Starting with:
\[ \frac{d}{dt} \int_{V^*(t)} F(\mathbf{x},t) dV = \lim_{\Delta t \to 0} \frac{1}{\Delta t} \left\{ \int_{V^*(t)} \Delta t \frac{\partial F(\mathbf{x},t)}{\partial t} dV + \int_{\Delta V} F(\mathbf{x},t) dV \right\} \]
Substituting Equation:
\[ \int_{\Delta V} F(\mathbf{x},t) dV \approx \int_{A^*(t)} F(\mathbf{x},t) (\mathbf{b} \Delta t \cdot \mathbf{n}) dA. \]
Substituting:
\[ \frac{d}{dt} \int_{V^*(t)} F(\mathbf{x},t) dV = \lim_{\Delta t \to 0} \frac{1}{\Delta t} \left\{ \int_{V^*(t)} \Delta t \frac{\partial F(\mathbf{x},t)}{\partial t} dV + \fbox{$\int_{\Delta V} F(\mathbf{x},t) dV$} \right\} \]
Replacing the boxed term:
\[ \frac{d}{dt} \int_{V^*(t)} F(\mathbf{x},t) dV = \lim_{\Delta t \to 0} \frac{1}{\Delta t} \left\{ \int_{V^*(t)} \Delta t \frac{\partial F(\mathbf{x},t)}{\partial t} dV + \fbox{$\int_{A^*(t)} F(\mathbf{x},t) (\mathbf{b} \Delta t \cdot \mathbf{n}) dA$} \right\} \]\( \Delta t \) is a common factor, divide by \( \Delta t \):
It produces the statement of Reynolds Transport Theorem.
This final result follows the pattern set by Liebniz’s theorem that the total time derivative of an integral with time-dependent limits equals the integral of the partial time derivative of the integrand plus a term that accounts for the motion of the integration boundary.