Mathematical model of a fluid: which viewpoint is the standard one? I've been studying fluid mechanics and I have some doubts on the mathematical model of a fluid. First we let $D\subset \mathbb{R}^3$ be a region in three-space filled with a fluid. This region has as many points as the real line. Since in Classical Mechanics when studying systems with a finite number of particles we consider each particle as a point in $\mathbb{R}^3$ I thought that each point of this region $D$ should be considered a particle of fluid. In that setting the fluid would be composed of uncountably many particles.
With this logic, I thought on labeling each particle by it's position at time $t = 0$, so that a state of the fluid would be a bijection from $D$ onto $D$. The time evolution would be then $t \mapsto \varphi_t : D \to D$ and we could summarize this with a map $\varphi : D\times \mathbb{R}\to D$ given by $\varphi(a,t) = \varphi_t(a)$. This is exactly the Lagrangian description of the fluid, so I thought I was right in considering each particle of the fluid as a point in $D$.
Now, I've been told it's wrong to think it this way, because if it were so, each particle would have zero mass. I've also been told that the right way to think is to imagine each particle as a infinitesimal volume $dV$ containing some real molecules of fluid with mass density $\rho : D\times \mathbb{R}\to \mathbb{R}$ so that each particle would have mass $\rho dV$. Since each piece of volume is infinitesimal we could still identify them with points.
My doubt is, when modeling a fluid with math which viewpoint is the standard one and why? The view point as in Newtonian and Lagrangian Mechanics of considering the fluid particles as points of $D$ or the view point of considering the fluid particles as infinitesimal chunks of fluid with volume $dV$ one at each point of $D$?
 A: The latter is definitely more standard because suppose you have a collection of different particles (e.g., $H_2O$ and $H_2O_2$). The masses of the individual molecules are different, but over an infinitesimal volume, the density could be taken as an average value.
I suppose, since $dV$ is an infinitesimal, then $\rho dV=m$ and the two pictures are equivalent, but the latter image gives a bit more intuitive sense of the fluid picture since we don't deal with particles in this model. 
So, rather than using 
$$
m\frac{d\mathbf v}{dt}=\mathbf F \\
\frac{d\mathbf x}{dt}=m\mathbf v
$$
to model $n$ particles, we can simply use the continuity + Navier-Stokes equations*:
$$
\partial_t\rho+\nabla\cdot\rho\mathbf v=0 \\
\rho\left(\frac{\partial\mathbf v}{\partial t}+\mathbf v\cdot\nabla\mathbf v\right)=-\nabla p+\nabla\cdot\mathsf T+\mathbf f
$$
and get our fluid model exactly from the properties we started off assuming: a roughly constant density, $\rho$, over an infinitesimal volume element, $dV$ (combined with the conservation law that flow in = flow out)--similarly for the velocity, $\mathbf v$, to get the Navier-Stokes equations.
Computationally speaking, the first method can only work for $10^{10}$ particles before cluster admins will be yelling at you, whereas the fluid picture is looking at way more particles than that. 

 Alternatively, you can use the Euler equations if you can assume zero viscosity.
