Why is the kinetic energy of a fluid given as an integral? The kinetic energy of a fluid occupying a region $\Omega \subset \mathbb{R}^3$ is given by
$$T = \frac{1}{2}\int_\Omega |v(x)|^2 dx.$$
I am looking for some physical intuition on where the above comes from given that the classical definition of kinetic energy is $T = \frac{1}{2}mv^2$. More specifically, I have two questions:

*

*What happened to the mass $m$?

*Why are we integrating?

My best guess is that the mass is taken to instead be a density, $\rho$, and we further assume that $\rho = 1$ (why is this justified?). In which case, for an infinitesimal amount of fluid, we find its average velocity $v(x)$ in that region and multiply the volume to get something like $T = \frac{1}{2}mv^2$. Repeating this over all infinitesimally small areas in $\Omega$ and summing them up amounts to taking the integral. If this is indeed correct, then my question reduces to why is taking $\rho = 1$ justified?
 A: Why are we integrating?
According to the no-slip boundary condition:

In fluid dynamics, the no-slip condition for viscous fluids assumes that at a solid boundary, the fluid will have zero velocity relative to the boundary.

Which means that if water flows in a long pipe, the outer droplets will have 0 velocity and 0 kinetic energy, while the center droplets will have the highest velocity and kinetic energy:

The only way to calculate the total kinetic energy of a given volume of water is to calculate the kinetic energy of every droplet. In other words: to integrate.
A: The kinetic energy of a fluid is the same as normal mechanics, $T=mv^2/2$. However, that's not generally useful as we don't usually have masses but densities, so we instead consider the kinetic energy density,
$$\mathcal{T}\equiv\frac{1}{2}\rho v^2.$$
Then in order to get the total kinetic energy, you must integrate over all space,
$$T=\int\mathcal{T}\,\mathrm{d}\mathcal{V}=\frac{1}{2}\int\rho v^2\,\mathrm{d}\mathcal{V}.$$
So to answer your enumerated questions, $m$ is absorbed by $\rho$ and you integrate over all space to get the total kinetic energy of the domain.
As to the subsequent question, I do not know why the source of your equation has neglected $\rho$. Presumably it was a typographical error, but that's not something someone can determine without seeing the source first hand. You may want to check other sources to confirm the equation.
A: *

*The elementary kinetic energy, i.e. the kinetic energy of a small element of fluid is
$dK = \dfrac{1}{2} dm |\mathbf{v}|^2 =  \dfrac{1}{2} \rho |\mathbf{v}|^2 dV$.


*Kinetic energy is an additive physical quantity. To get the total kinetic energy of the fluid in the volume $V$, we need to add all the elementary contributions. The fluid in the volume $V$ is a continuous medium, and integration is the operation of summing over continuous functions.
Thus, the total kinetic energy
of the fluid in volume $V$ is
$K = \displaystyle \int_V \dfrac{1}{2} \rho |\mathbf{v}|^2 dV$.
