= Solution
Let $V(t)$ be a smoothly moving material <volume>, carried by a smooth <velocity field> $\mathbf u$, and let $f$ be continuously differentiable. The material-volume form of the <Reynolds transport theorem> states
$$
\boxed{\frac d{dt}\int_{V(t)}f\,dV=\int_{V(t)}\left(\frac{Df}{Dt}+f\nabla\cdot\mathbf u\right)dV.}
$$
Equivalently, the <divergence theorem> gives
$$
\frac d{dt}\int_{V(t)}f\,dV=\int_{V(t)}\partial_t f\,dV+\int_{\partial V(t)}f\mathbf u\cdot\mathbf n\,dS,
$$
where $\mathbf n$ is the outward <unit normal vector>. The second integral measures the <flux> carried across the instantaneous boundary by its motion.
For the proof, use the particle map $\mathbf x=\mathbf X(\mathbf a,t)$ from $V(0)$, with positive <Jacobian determinant> $J=\det(\partial\mathbf X/\partial\mathbf a)$. <Euler's volume identity> is
$$
\partial_tJ=J\nabla\cdot\mathbf u.
$$
It is the <Jacobian evolution of a smooth flow>: if $F=\partial\mathbf X/\partial\mathbf a$, differentiation gives $F_t=(\nabla\mathbf u)F$, and differentiation of the <determinant> gives $J_t=J\operatorname{tr}(F^{-1}F_t)=J\nabla\cdot\mathbf u$. The <change of variables> formula now moves the integral to the fixed reference <volume>:
$$
\begin{aligned}
\frac d{dt}\int_{V(t)}f\,dV
&=\frac d{dt}\int_{V(0)}f(\mathbf X(\mathbf a,t),t)J\,d\mathbf a\\
&=\int_{V(0)}\left(\frac{Df}{Dt}J+fJ_t\right)d\mathbf a\\
&=\int_{V(t)}\left(\frac{Df}{Dt}+f\nabla\cdot\mathbf u\right)dV.
\end{aligned}
$$
Finally, $Df/Dt+f\nabla\cdot\mathbf u=\partial_t f+\nabla\cdot(f\mathbf u)$, giving the equivalent boundary form.
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