= 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:
$$
\frac d{dt}\int_{V(t)}F\,dV=\frac d{dt}\int_{V(0)}F(\chi(\mathbf a,t),t)J\,d\mathbf a.
$$
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 theorem>
$$
\boxed{\frac d{dt}\int_{V(t)}F\,dV=\int_{V(t)}[F_t+\nabla\cdot(F\mathbf u)]\,dV.}
$$
Equivalently 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>.
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