 Solution

Label fluid particles by $\mathbf{a}$, with flow map $\mathbf{x}=\chi (\mathbf{a},t)$ satisfying ${\chi }_t=\mathbf{u}(\chi ,t)$. If $J=\det D_{\mathbf{a}}\chi$, differentiation of the determinant gives $J_t=J\nabla \cdot \mathbf{u}$. Pull the material-volume integral back to a fixed initial volume:The chain rule and the Jacobian identity give the integrand $J(F_t+\mathbf{u}\cdot \nabla F+F\nabla \cdot \mathbf{u})$. Transforming back yields the Reynolds transport theoremEquivalently the additional surface flux is ${\int }_{\partial V(t)}F\mathbf{u}\cdot \mathbf{n}dS$. The material boundary moves with the same normal velocity as the fluid, which is the key distinction from a fixed control volume.