First, here's a picture of Greg's idea of fixing the problem with Moser's trick. In his notation, we have volume forms $mu_{alpha,t} = mu^t mu_alpha^{1-t}$. They are defined because the two volume form must have the same sign everywhere (since everything is orientation-preserving; note that because of this I doubt that this argument can be generalized to symplectic forms in higher dimensions).

The obvious, but wrong, solution (described in my comments to his answer) would be to apply Moser's theorem to $mu_{alpha,0}$ in this notation. This would correspond to flowing along the horizontal axes of the picture below. However, Greg's idea is to flow in a different direction: we fix $alpha$ and vary $t$. Strictly speaking, we should also show that the resulting flow $phi_{alpha,1}$ will be smooth, but this should follow from the proof of Moser's theorem quite easily.

                                         Moser flow,
                                t        defines phi_alpha,t
        (mu_{alpha,1} = mu)   1 ^          ^
                                |          .
                                |          .
                                |          .
                                |          .
   (mu_{alpha,0} = mu_alpha)  0 ----------------> alpha
 (phi_{alpha,0} = phi_alpha)    

(Note that, for all t, $mu_{0,t}=mu_{1,t}=mu$)

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Secondly, Greg's answer implies that there is an isotopy $psi_t: D^2 to D^2$ such that each $psi_t$ is volume preserving (or, equivalently, a symplectomorphism), $psi_0$ is the identity, and $psi_1=psi$. Here's how we find the time-dependent Hamiltonian such that this is the Hamiltonian flow. Let X_t be the time-dependent vector field that is the derivative of the flow $psi_t$. The fact our flow is volume-preserving is equivalent to the fact that the 1-form $iota_{X_t} omega$ is be closed for all $t$. Since we are on a disk, this form will also be exact. So, let $H_t$ be the function such that $dH_t = iota_{X_t} omega$. The Hamiltonian flow of the function $H_t$ is precisely $psi_t$; this is immediate from the definitions.