So, I was trying to obtain the point form of the conservation of linear momentum equation in integral form, namely:
$\int_{\partial \Omega} \vec{V} \rho \vec{V} \cdot \vec{dS} + \int_{\partial \Omega} p \vec{dS} = 0$
According to the Gauss theorem for a closed surface $S$:
$\iint_S \vec{A} \cdot \vec{dS} = \iiint_V \nabla \cdot \vec{A} dV $
But if I apply that to the above equation I get
$\int_{\partial \Omega} \vec{V} \rho \vec{V} \cdot \vec{dS} = \int_{\Omega} \nabla \cdot (\vec{V} \rho \vec{V}) dV =\\ \quad \int_{\Omega} (\nabla \cdot \vec{V}) \rho \vec{V} dV $
Which can't be right, since for an incompressible flow $\nabla \cdot \vec{V} = 0$.
Isn't the dot product supposed to be commutative? What am I missing?
I apologize for any misuse of mathematical notation, let me know of any mistakes.
The correct identity is
$$\nabla\cdot (\rho \mathbf{v} \mathbf{v}) = \rho \mathbf{v} \cdot \nabla\mathbf{v} \, + \, (\nabla \cdot \mathbf{v}) \rho \mathbf{v},$$
where only the second term on the RHS vanishes in incompressible flow.
To check, this should all be consistent with the Navier-Stokes equations for steady flow (minus viscous terms that you have excluded):
$$\rho\mathbf{v} \cdot \nabla \mathbf{v} = - \nabla p,$$
which follows when $\nabla \cdot \mathbf{v} = 0$ from
$$\begin{align}0 &= \int_{\partial \Omega}\rho \mathbf{v} \mathbf{v} \cdot d\mathbf{S} + \int_{\partial \Omega} p \,d\mathbf{S} \\&= \int_{\Omega}\nabla \cdot (\rho \mathbf{v} \mathbf{v}) \, dV + \int_{\Omega} \nabla p \,dV\\ &= \int_{\Omega} \{ \rho \mathbf{v} \cdot \nabla \mathbf{v} + (\nabla \cdot \mathbf{v})\rho \mathbf{v} + \nabla p\} \, dV \\ &= \int_{\Omega} \{ \rho \mathbf{v} \cdot \nabla \mathbf{v} + \nabla p\} \, dV \end{align} $$