From Leibniz's Theorem to the Reynolds Transport Theorem

Integrate over the moving control volume \( V^*(t+\Delta t) \): \( \int_{V^*(t+\Delta t)} F(\mathbf{x},t+\Delta t) dV \)

Using Taylor expansion: \( \int_{V^*(t+\Delta t)} \left( F(\mathbf{x},t) + \frac{F'(\mathbf{x},t)}{1!} \Delta t \right) dV \)

Distribute the integral: \( \int_{V^*(t+\Delta t)} F(\mathbf{x},t) dV + \int_{V^*(t+\Delta t)} \frac{F'(\mathbf{x},t)}{1!} \Delta t dV \)

Since \( V^*(t+\Delta t) \) can be rewritten as \( V^*(t) + \Delta V \), we split the integral into the original volume \( V^*(t) \) and the new volume increment \( \Delta V \):

\( \left( \int_{V^*(t)} F(\mathbf{x},t) dV + \int_{\Delta V} F(\mathbf{x},t) dV \right) + \left( \int_{V^*(t)} \frac{F'(\mathbf{x},t)}{1!} \Delta t dV + \int_{\Delta V} \frac{F'(\mathbf{x},t)}{1!} \Delta t dV \right) \)

Expanding the first term inside the \(\{\}\)-braces into four terms:

\[ \int_{V^*(t+\Delta t)} F(\mathbf{x},t+\Delta t) dV = \underbrace{\int_{V^*(t)} F(\mathbf{x},t) dV}_{\text{Original}} + \underbrace{\int_{V^*(t)} \Delta t \frac{\partial F(\mathbf{x},t)}{\partial t} dV}_{\text{Time Derivative}} + \underbrace{\int_{\Delta V} F(\mathbf{x},t) dV}_{\text{New Volume}} + \underbrace{\int_{\Delta V} \Delta t \frac{\partial F(\mathbf{x},t)}{\partial t} dV}_{\text{New Volume Change}} \]
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