Laminar flow around a cylinder using conservation of momentum I'm trying to calculate the drag force for a laminar flow around a cylinder. 
We make the following assumptions : 


*

*the flow is horizontal

*the flow is incompressible ($\vec{\nabla} \cdot \vec{v} = 0$) 

*the flow is stationnary

*We are in 2D


Here's a picture of the situation : 

I would like to start with the conservation of the momentum using Reynold Transport Theorem : 
$$ 
\int_\Omega \rho v dS + \int_{\partial{\Omega}} \rho v^2 dl = -\vec{F}_d
$$
However, in the lectures notes, there is no $ \int_\Omega \rho v dS $ term. 
In fact, it says that 
$$
  \int_{\partial{\Omega}} \rho v^2 dl = -\vec{F}_d \implies 2(\int_0^\xi -pU^2_\infty dl + \int_0^\xi pu^2 dl + \int_0^L \rho u v dl) = - \vec{F}d
$$ 
I don't understand why ?
 A: Your formula for the Reynold Transport theorem is false (you have forgotten a partial derivative in the first integral).
The correct version is shown below:
$$-\overrightarrow{F_d} = \frac{D}{Dt} \int \limits_{\Omega} \rho \overrightarrow{v}(\mathbf{x},t) dS = \int \limits_{\Omega} \rho \frac{\partial \overrightarrow{v}(\mathbf{x},t)}{\partial t} dS+ \int \limits_{\partial \Omega} \rho \overrightarrow{v}(\mathbf{x},t) (\overrightarrow{v(\mathbf{x},t)}\cdot \overrightarrow{n_{\partial \Omega}}) dl,$$
where $\overrightarrow{n_{\partial \Omega}}$ is the vector normal to the boundary $\partial \Omega$ (see for instance this Wikipedia page). If the flow is stationnary, the first term is zero, and you just have the second term remaining, that can be simplified if $\overrightarrow{v}$ is orthogonal to the surface $\partial \Omega$ into:
$$-\overrightarrow{F_d} = \int \limits_{\partial \Omega} \rho v(\mathbf{x},t) \overrightarrow{v}(\mathbf{x},t) dl$$
This can be seen as a macroscopic version of Newton's second law: for a stationary flow, the fluid exiting the surface $\Omega$ has a smaller momentum than the one entering it, so it must be that the cylinder is exerting an equivalent force $\overrightarrow{F_d}$ on the fluid such that $\overrightarrow{F_d} = \frac{d \overrightarrow{P_{\mathrm{fluid}}}}{dt}$ (and using the third law of motion, the fluid must exert a opposite force $-\overrightarrow{F_d}$ on the cylinder).
A: Is the drag not zero by d'Alembert's paradox? https://en.wikipedia.org/wiki/D%27Alembert%27s_paradox
