navier stokes question

hey pf!

so when deriving navier stokes we have, from newton's second law, [itex]\sum \vec{F} = m\frac{d \vec{V}}{dt}[/itex] when deriving the full navier stokes (constant density) the acceleration term can be thought of as two pieces: a body change of velocity within the control volume and a mass flow leaving the control volume through the a small surface element [itex]d\vec{S}[/itex]. my question is, when momentum leaves we have: [tex]\iint \rho \vec{V}\vec{V}\cdot d\vec{S}[/tex] and through the divergence theorem we have, as an velocity [itex]\vec{V}_x[/itex] component, [tex]\nabla \cdot
(\rho V_x \vec{V})[/tex]

can someone explain how? also, why is there no negative sign (momentum leaving, not entering)?

thanks!