Let $S_{n-1}(R)=\{ {\bf x}\in {\mathbb{R}}^n : ||{\bf x}||^2=R^2\}$ be the sphere in ${\mathbb{R}}^n$ with radious $R>0.$ Let the projection map $\tau_m({\bf x})=(x_1,...,x_m)$ with $m\leq n.$ Now, if ${\bf x}=(x_1,...,x_n)$ is uniformly distributed on the unit sphere $S_{n-1}(1)$ (we write it as ${\bf x}\sim_{\mathcal{U}} S_{n-1}(1) )$ then the probability $||\tau_m({\bf x})||^2\leq r^2 \ (r<R)$ is $${\mathbb{P}}_n(1,r) = I_{a,b}(r^2),$$ where $a=m/2, b=(n-m)/2$ and $I_{a,b}(x)$ is the regularized incomplete beta function. This is because $||\tau_m({\bf x})||^2=\sum_{j=1}^{m} x_j^2$ follows the beta distribution ${\rm Beta}({a=\frac{m}{2},b=\frac{n-m}{2}})$ see [Lemma 1,Ref. 1].
My question concerns the computation of the following probabilities : First what will happen if ${\bf x}$ follows uniform distribution in $S_{n-1}(R)$ for some $R>0,$ i.e. $${\mathbb{P}}_n(R,r) = Pr\Big({\bf x}=(x_i)\sim_{{\mathcal{U}}} S_{n-1}(R) : ||\tau_m({\bf x})||^2\leq r^2 \Big)$$ and ${\mathbb{P}}_n(1;r_1,r_2)=$ $$Pr\Big({\bf x}=(x_i)\sim_{{\mathcal{U}}} S_{n-1}(1) : ||\tau_m({\bf x})||^2\leq r_1^2, ||\tau_{\ell}({\bf x})||^2\leq r_2^2\ \text{where} \ (m>\ell, r_1>r_2)\Big)$$ How can I compute them?
An example Say that ${\bf x}$ is uniformly distribute in $S_{2}(1),$ i.e. the unit sphere in ${\mathbb{R}}^3.$ We ask, what is the probability $x_1^2+x_2^2\leq 2/3$ and $x_1^2\leq 1/3.$ Monte Carlo simulation shows that this event has probability $\frac{1}{3}.$ That is ${\mathbb{P}}_3(1;\sqrt\frac{2}{3},\sqrt\frac{1}{3})=\frac{1}{3}.$ In fact by inspection we get $${\mathbb{P}}_n\Big(1;\sqrt\frac{n-1}{n},\sqrt\frac{n-2}{n},\cdots,\sqrt\frac{1}{n}\Big) = \frac{1}{n}.$$ EDIT 1 From a comment of Whuber easily we can compute the first probability $${\mathbb{P}}_n(R;r) = {\mathbb{P}}_n(1;r/R).$$
