Suppose we have a cylindrical pipe let's say length L = 10 m and radius R = 1 m through which water is flowing. The velocity distribution is given by Poiseuille's Law: $v(r)=\frac{\Delta P}{4L\eta}\left(R^2-r^2\right)$ with $\eta = cte$ its viscosity.
The question is how can I calculate the total momentum $v$ of the fluid? $p=mv$ but v is not a constant so what should I do? (maybe integrate?)
You need to evaluate the integral defining the momentum,
$$ \mathbf{Q} := \int_V \rho \mathbf{u} $$
performed over the volume of fluid $V$ of your interest. Here $\rho$ and $\mathbf{u}$ are the density and velocity fields.
A cylindrical shell with small thickness $dr$ at a distance $r$ from the centre of the pipe has volume $dV = 2 \pi L r \space dr$ and so contains a mass $dm = \rho dV = 2 \pi \rho L r \space dr$ of water, where $\rho$ is the density of water. All of the water in this shell has the same velocity $v(r)$, so the shell has momentum $dp = v(r) \space dm = 2 \pi \rho L r v(r) \space dr$. So the momentum of the water in the whole pipe is
$p = \int dp = 2 \pi \rho L \int_0^R rv(r) \space dr$
