= Solution
For a <material volume> $V(t)$, Newton's second law equates the rate of change of momentum to <pressure> and body forces:
$$
\frac d{dt}\int_{V(t)}\rho\mathbf u\,dV
=-\int_{\partial V(t)}p\mathbf n\,dS+\int_{V(t)}\rho\mathbf g\,dV.
$$
Apply the supplied <Reynolds transport theorem> to each component of $\mathbf u$, and apply the <divergence theorem> to the <pressure> force. The result is
$$
\int_{V(t)}\left[\rho\left(\partial_t\mathbf u+(\mathbf u\cdot\nabla)\mathbf u\right)
+\nabla p-\rho\mathbf g\right]dV=0.
$$
Since the <material volume> at the chosen time is arbitrary, the integrand vanishes pointwise for smooth fields. Dividing by positive <density> gives the <Euler equations>
$$
\boxed{\partial_t\mathbf u+(\mathbf u\cdot\nabla)\mathbf u
=-\rho^{-1}\nabla p+\mathbf g.}
$$
At a fixed time $t$, a <streamline> is a curve tangent everywhere to the instantaneous velocity. Parametrize it by $d\mathbf x/ds=\mathbf u(\mathbf x(s),t)$. Then
$$
\frac d{ds}f(\mathbf x(s),t)=\nabla f\cdot\mathbf u=0,
$$
so $f$ is constant along it when the displayed directional derivative vanishes. Here time is held fixed: streamlines need not coincide with particle paths in an unsteady flow.
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