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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.