How do we find the volume of an Euclidean ball in $\mathbb{R}^n$?

Question

Following up on this post, I am not sure how the volume of a Euclidean ball has been found in the following extract:

Questions:

1. What does it mean when we say "the mass of a measure"?
2. What does rotational invariance mean here, and how was it established?
3. Why do we integrate the function $$x \mapsto \exp \left(-\frac{1}{2}\sum_1^n x_i^2\right)$$ to find $v_n$?
4. How is the function invariant under rotations?
5. How did we pull out $nv_n$ from the integral in (1.1)?

I apologize for asking too many questions but I'd really appreciate any help I can get in understanding the discourse above! Thanks a lot.

Answer 1

The order 12435 would have been more helpful, but:

1. A mass-$1$ measure on the surface is basically a probability density, so the mass of a region therein is the probability of a random point lying therein ("random" in the sense of whatever distribution we choose to sample). A mass-$M$ measure would be $M$ times that.
2. Rotation about the centre, which is here taken as the origin, doesn't change the surface.
3. The motive is that $\prod_i\exp(-\tfrac12x_i^2)dx_i=\exp(-\tfrac12\underbrace{\sum_ix_i^2}_{r^2})d^nx=r^{n-1}\exp(-\tfrac12r^2)drd\Omega$, with $d\Omega$ the infinitesimal element of solid angle.
4. Rotations also don't change sums of the form $\sum_ix_iy_i$ because$$R^TR=I\implies(Rx)^TRy=x^TR^TRy=x^Ty.$$
5. Integrating over the surface,$$\left(\int_{\Bbb R}\exp(-\tfrac12x^2)dx\right)^n=nv_n\int_0^\infty r^{n-1}\exp(-\tfrac12r^2)dr$$(because, while $v_n$ is the volume, $nv_n$ is the surface $\int_{S^{n-1}}d\Omega$).