From Leibniz's Theorem to Reynolds Transport Theorem

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 $:

\[ \frac{d}{dt} \int\limits_{V^*(t)} F(\mathbf{x},t) dV = \int\limits_{\substack{V^*(t)}} \frac{\partial F(\mathbf{x},t)}{\partial t} dV + \int\limits_{\substack{A^*(t)}} F(\mathbf{x},t) \mathbf{b} \cdot \mathbf{n} dA \] 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.