Consider a two dimensional vector field $X$ defined by $x'(t)=p(x,y), y'(t)=q(x,y)$, where $p$ and $q$ are differentiable functions.
Specifically I am trying to show that $\frac{d}{dt}A(\Omega_t)|_{t=0}=\int\int_{\Omega}(\frac{\partial p}{\partial x}+\frac{\partial q}{\partial y})dxdy$
I am using the area integral that states $A(\Omega_t)=\int\int_{\Omega_t} dxdy=\int\int_{\Omega}det(D\phi_t)dxdy$
My main issue is that I don't know what $D\phi_t$ is, which is the flow.
$D\phi_t$ would be the deformation gradient tensor. To obtain it you would have to integrate the velocity vector field to get the deformation, then take its gradient. But actually you can do this problem without considering $D\phi_t$ at all.
Using the Reynolds transport theorem: $$\frac{d}{dt}A(\Omega_t)=\frac{d}{dt}\iint_{\Omega_t}dA=\iint_{\Omega_t}\frac{\partial}{\partial t}(1)dA+\int_{\partial\Omega_t}\vec v\cdot d\vec n$$
Where $d\vec n$ is $\vec ndl$, the normal vector to the boundary times a differential boundary element. The first term is zero since 1 is constant. Using the divergence theorem: $$\int_{\partial\Omega_t}\vec v\cdot d\vec n=\iint_{\Omega_t}\nabla\cdot\vec v\,dA$$
So: $$\frac{d}{dt}A(\Omega_t)=\iint_{\Omega_t}\nabla\cdot\vec v\,dA$$
Since $\nabla\cdot\vec v=\frac{\partial p}{\partial x}+\frac{\partial q}{\partial y}$ and $\Omega_0=\Omega$, evaluating this last line at $t=0$ gives you what you wanted to show.