Rotation of a fluid element

Consider the following two-dimensional fluid element:

After some time $dt$, the angle sides $AB$ and $AD$ will have rotated a small angle $d\alpha$ and $d\beta$, respectively, due to the velocity difference that exists between points $A-B$ and point $A-D$. The differential angles are:

$tan(d\alpha) \approx d\alpha=\frac{\frac{\partial v}{\partial x}dxdt}{dx}$

$tan(d\beta) \approx d\beta=\frac{\frac{\partial u}{\partial y}dydt}{dy}$

Rearranging the previous relations

$\frac{d\alpha}{dt}=\frac{\partial v}{\partial x}$

$\frac{d\alpha}{dt}=\frac{\partial u}{\partial y}$

Given that the rotation is defined positive if it's counterclockwise, the angle $d\beta$ is actually:

$\frac{d\beta}{dt}=-\frac{\partial u}{\partial y}$

The angular velocity of the element, along the $z$ axis, is defined as the average angular velocity of sides $A-B$ and $A-C$. Thus

$\Omega_{z}=\frac{1}{2}(\frac{d\alpha}{dt}+\frac{d\beta}{dt})=\frac{1}{2}(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y})$

Similarly, for the other two directions:

$\Omega_{x}=\frac{1}{2}(\frac{\partial w}{\partial y}-\frac{\partial v}{\partial z})$

$\Omega_{y}=\frac{1}{2}(\frac{\partial u}{\partial z}-\frac{\partial w}{\partial x})$