Understaind configuration space of rigid body, necessity of integral frame

Question

I'd like to undertand more deeply the concept of configuration space of a rigid body. The question arises from the formula's regarding know physics quantities for rigid bodies. Let's took in exam a random one, for example the angular moment $M_Q$, with respect to $Q \in \mathbb{E}^3$, where $\mathbb{E}^3$ is the underlying affine space :

$$M_Q = \int_C \rho(x')(\chi(q;x')-x_Q) \times v(q,\dot{q},x')dx' = \int_C \rho(x')(x_{O'}+Rx'-x_Q)\times(v_{O'}+\omega \times R x')dx' $$

Here $C$ denotes the rigid body,$\Sigma'$, $\rho$ the density (integrable on the coordinate of $C$ in $\Sigma'$), $q = (x_{O'},\alpha)$, $R = R(\alpha)$ where $\alpha = (\varphi,\theta,\psi)$ are the Euler's angles.

What I asked myself is why do we need to integrate on a solidal frame (I don't the terminology here could be integral with the rigid body, what I mean is that the velocity of the points of the rigid body in $\Sigma'$ are $0$, so costant coordinates). Running through my notes I think the reason is due to the following theorem :

Theorem : Consider a rigid body with three points $P_1,P_2,P_3$ not aligned. The map $$\phi : \mathbb{R}^3 \times SO(3) \longrightarrow \mathbb{R}^9$$ such that $\phi(x_{O'},R) = (x_{O'}+Rx_1',x_{O'}+Rx_2',x_{O'}+Rx_3')$ where $x_j'$ are the coordinates of $P_j$ in the solidal frame, is a diffeomorphism on the image $\phi(\mathbb{R}^3 \times SO(3))$.

Doubts I'd like to clear :

$\bullet \hspace{0.1cm}$ I do understand the proof of the theorem, which depends on taking a solidal frame $\Sigma'$, what I don't understand is why is necessary. In general can't be possible to determine the coordinates of the rigid body in a fixed frame $\Sigma$, without passing through a solidal frame $\Sigma'$? If so, does a manageable counterexample exists ?

(Specifically I don't see how the Euler angles play a special role when integral frames are involved, I think the same Euler angles could describe passing from two orthonormal basis to another, without the second be "fixed" to the body, am I wrong ?)

$\bullet \hspace{0.1cm}$ In general, if I don't take a solidal frame, the space of configuration could not be a manifold ? So it shouldn't make sense integrating on that ? Is this correct ? Otherwise, why bother to define the integrals passing through an integral frame ?

I'm new to Physics Stack Exchange so I apologize if I made some mistake about the policy of the questions, any help would be appreciated.

Edit : The definition of "solidal" frame I'm using is : given a frame $\Sigma' = O' e_1',e_2',e_3'$, this is called solidale to the rigid body $C$ is all the points $P_j$ of the body have $0$ velocity respect to $\Sigma'$, i.e the coordinates of the points $P_j$ are costant in $\Sigma'$.

Answer 1

1. You don't need a body frame to integrate any of the physical quantities (angular momentum, moment of inertia, energy, etc). Manageable counterexample? Consider a wheel rotating around its axis. I bet you can calculate its momentum in both fixed and body frames. Euler angles don't need to be attached to the body. Sometimes, when you solve a complex motion (for example a cone rolling on a table), you will consider an intermediate rotating frame that isn't associated with any body and describe it with Euler angles.

2. The space of configuration couldn't be a manifold in general. But in the case of a rigid body, it is. It's your mentioned $\mathbb R^3\times SO(3)$.

The reason why it's sometimes useful to integrate in a body frame, is that it splits the problem into parts like 1) calculating quantities in a body frame, which doesn't depend on time or orientation of a body 2) using known laws of transformation to get it back to the lab system if needed. In other words, it utilizes the symmetry of the rigid body motion.