Angular Velocity via Extrinsic Euler Angles I am wondering if the angular velocity of a rotating coordinate system, if expressed through extrinsic Euler angles, is $(\dot{\alpha},\dot{\beta}, \dot{\gamma})$ since extrinsic Euler angles are rotations about fixed axes so the rates should be orthogonal to each other.
 A: The Rotation matrix is created with those 3 matrices
Rotation about x-axes with the angle $~\alpha~$
$$\mathbf R_x= \left[ \begin {array}{ccc} 1&0&0\\0&\cos \left( 
\alpha \right) &-\sin \left( \alpha \right) \\ 0&
\sin \left( \alpha \right) &\cos \left( \alpha \right) \end {array}
 \right]
$$
Rotation about y-axes with the angle $~\beta~$
$$\mathbf R_y= \left[ \begin {array}{ccc} \cos \left( \beta \right) &0&\sin \left( 
\beta \right) \\ 0&1&0\\ -\sin
 \left( \beta \right) &0&\cos \left( \beta \right) \end {array}
 \right] 
$$
Rotation about z-axes with the angle $~\gamma~$
$$\mathbf R_z=\left[ \begin {array}{ccc} \cos \left( \gamma \right) &-\sin \left( 
\gamma \right) &0\\ \sin \left( \gamma \right) &\cos
 \left( \gamma \right) &0\\ 0&0&1\end {array}
 \right]
$$
Example

*

*first rotation about the z-axes $~\mathbf R_z(\gamma)$

*second rotation about the new axes y'  $~\mathbf R_{y'}(\beta)$

*third rotation about the new axes z'  $~\mathbf R_{z'}(\alpha)$
hence the rotation matrix $\mathbf R~$ is
$$\mathbf R=\mathbf R_z(\gamma)\,~\mathbf R_{y'}(\beta)\,\mathbf R_{z'}(\alpha)$$
from here you obtain that angular velocity $~\mathbf\omega$
$$\mathbf\omega=\mathbf A(\alpha~,\beta~,\gamma)\,\begin{bmatrix}
   \dot\alpha \\
   \dot\beta\\
   \dot\gamma\\
  \end{bmatrix}
\Rightarrow\quad
\begin{bmatrix}
   \alpha \\
    \beta\\
    \gamma\\
  \end{bmatrix}=\int\,\mathbf A^{-1}(\alpha~,\beta~,\gamma)\,\mathbf\omega\,dt
$$
your question.
for a "small" rotation angle $~\varphi~,$ $~\cos(\varphi)=1~,\sin(\varphi)=\varphi~$
the  rotation matrix is now:
$$\mathbf R=\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
  \end{bmatrix}+
 \left[ \begin {array}{ccc} 0&-\gamma&\beta\\ \gamma
&0&-\alpha\\ -\beta&\alpha&0\end {array} \right] 
$$
and the angular velocity
$$\mathbf\omega=\begin{bmatrix}
   \dot\alpha \\
    \dot\beta\\
    \dot\gamma\\
  \end{bmatrix}
\Rightarrow\quad
\underbrace{\begin{bmatrix}
   \alpha \\
    \beta\\
    \gamma\\
  \end{bmatrix}}_{\mathbf \phi}=\int\mathbf\omega\,dt
$$
$\mathbf \phi~$ is now a pseudo vector.
