Let $(X_1,X_2,X_3)$ be a random variable that has the uniform distribution on the unit sphere $S_2(1)=\{{\bf x}=(x_1,x_2,x_3)\in{\mathbb{R}}^3: ||{\bf x}||=1\}$. As also noted here, the marginals are such that $X_i^2\sim Beta(\frac{1}{2},1). $ Further analysis shows that $X_1^2+X_2^2 \sim Beta(1,\frac{1}{2}).$ Thus, we can easily find the probability $Pr(x_1^2+x_2^2<\alpha)$ using the beta distribution. This probability is equal $I(1,\frac{1}{2};\alpha)$ (the regularized incomplete beta function with parameters 1, 1/2) My question concerns how to find the joint density for $X_1^2$ and $X_1^2+X_2^2.$ As far as I understand, these random variables are not independent (right?), so it is not straightforward to find the joint density. I.e. how to compute $Pr(x_1^2<\alpha,\ \ x_1^2+x_2^2<\beta).$
Also, the joint distribution of $(x_1,x_2)$ has density $$\frac{\Gamma(3/2)}{\Gamma(1/2)\pi}\Big(1 - x_1^2 - x_2^2\Big)^{-1/2}=\frac{1}{2\pi}\Big(1 - x_1^2 - x_2^2\Big)^{-1/2},$$ see [example 3.1.3] of K. T. Fang, s. Kotz and Kai-Wang Ng, Multivariate and related distributions.
First I was thinking somehow to compute this probability using the conditional probability for $X_1^2|X_1^2+X_2^2.$ I made some experiments and found that the histogram for $Pr(X_1^2<1/3|X_1^2+X_2^2<2/3)$ is the following
Another way is to compute the suitable double integrals (since this probability translates to an area section of a spherical cap) which is tedious. Although is doable and I managed for instance to prove that $Pr(X_1^2<1/3,\ X_1^2+X_2^2<2/3)=1/3.$ Also, this approach, if we let the dimension of the sphere $n$ to become larger does not seem good. So I am looking for an apporach using only properties of distributions.
