The differential equation of the balance of the momentum is $$\rho \frac{\partial{\overrightarrow{u}}}{\partial{t}}=-\rho (\overrightarrow{u} \cdot \nabla )\overrightarrow{u}-\nabla p+\rho \overrightarrow{b}$$
The continuity equation is $$\frac{\partial{\rho}}{\partial{t}}+\nabla \cdot (\rho \overrightarrow{u})=0$$
From these two equations we get $$\frac{\partial}{\partial{t}}(\rho \overrightarrow{u})=-div (\rho \overrightarrow{u})\overrightarrow{u}-\rho (\overrightarrow{u} \cdot \nabla )\overrightarrow{u}-\nabla p+\rho \overrightarrow{b}$$
I want to derive the integral form of the equation of the balance of the momentum.
In my notes there is the following way:
If $\overrightarrow{e}$ is a constant vector in space, we have $$\overrightarrow{e} \cdot \frac{\partial}{\partial{t}}(\rho \overrightarrow{u})=-div (\rho \overrightarrow{u})\overrightarrow{u}\cdot \overrightarrow{e}-\rho (\overrightarrow{u} \cdot \nabla )\overrightarrow{u}\cdot \overrightarrow{e}-(\nabla p)\cdot \overrightarrow{e}+\rho \overrightarrow{b}\cdot \overrightarrow{e}\\ =-div (p\overrightarrow{e}+\rho \overrightarrow{u}(\overrightarrow{u} \cdot \overrightarrow{e}))+\rho \overrightarrow{b} \cdot \overrightarrow{e}$$
If $W$ is a constant space (independent of $t$) the rate of change of the momentum at $W$ at the direction of $\overrightarrow{e}$ is (by applyinh the divergence theorem) $$\overrightarrow{e} \cdot \frac{d}{dt} \int_W \rho \overrightarrow{u}dV=-\int_{\partial{W}} (p\overrightarrow{e}+\rho \overrightarrow{u}(\overrightarrow{u} \cdot \overrightarrow{e}))\overrightarrow{n} dA+\int_{W}\rho \overrightarrow{b} \cdot \overrightarrow{e}dV$$
So, the integral form is $$ \frac{d}{dt} \int_W \rho \overrightarrow{u}dV=-\int_{\partial{W}} (p\overrightarrow{n}+\rho \overrightarrow{u}(\overrightarrow{u} \cdot \overrightarrow{n})) dA+\int_{W}\rho \overrightarrow{b} dV$$
Could you explain to me why we have to take the vector $\overrightarrow{e}$ to derive the integral form of the equation of the balance of the momentum??