= Solution
{c}
Let $V(t)$ have a boundary moving with <velocity> $\mathbf v_b$, and let $\mathbf n$ be its outward unit <normal vector>. For a smooth scalar field $q$, the <Reynolds transport theorem> is
$$
\frac{d}{dt}\int_{V(t)}q\,dV
=\int_{V(t)}\partial_tq\,dV
+\int_{\partial V(t)}q\mathbf v_b\cdot\mathbf n\,dS
=\int_{V(t)}[\partial_tq+\nabla\cdot(q\mathbf v_b)]\,dV.
$$
In the last expression, extend $\mathbf v_b$ smoothly into $V(t)$ and use the <divergence theorem>. For a <material volume>, its boundary moves with the fluid <velocity> $\mathbf u$, so $\mathbf v_b=\mathbf u$. Equivalently, the <Reynolds transport theorem> for a <material volume> reads
$$
\frac{d}{dt}\int_{V(t)}q\,dV
=\int_{V(t)}\left(\frac{Dq}{Dt}+q\nabla\cdot\mathbf u\right)dV,
\qquad \frac D{Dt}=\partial_t+\mathbf u\cdot\nabla,
$$
where $D/Dt$ is the <material derivative>.
For an arbitrary moving region, <mass conservation> includes the relative <mass flow rate> across its boundary:
$$
\frac{d}{dt}\int_{V(t)}\rho\,dV
=-\int_{\partial V(t)}\rho(\mathbf u-\mathbf v_b)\cdot\mathbf n\,dS.
$$
Substitution of the <Reynolds transport theorem> and the <divergence theorem> gives
$$
\int_{V(t)}[\partial_t\rho+\nabla\cdot(\rho\mathbf u)]\,dV=0.
$$
Since the region is arbitrary, a continuous integrand must vanish: if it had a positive or negative value at a point, a sufficiently small surrounding region would have an <integral> of the same sign. This gives the <continuity equation>. With constant positive <mass density> $\rho$, it reduces to
$$
\boxed{\nabla\cdot\mathbf u=0.}
$$
Thus \b[the flow is incompressible]. For a <material volume>, the same argument starts directly from conservation of its total <mass>, with zero relative boundary flux.
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