Suppose I have a Hamiltonian $H=H(p,q)=H(p,q+1)$ defined on a cylinder $\mathbb{T} \times \mathbb{R}^{+}$, such that $$H(p,q) = \sqrt{p}(5+\sin 2\pi q)$$ i.e. we have $p \in \mathbb{R}^{+}$ and $q \in \mathbb{T}$.
Hamilton's equations are $$\dot{p} = - \frac{\partial H}{\partial q} = -2\pi\sqrt{p} \cos 2\pi q$$ $$\dot{q} = \frac{\partial H}{\partial p} = \frac{5+\sin 2\pi q}{2 \sqrt{p}}$$
We see from the form of $H$ that all orbits are closed curves. I want to convert $H(p,q)$ to action-angle variables $(I, \theta)$ and verify that the corresponding time-1 map on the cylinder is twist. Now action $I$ is the area under the trajectory on the cylinder, so $I = \int^{1}_{0} p dq$ which gives, using that $p = \frac{H^{2}}{(5+\sin 2 \pi t)^{2}}$
$$I = \int^{1}_{0} \frac{H^{2}}{(5+\sin 2 \pi t)^{2}} dq = k H^{2}$$
for some constant $k$ that corresponds to the evaluated definite integral. Thus $H = \sqrt{\frac{I}{k}}$ and since Hamilton's equations in action angle variables are $$\dot{I} = 0, \qquad \dot{\theta} = \frac{\partial H}{\partial I} = \omega(I) = \frac{1}{2\sqrt{kI}} $$
we have indeed found that the time-1 map is twist, since the twist condition implies monotone dependence of period on the area, or that $\omega'(I) \neq 0$.
Can anyone shed light if the above derivation is correct? Thanks!