= Solution
Let $V(t)$ be any material <volume>, moving with the <velocity field> $\mathbf u$. The <Reynolds transport theorem> in the required form is
$$
\frac d{dt}\int_{V(t)}f\,dV=\int_{V(t)}\left[\partial_t f+\nabla\cdot(f\mathbf u)\right]dV,
$$
applied componentwise when $f$ is vector-valued. <Mass conservation> gives
$$
0=\frac d{dt}\int_{V(t)}\rho\,dV
=\int_{V(t)}\left[\rho_t+\nabla\cdot(\rho\mathbf u)\right]dV.
$$
Since the material <volume> is arbitrary, the local <continuity equation> follows:
$$
\boxed{\rho_t+\nabla\cdot(\rho\mathbf u)=0.}
$$
For an inviscid fluid with no <body force>, the <balance of linear momentum> equates the rate of change of total <momentum> to the boundary <pressure> force:
$$
\frac d{dt}\int_{V(t)}\rho\mathbf u\,dV
=-\int_{\partial V(t)}p\mathbf n\,dS=-\int_{V(t)}\nabla p\,dV,
$$
using the <divergence theorem>. The integrand on the left expands as
$$
\partial_t(\rho\mathbf u)+\nabla\cdot(\rho\mathbf u\otimes\mathbf u)
=\rho\frac{D\mathbf u}{Dt}+\mathbf u\left[\rho_t+\nabla\cdot(\rho\mathbf u)\right].
$$
The bracket vanishes by <mass conservation>, leaving the <Euler momentum equation>
$$
\boxed{\rho\frac{D\mathbf u}{Dt}=-\nabla p,\qquad
\frac{D}{Dt}=\partial_t+\mathbf u\cdot\nabla.}
$$
Unlike the <incompressible flow> case, the <mass density> need not be constant along a <fluid element>.
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