Meaning of scalar product with normal in fluid flow I'm having trouble with understanding the meaning of the following sentence:

Let $n$ be the outward normal defined at points of the boundary of a region $W$  and let $dA$ denote the area element on this boundary. 
  The volume flow rate across $\partial W$ per unit area is $$u \cdot n$$  

Can someone please explain to me why does this scalar product interpretation is the volume flow rate? Is there any way I can see this?
[edit: $u$ is the velocity field of the fluid ] 
Thanks !!!
 A: At time $t$ the fluid (moving at velocity $v$) moves a distance $v\Delta t$. If the cross-section of the pipe has area $A$, then the volume that moves past a given flat surface
is $\Delta V = Av\Delta t$. The flow rate is the volume per time, ${\Delta V\over \Delta t}=Av$.

However, if the cross-section isn't perpendicular to the fluid flow (shown below),

we adjust our calculations accordingly. The fluid still moves a distance $v\Delta t$, and the volume that moves through the cross section is the area $A$ times $v\Delta t$. The area of a parallelogram is the length of one side times the perpendicular distance $h$ from
that side to its opposite side. (See below.) 

Similarly the volume of a parallelepiped is the area of one side times
the perpendicular distance from that side to the side opposite. The perpendicular distance is $v\Delta t\cos\alpha$. It can be described by the angle that the normal to the
plane makes with the direction of the 
fluid velocity, which yields
$$
\Delta V = Ah = A(v\Delta t)\cos\alpha.
$$
The flow rate is then 
$${\Delta V\over \Delta t}=A v\cos\alpha.\tag{1}$$
If we introduce the normal vector $\hat{n}$, then $(1)$ can be rewritten in terms of a dot product involving $\hat{n}$:
$$
{\Delta V\over \Delta t}=A v\cos\alpha=A\vec{v}\cdot\hat{n}.
$$
Finally, the volume flow rate per unit area that you seek is given by
$$
{\Delta V/\Delta t\over A}=\vec{v}\cdot\hat{n},
$$
where $\vec{v}$ is the fluid velocity.
Source for diagrams (and similar explanation).
