Area of a billiard and its preservation From S. Tabachnikov's Geometry and billiards

Consider a plane billiard table $D$ whose boundary is
  a smooth closed curve $γ$. Let $M$ be the space of unit tangent vectors
  $(x, v)$ whose foot points $x$ are on $γ$ and which have inward directions.
  A vector $(x, v)$ is an initial position of the billiard ball. The ball moves
  freely and hits $γ$ at point $x_1$; let $v_1$ be the velocity vector reﬂected
  oﬀ the boundary. The billiard ball map $T\colon M \to M$ takes $(x, v)$ to
  $(x_1 , v_1)$. [...]
Parameterize $γ$ by arc length $t$ and let $α$ be the angle between
  $v$ and the positive tangent line of $γ$. Then $(t, α)$ are coordinates on
  $M$; in particular, $M$ is the cylinder. A fundamental property of the
  billiard ball map is the existence of an invariant area form.
Theorem 3.1. The area form $ω = \sin α \,dα \wedge dt$ is $T$-invariant.

This seems really important as it let us apply Poincaré's recurrence theorem, but I don't understand what this area is that is being preserved.
My question would probably be What is an area form?. I've found some definitions online, but they are just too advanced for me. Is it possible to rephrase the theorem above in simpler terms or should I wait until I get necessary tools to understand that?
 A: Let $\gamma(s)$ for $s \in [0, L]$ be the archlength parametrized boundary of the domain $D$. For each point $\gamma(s)$ on $\gamma$ consider the unit (tangent) circle centered at $\gamma(s)$ parametrized by angle $\alpha \in [0, 2\pi]$ measured from $\dot{\gamma}(s)$ in a counterclockwise direction. Hence, you get the set of points $(s,\alpha) \in [0,L] \times [0,2\pi]$ and because $0$ and $L$ are identified and $0$ and $2\pi$ are also identified, you can think that (by abusing some notation) $(s,\alpha) \in S^1 \times S^1$ where $S^1$ is the unit circle, which we can think of as $S^1 \cong \gamma \cong \Big( [0,L]\mod L \Big) \cong \Big([0,2\pi] \mod 2\pi\Big)$. 
The discrete time evolution of the billiard is realized as follows: we start from a point $(x, v)$ where $x \in \gamma$ and $v$, which we can think of as a point on $S^1$ centered at $x$, is a unit vector in $\mathbb{R}^2$ pointing from point $x$ in a direction of the billiard trajectory inside the domain $D$. During the motion of the point inside $D$, the trajectory is a striaght line and the velocity is always $v$ (so uniform motion) until the point reaches again the boundary $\gamma$ of $D$ at a point $x_1 \in \gamma$ and after reflection in $\gamma$ we obtain a new direction vector $v_1$. Thus we have a map $$\Phi : \gamma \times S^1 \to \gamma \times S^1$$ $$(x,v) \mapsto (x_1,v_1)$$
Since $(s,\alpha)$ are natural coordinates on the space (it is a torus) $\gamma \times S^1 \cong S^1 \times S^1$, we can represent the point $x$ as the coordinate $s$ and the direction vector $v$ by its angle $\alpha$ with the tangent $\dot{\gamma}(s)$. Hence 
$$\Phi : S^1 \times S^1 \to S^1 \times S^1$$ $$(s,\alpha) \mapsto (s_1,\alpha_1)$$ is a map defined on a subset of the torus mapping it into the torus. 
However, the vector $v$ can only point inside the domain $D$, so the angle $\alpha$ can change only between $0$ and $\pi$ (only the directions pointing on one side of the tangent $\dot{\gamma}(s)$). So technically, the map $\Phi$ is defined as follows 
$$\Phi : S^1 \times (0,\pi) \to S^1 \times (0,\pi)$$ $$(s,\alpha) \mapsto (s_1,\alpha_1)$$ 
where $ S^1 \times (0,\pi)$ is a cylinder. The form $\omega = \sin(\alpha)\, d\alpha \wedge ds$ is a symplectic form on $ S^1 \times (0,\pi)$ and the map $\Phi$ preserves it (it is a symplectic map), i.e. $\Phi^*\omega = \omega$. Thus 
$ S^1 \times (0,\pi), \, \omega$ is a two dimensional symplectic manifold. But in dimension two, $\omega$ is also an area form. Therefore, $\Phi$ preserves area. And here is where the Poincare recurrence theorem applies. 
Area form is a form that allows you to introduce the notion of area on a manifold, in our case, the surface $ S^1 \times (0,\pi)$. If $K \, \subset \, S^1 \times (0,\pi)$ is a measurable subset, then
$$\text{Area}(K) = \int_{K} \omega = \int \int_{K} \sin(\alpha)\, d\alpha \,ds$$ You may have different notions of ares on the same space.       
