Angular velocity of a fluid element If we have a fluid element that is subjected to:

*

*Translation

*Rotation

*Extensional strain (dilatation)

*Shear strain

As in this picture starting from time $t$

it can be shown that angles $d\alpha$ and $d\beta$ equal $\frac{\partial v}{\partial x} dt$ and $\frac{\partial u}{\partial y} dt$ respectively, most textbooks I've encountered defines the rate of rotation (angular velocity) of this fluid element as the average of the rate of rotation of the two angles (the minus sign due to the difference in rotation directions):
$$\omega = \frac{1}{2}\left (\frac{\partial v}{\partial x} -  \frac{\partial u}{\partial y}\right )$$
My question is about the accuracy of this definition, how do we know that the rate of rotation of the fluid element is simply the rate of the arithmetic mean of $d\alpha$ and $d\beta$? What if one side is deforming at very much high speed than the other?
 A: It's important to note that these displacements are all infinitesimal, so while one may be thousands of times the other, during an infinitesimal time slice the displacement is still infinitesimal. If one displacement is much much greater than the other, then that displacement will effectively be half rotation and half distortion, if you rotated the fluid back, then the two axes would displace equally. The reason there's no square term or sines or cosines is that this is a linear approximation, that works because all of the displacements are infinitesimal.
A: Here is an alternative way to relate the vorticity (denoted as $\vec{\omega}$) and the average angular velocity that I think is more transparent.  
Consider the average angular velocity around an infinitesimal loop of radius $l$: 
$$ \frac{1}{2\pi l}\oint_{\partial A} \frac{\vec{u}}{l} \cdot d\vec{s}$$
By Stokes's theorem, we find 
$$ \frac{1}{2\pi l}\oint_{\partial A} \frac{\vec{u}}{l} \cdot d\vec{s} =\frac{1}{2\pi l^2}\int_A \vec{\omega}\cdot d\vec{A}$$
Taking $l\to 0$, we find 
$$ \frac{1}{2\pi l}\oint_{\partial A} \frac{\vec{u}}{l} \cdot d\vec{s}  \to \frac{1}{2}\vec{\omega} \quad \text{as}\quad  l \to 0.$$
So, we arrive as the same result that you have found for the square you considered, but in this case we have averaged over a continuous set of values of the angular momentum, instead of just the angular momentum of the two line elements. 
Note, it is the vorticity, not the angular momentum, that we care about in fluid mechanics (in fact, vorticity is the property that characterizes a fluid). 
I hope this helps. Feel free to ask any questions for clarity. 
A: In the time $dt$ your diagram shows the infinitesimal transformation M was done to the box (eg: the corners of the box):
$$
M=I+\begin{bmatrix} 0 & d\beta \\ d\alpha & 0 \end{bmatrix}
$$
$$
=I+\begin{bmatrix} 0 & {1 \over 2}(d\beta + d\alpha) \\ {1 \over 2}(d\beta + d\alpha) & 0 \end{bmatrix} + \begin{bmatrix} 0 & {1 \over 2}(d\beta - d\alpha) \\ -{1 \over 2}(d\beta - d\alpha) & 0 \end{bmatrix}
$$
This says the box was parallelepiped strained by ${1 \over 2}(d\beta + d\alpha)$ radians and rotated by ${1 \over 2}(d\beta - d\alpha)$ radians.  The strain and rotation can be done simultaneously as written or done in either order (because the angles are infinitesimal):
$$
=(I+\begin{bmatrix} 0 & {1 \over 2}(d\beta + d\alpha) \\ {1 \over 2}(d\beta + d\alpha) & 0 \end{bmatrix})
(I+\begin{bmatrix} 0 & {1 \over 2}(d\beta - d\alpha) \\ -{1 \over 2}(d\beta - d\alpha) & 0 \end{bmatrix})
$$
$$
=(I+\begin{bmatrix} 0 & {1 \over 2}(d\beta - d\alpha) \\ -{1 \over 2}(d\beta - d\alpha) & 0 \end{bmatrix})
(I+\begin{bmatrix} 0 & {1 \over 2}(d\beta + d\alpha) \\ {1 \over 2}(d\beta + d\alpha) & 0 \end{bmatrix})$$
Since the rotatation was done in a time $dt$ seconds, the box has an angular velocity of ${1 \over 2}(d\beta - d\alpha)$ radians per second.