hence: only for a small angles  $~\alpha~,\beta~,\gamma~$  the angles are rotation about the axes $~x~,y'~,z'~$
Edit
how to obtain the angular velocity from the rotation matrix $~\mathbf R$
with
$$\mathbf{\dot{R}}=\mathbf R\, \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}}
\\  \omega_{{z}}&0&-\omega_{{x}}
\\  -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \quad 
\Rightarrow\quad
 \left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}}
\\  \omega_{{z}}&0&-\omega_{{x}}
\\  -\omega_{{y}}&\omega_{{x}}&0\end {array} \right]=\mathbf R^T\,\mathbf{\dot{R}}
$$
I use MAPLE program to obtain the result , for the example above ,you obtain
$$ \left[ \begin {array}{c} \omega_{{x}}\\ \omega_{{y}
}\\\omega_{{z}}\end {array} \right] 
=  \underbrace{\left[ \begin {array}{ccc} 1&0&-\sin \left( \beta \right) 
\\ 0&\cos \left( \alpha \right) &\sin \left( \alpha
 \right) \cos \left( \beta \right) \\ 0&-\sin
 \left( \alpha \right) &\cos \left( \alpha \right) \cos \left( \beta
 \right) \end {array} \right]}_{\mathbf A(\alpha,\beta)} \,\underbrace{\begin{bmatrix}
  \dot\alpha \\
  \dot\beta \\
  \dot\gamma \\
\end{bmatrix}}_{\mathbf{\dot{\phi}}}\\
\begin{bmatrix}
  \dot\alpha \\
  \dot\beta \\
  \dot\gamma \\
\end{bmatrix}=\left[ \begin {array}{ccc} 1&{\frac {\sin \left( \alpha \right) \sin
 \left( \beta \right) }{\cos \left( \beta \right) }}&{\frac {\cos
 \left( \alpha \right) \sin \left( \beta \right) }{\cos \left( \beta
 \right) }}\\ 0&\cos \left( \alpha \right) &-\sin
 \left( \alpha \right) \\ 0&{\frac {\sin \left( 
\alpha \right) }{\cos \left( \beta \right) }}&{\frac {\cos \left( 
\alpha \right) }{\cos \left( \beta \right) }}\end {array} \right]
\begin{bmatrix}
  \omega_x \\
  \omega_y \\
  \omega_z \\
\end{bmatrix}
$$
the components of the angular velocity are given in the rotating system, not in inertial system. the components of the angular velocity in inertial system are
$$\mathbf\omega_I=\mathbf R\,\mathbf\omega$$
notice the singularity at $~\beta=\pi/2~$  . each rotation matrix has  singularity at some rotation angle.
A: So there is a formal way of deriving $\boldsymbol{\omega}$ from any sequence of rotations $\mathrm{R}_i$ given their angles and speeds.
1. Definitions
You have a sequence of three elementary rotations $\mathrm{R}_\alpha$, $\mathrm{R}_\beta$, $\mathrm{R}_\gamma$,  each about their axis $\boldsymbol{z}_\alpha$, $\boldsymbol{z}_\beta$, $\boldsymbol{z}_\gamma$, and with angles $\alpha$, $\beta$ and $\gamma$. The final orientation is given by $$ \mathrm{R} = \mathrm{R}_\alpha \mathrm{R}_\beta \mathrm{R}_\gamma \tag{1}$$
2. Time Derivatives
The time derivative of each elementary rotation is given by the differentiation on a rotating frame formula, considering that each column of an $\mathrm{R}_i$ matrix represents a frame fixed basis vector.
$$ \begin{aligned}
   \dot{\mathrm{R}}_\alpha & = (\boldsymbol{z}_\alpha \dot{\alpha}) \times \mathrm{R}_\alpha \\
   \dot{\mathrm{R}}_\beta & = (\boldsymbol{z}_\beta \dot{\beta}) \times \mathrm{R}_\beta \\
   \dot{\mathrm{R}}_\gamma & = (\boldsymbol{z}_\gamma \dot{\gamma}) \times \mathrm{R}_\gamma \\
\end{aligned} \tag{2}$$
3. Rotational Velocity
Similarly, the time derivative of the rotation matrix is used to define the rotational velocity vector $\boldsymbol{\omega}$
$$ \dot{\mathrm{R}} = \boldsymbol{\omega} \times \mathrm{R} \tag{3} $$
Now take (3) and apply (1) and the product rule of differentiation
$$ \begin{aligned}\boldsymbol{\omega}\times\mathrm{R} & =\tfrac{{\rm d}}{{\rm d}t}\mathrm{R}\\
 & =\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\
 & =\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\alpha}\right)\mathrm{R}_{\beta}\mathrm{R}_{\gamma}+\mathrm{R}_{\alpha}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\
 & =\left(\boldsymbol{z}_{\alpha}\dot{\alpha}\right)\times\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\mathrm{R}_{\gamma}+\mathrm{R}_{\alpha}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)
\end{aligned} \tag{4a}$$
Here we take a detour to deal with the last term. Note that we use the distributed property for rotation matrices $\mathrm{R} ( a \times b) = (\mathrm{R} a) \times (\mathrm{R} b)$.
