Find the centroid of a symmetric compact manifold. What is shown below is a reference from the text Analysis on Manifolds by James Munkres

So we say that the centroid $C_M$ of a compact $k$-manifold $M$ of $\Bbb R^n$ is the point of $\Bbb R^n$ whose $i$-th coordinate is given by the identity
$$
C_M:=\frac{1}{v(M)}\int_M \pi_i
$$
where $\pi_i$ is the $i$-th projection function.
Now I would like to discuss the solution of the following problem.
Show that if $M$ is symmetric with respect to the subspace $x_i=0$ of $\Bbb R^n$ then $c_i(M)=0$
To follow the solution given by Yan Zeng

So to use the theorem $25.4$ is necessary to prove that the set
$$
M\cap\{x\in\Bbb R^n:x_i=0\}
$$
has measure zero in $M$ but unfortunately I did not able to do this; moreover in the proposed solution I do not see which are the set $M_i$ of the theorem $25.4$. So could some one help me, please?
 A: I don't know what Yan Zeng had in mind when referring to Theorem 25.4, but it doesn't justify the formula it's supposed to justify. Apparently the idea was to use $L_0,L_1$ as the sets $M_i$ and set $K = M\cap\{x\in\mathbb R^n: x_i=0\}$. But $L_0$ and $L_1$ need not be images of coordinate patches, and it's not true that $K$ has measure zero in general. For example, in $\mathbb R^3$, if $M$ is the unit circle in the $(x_1,x_2)$-plane, then $M$ is symmetric with respect to the $x_3=0$ subspace, but the portion of $M$ lying in that subspace is all of $M$. (Some version of the argument could be made to work anyway, because the integrand is zero on $K$; but it's not a simple consequence of Theorem 25.4.)
In general, this argument is much more labored than it needs to be. The result you're trying to prove is a simple consequence of the following fact, which you should be able to prove easily: If $A\colon \mathbb R^n\to \mathbb R^n$ is an orthogonal linear transformation, $M\subset \mathbb R^n$ is a compact $k$-manifold such that $A(M)=M$, and $f\colon M\to \mathbb R$ is continuous, then
$$
\int_M f = \int_M f\circ A.
$$
