Reynolds transport theorem: link with the Lie derivative? In this Wikipedia article (see "Higher dimensions") there seems to be a connection between the Reynolds transport theorem (here) and the Lie derivative:
$$\frac{d}{dt}\int_{\Omega(t)}\omega=\int_{\Omega(t)} i_{\vec{\textbf v}}(d\omega)+\int_{\partial \Omega(t)} i_{\vec{\textbf v}} \omega+\int_{\Omega(t)}\dot{\omega} \qquad(1)$$
were $\omega$ is a p-form and the domain is time-varying. The symbol $i_X$ denotes the contraction with the vector field $X$.
By using the Cartan magic formula 
$L_{X} \omega = d (i_X \omega) + i_X (d \omega)$, it seems that 
$$\frac{d}{dt}\int_{\Omega(t)}\omega=\int_{\Omega(t)} i_{\vec{\textbf v}}(d\omega)+\int_{\Omega(t)} d(i_{\vec{\textbf v}} \omega)+\int_{\Omega(t)}\dot{\omega} \, ,$$
namely 
$$\frac{d}{dt}\int_{\Omega(t)}\omega
=
\int_{\Omega(t)} L_{\vec{\textbf v}}\omega+\int_{\Omega(t)}\dot{\omega} \, .$$
It is clear that the Reynolds transport theorem is equivalent to the conservation of mass in hydrodynamics:
$$\frac{d}{dt}\int_{\Omega(t)}\rho
=
\int_{\Omega(t)} div(\rho \vec{\bf{v}}) +\int_{\Omega(t)}\dot{\rho} \, .$$
What is the relation between the two formulations? Clearly $\omega =\rho$ is not the correct thing to do: we can only integrate 3-forms and $\rho$ is a 0-form (the domain $\Omega(t)$ is three dimensional). How to derive the mass conservation from equation (1)?
 A: Instead of $\rho$, think about the three form $\rho ~\mathrm{d}x$. (Like I tell my calculus students, you should always write the integration element in an integral.) Here $\mathrm{d}x$ is the standard volume form. 
By definition of the divergence of a vector field you have that 
$$ \mathrm{div}(\rho \vec{v}) ~\mathrm{d}x = \mathrm{d}( \iota_{\rho\vec{v}} ~\mathrm{d}x ) = \mathrm{d} (\iota_{\vec{v}} \rho ~\mathrm{d}x ) = L_{\vec{v}} \left(\rho~\mathrm{d}x\right) $$
which is what you expect. 
A: Edit: I wrote this answer a long time ago. Coming back, I realised there were a lot of things that could have been much clearer. I hope it's better now.
There is a connection, indeed. Recall the definition of the Lie derivative for a $k$-form $\omega$ along a vector field $X$. If $\chi$ is the one-parameter family of diffeomorphisms defined (at least locally) by $X$, then
$$L_X\omega = \left.\frac{d}{d\tau}\right|_0(\chi_{\tau}^{*}\omega)$$
Integrating over some $k$-submanifold $\Omega$, we get
$$ \int_\Omega L_X\omega = \int_{\Omega}\left.\frac{d}{d\tau}\right|_0(\chi_{\tau}^{*}\omega)
= \left.\frac{d}{d\tau}\right|_0\int_{\Omega}\chi_{\tau}^{*}\omega
= \left.\frac{d}{d\tau}\right|_0\int_{\chi_\tau[\Omega]}\omega $$
Note that we first used the Leibniz integral rule in one variable, and then the fact that diffeomorphisms preserve the integral. I claim Reynold's transport theorem is a special case of the formula
$$\left.\frac{d}{d\tau}\right|_0\int_{\chi_\tau[\Omega]}\omega = \int_\Omega L_X\omega \tag{1}$$
we just obtained.
Say your space is an $n$-dimensional manifold $M$, and consider an interval $I\subseteq \mathbb R$ (say with $0\in I$) which for us will represent time.
In the context of Reynold's transport theorem you have a submanifold $\Omega$ of $M$ (say of dimension $k$) which "varies (smoothly) over time".
To formalize this, take a $k$-dimensional manifold $\Omega$ and an embedding $i:I\times\Omega\to I\times M$ where the first component is the identity. Then the submanifold $\Omega_t=i[\{t\}\times \Omega]\subseteq \{t\}\times M$ is the "position of $\Omega$ inside $M$ at time $t$".