$$\begin{aligned}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right) & =\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\right)\mathrm{R}_{\gamma}+\mathrm{R}_{\beta}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\gamma}\right)\\
 & =\left(\boldsymbol{z}_{\beta}\dot{\beta}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)+\mathrm{R}_{\beta}\left(\left(\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\mathrm{R}_{\gamma}\right)\\
 & =\left(\boldsymbol{z}_{\beta}\dot{\beta}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)+\left(\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\
 & =\left(\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)
\end{aligned} \
\tag{5}$$
Now continuing from (4a)
$$ \begin{aligned}\boldsymbol{\omega}\times\mathrm{R} & =\left(\boldsymbol{z}_{\alpha}\dot{\alpha}\right)\times\mathrm{R}+\mathrm{R}_{\alpha}\tfrac{{\rm d}}{{\rm d}t}\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\
 & =\left(\boldsymbol{z}_{\alpha}\dot{\alpha}\right)\times\mathrm{R}+\mathrm{R}_{\alpha}\left(\left(\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\left(\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\right)\\
 & =\left(\boldsymbol{z}_{\alpha}\dot{\alpha}\right)\times\mathrm{R}+\left(\mathrm{R}_{\alpha}\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)\times\left(\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\mathrm{R}_{\gamma}\right)\\
 & =\underbrace{\left(\boldsymbol{z}_{\alpha}\dot{\alpha}+\mathrm{R}_{\alpha}\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\right)}_{\text{defn. }\boldsymbol{\omega}}\times \mathrm{R}
\end{aligned} \tag{4b} $$
and since $\mathrm{R} = \mathrm{R}_{\alpha}\mathrm{R}_{\beta}\mathrm{R}_{\gamma}$ the above is used to define the rotational velocity vector and the jacobian matrix.
$$ \begin{aligned}\boldsymbol{\omega} & = \boldsymbol{z}_{\alpha}\dot{\alpha}+\mathrm{R}_{\alpha}\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\alpha}\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\dot{\gamma}\\
\boldsymbol{\omega} & =\underbrace{\begin{bmatrix}\vdots & \vdots & \vdots\\
\boldsymbol{z}_{\alpha} & \mathrm{R}_{\alpha}\boldsymbol{z}_{\beta} & \mathrm{R}_{\alpha}\mathrm{R}_{\beta}\boldsymbol{z}_{\gamma}\\
\vdots & \vdots & \vdots
\end{bmatrix}}_\text{Jacobian Matrix}\begin{pmatrix}\dot{\alpha}\\
\dot{\beta}\\
\dot{\gamma}
\end{pmatrix}
\end{aligned} \tag{6} $$
4. Recursive Formula
I like to arrange (6) as follows
$$\boldsymbol{\omega}=\boldsymbol{z}_{\alpha}\dot{\alpha}+\mathrm{R}_{\alpha}\left(\boldsymbol{z}_{\beta}\dot{\beta}+\mathrm{R}_{\beta}\left(\boldsymbol{z}_{\gamma}\dot{\gamma}+\ldots\right)\right)$$
Indicating that each successive rotation provides a $\boldsymbol{z}_i \dot{q}_i$ term on the local (intermediate) coordinate frame of its previous operation. The formula would continue on in a similar fashion if there were more than 3 rotations, although that is only seen on >6 DOF robotic arms or serial chains of rigid bodies.
