( Proof Explanation ) Show that a certain system preserves the weighted area $ (dx \wedge dy)/xy$ I already told few questions ago that I'm currently reading an abstract about the Lotka Volterra differential equations. But now I have a proof, where I need explanations. Consider:
$$ \dot{x} = -xy\frac{\delta H}{ \delta y} , x(0) = \hat{x} $$
$$  \dot{y} = xy\frac{\delta H}{ \delta x} , y(0) = \hat{y} $$where $H(x,y) = x + y - ln(x) -ln(y)$.
I have to show that this System preserve the weighted area $(dx \wedge dy)/xy$.
I marked my Questions in the proof below.
Proof:
Let $\Omega_0$ be a subset of $\mathbb{R}^2$ at time $t_0$ and $ \Omega_1$ the set into which $\Omega_0$ is mapped by the system above at time $t_1$. Preservation of $(dx \wedge dy)xy$ is equivalent to $$ \int_{\Omega_0} \frac{1}{xy}dxdy = \int_{\Omega_1} \frac{1}{xy} dxdy $$ first Question: why is this equivalent? We now look at the Domain $D$ in x,y,t space with bondary $\delta D$ given by $\Omega_0$ at $t_0$, $\Omega_1$ at $t_1$ and the set of trajectories emerging from the boundary of $\Omega_0$ and ending on the boudnary of $\Omega_1$. Consider the vector field $$ v := \frac{1}{xy}(\dot{x},\dot{y},1)^T $$ in $x,y,t$ space. Integrating this vector field over the boundary $\delta D$ of $D$, we obtain    $$ \int_{\delta D} v \cdot n = \int_{\Omega_0} v \cdot n_0 + \int_{\Omega_1} v \cdot n_1 = \int_{\Omega_0} \frac{1}{xy} dxdy - \int_{\Omega_1} \frac{1}{xy}dxdy $$ where $n_0 =(0,0,-1)^T$ denote the unit outward normal of $\Omega_0$ and $\Omega_1$.Second question & Third question: Can you explain why we integrate  $v \cdot n$ ? I thought we integrate $v$ and can you explain the first equation above? There is no other contribution to the surface integral, because the vector field $v$ is by contruction parallel to the trajectories, which form the rest of the bondary $\delta D$.  Forth question: Can you explain why vector field is parallel to the trajectories? Applying the divergence theorem to the left hand side of the same equation, we get $$ \int_{\delta D} v \cdot n = \int_D \nabla v = \int_D - \frac{\delta H^2}{\delta x \delta y} + \frac{ \delta H^2}{\delta x \delta y} + 0 = 0 $$ which concludes the proof.
I hope that my questions are not to easy, but I'm a beginner. 
 A: I believe the proof is taken from Mickens, Applications of Nonstandard Finite Difference Schemes. Preservation of the area weighted by the factor $\rho$ by definition means that
$$\int_{\Omega(t_0)} \rho \, dS = \int_{\Omega(t)} \rho \, dS$$
if the set $\Omega(t_0)$ is mapped to $\Omega(t)$ by the system (the omegas are sets of $(x, y)$ points). We want to show that this holds for the given system and for $\rho(x, y) = 1/(x y)$.
Suppose we parametrize $\partial \Omega(t_0)$ by $\phi$. A point $(x, y, t)$ on the surface $\mathcal S$ is given by specifying $\phi$ and $t$: $x$ and $y$ are the solution of the system at time $t$ with the initial conditions given by $\phi$. If we fix $\phi$ and vary $t$, we'll get a curve $(x, y, t)$, which lies on $\mathcal S$ by construction. Since $\boldsymbol v = (\dot x, \dot y, 1)$ is tangent to the curve, it is also tangent to $\mathcal S$.
Then we take $\partial D = \mathcal S \cup \Omega(t_0) \cup \Omega(t)$ and take $\hat {\boldsymbol n}$ to be the outward unit normal to $\partial D$. Since $\boldsymbol v \cdot \hat {\boldsymbol n} = 0$ on $\mathcal S$ and $\dot \rho = 0$,
$$\int_{\partial D}
 \rho \hspace {1px} \boldsymbol v \cdot \hat {\boldsymbol n} \, dS =
-\int_{\Omega(t_0)} \rho \,dS + \int_{\Omega(t)} \rho \, dS, \\
\int_{\partial D}
 \rho \hspace {1px} \boldsymbol v \cdot \hat {\boldsymbol n} \, dS =
\int_D \nabla \cdot (\rho \hspace {1px}  \boldsymbol v) \, dV =
\int_D \nabla \cdot \left(
 -\frac {\partial H} {\partial y},
 \frac {\partial H} {\partial x},
 \rho \right) dV = 0.$$
