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A material volume is a region containing the same fluid particles at every time: its boundary moves with the fluid velocity field. Its mass is fixed by mass conservation, while its volume can change in compressible motion. Applying the Reynolds transport theorem to its volume gives $d|V(t)|/dt={\int }_{V(t)}\nabla \cdot \mathbf{u}dV$. A material volume element is the infinitesimal version.

A material volume element contains the same fluid particles as it moves and deforms. If its edges transform with the local velocity gradient, its volume changes at a rate equal to the trace of that gradient, giving $D(\delta V)/Dt=(\nabla \cdot \mathbf{u})\delta V$. Mass conservation states $D(\rho \delta V)/Dt=0$, not that its volume is always fixed. These two identities give the material form of the continuity equation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$.

For an inviscid material volume element, neglecting body forces, Newton's second law gives $D(\rho \mathbf{u}\delta V)/Dt=-\nabla p\delta V$. Since mass conservation keeps $\rho \delta V$ constant, the result is $\rho D\mathbf{u}/Dt=-\nabla p$. Cancelling $\delta V$ before differentiating would incorrectly omit its changing volume. A fixed control volume instead requires the advected momentum flux, yielding ${\partial }_t(\rho \mathbf{u})+\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=-\nabla p$.

The first argument mixes a material fluid element with a fixed spatial volume. For a material volume element following the fluid, its volume is generally time dependent:Therefore the material derivative cannot be taken outside the product with $\delta V$ as in the proposed cancellation. Instead, mass conservation statesThe momentum balance for a material fluid element, using Newton's second law, is consequentlyFor a fixed control volume, fluid crosses its boundary and carries momentum; a momentum flux term must be included. Its local conservative form is ${\partial }_t(\rho \mathbf{u})+\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=-\nabla p$, equivalent to the displayed Euler equations after using the continuity equation. The erroneous equation omits the term $\rho \mathbf{u}\nabla \cdot \mathbf{u}$ needed to convert a material derivative of momentum density into this conservative form.

For the second task, take a homogeneous fluid with constant mass density $\rho$, prescribed gravitational acceleration $\mathbf{g}$, incompressible flow and inviscid flow. Its equations areUnder the replacement $\rho \mapsto \tilde{\rho }=C\rho$, the same velocity solves these equations if $p\mapsto \tilde{p}=Cp$, up to a spatially constant pressure offset. This is density scaling invariance of incompressible Euler flow:The comparison assumes the same geometry and compatible velocity or gravitational free surface conditions; an independently specified nonuniform boundary pressure would supply additional information. The gravitational field is taken as prescribed. If the fluid's own self-gravity must be included, its field changes with mass density, so the invariance need not hold. A single freely moving particle's density-independent gravitational acceleration does not justify neglecting that distinction.