Now we want to get the "spacetime velocity" of $\Omega$, which will be a vector field defined on the submanifold $i[I\times \Omega]$ (the "worldsheet" of $\Omega$).
To do this, observe that for every $x\in\Omega$ we get a smooth curve $\gamma_x:I\to I\times M$ given by $\gamma_x(t)=i(t,x)$. The vectors tangent to this family of curves give us a vector field $X$ defined by
$$\tilde X_{i(t,x)}=\gamma'_x(t)$$
This is the spacetime velocity we wanted. Using the extension lemma for vector fields on submanifolds, we get a vector field $X$ on $I\times M$ which agrees with $\tilde X$ on $i[I\times \Omega]$.
Its spacelike part $V=\text{proj}_{TM}X$ is the vector field that appears in Reynold's transport theorem, and we have $X=\frac{\partial}{\partial t}+V$.
Now a time-dependent $k$-form $\omega$ on $M$ can be regarded as a $k$-form on $I\times M$
such that $\iota_{v}\omega=0$ for any "purely timelike" vector $v\in T(I\times\{x\})$.
Note that we may (and will) always choose charts where the projection $t:I\times M\to I$ is a coordinate, so that $\omega$ has no $dt$ component.
Now we have all the pieces.
Let $\chi$ be the one-parameter family of diffeomorphisms induced by $X$ (we only need $\chi$ to be defined in a neighbourhood of $\Omega_0$). Applying formula (1) to the submanifold $\Omega_0$, we have
$$\left.\frac{d}{dt}\right|_0\int_{\chi_t[\Omega_0]}\omega = \int_{\Omega_0} L_X\omega .$$
Since $X$ at time $t$ is the velocity of $\Omega_t$, we have $\chi_t[\Omega_0]=\Omega_t$. Also, by Cartan's magic formula, we have $L_X\omega = d(\iota_X\omega) + \iota_X(d\omega)$. Hence
$$\left.\frac{d}{dt}\right|_0\int_{\Omega_t}\omega = \int_{\Omega_0} d(\iota_X\omega) + \int_{\Omega_0}\iota_X(d\omega) \tag{2}$$
Now lets treat the two integrals on the right hand side of (2) separately.
For the first integral,
since $V$ is the spatial part of $X$, we have $X=\frac{\partial}{\partial t}+V$ and hence $\iota_X\omega=\iota_V\omega$. On the other hand, since the tangent space of $\Omega_0$ is contained in $\{0\}\times M$, it is annihilated by $dt$. These two observations give the first two equalities in
$$\int_{\Omega_0} d(\iota_X\omega)=\int_{\Omega_0} d(\iota_V\omega)=\int_{\Omega_0} d_M(\iota_V\omega)=\int_{\partial\Omega_0}\iota_V\omega \tag{3}$$
and the third one comes from Stokes' theorem.
Now let's look at the second integral on the right hand side of (2).
Since $\omega=\omega_Idx^I$ (where $I$ is a multiindex where $t$ does not appear) we have
$$\begin{aligned}
d\omega
&=\partial_t\omega_Idt\wedge dx^I+\sum_i \partial_i\omega_Idx^i\wedge dx^I \\
&=\partial_t\omega_Idt\wedge dx^I+d_M\omega
\end{aligned}$$
and since $X=\frac{\partial}{\partial t}+V$, the second integral on the right hand side becomes
$$\begin{aligned}
\int_{\Omega_0}\iota_X(d\omega)
&= \int_{\Omega_0}\iota_{\frac{\partial}{\partial t}+V}(\partial_t\omega_Idt\wedge dx^I+d_M\omega) \\&= \int_{\Omega_0}\partial_t\omega_Idx^I+\iota_V(d_M\omega) \\
&= \int_{\Omega_0}\dot\omega+\int_{\Omega_0}\iota_V(d_M\omega)
\end{aligned} \tag{4}$$
The the crossed terms of the interior product vanish because they match $dt$ with $V$ and purely spatial basis forms with $\frac{\partial}{\partial t}$.
At last, plugging (3) and (4) into (2), we have
$$\left.\frac{d}{dt}\right|_0\int_{\Omega_t}\omega = \int_{\partial\Omega_0}\iota_V\omega+\int_{\Omega_0}\dot\omega+\int_{\Omega_0}\iota_V(d_M\omega)$$
as desired.
