Let the surface $E_2 = \{(x_1,x_2,x_3)\in{\mathbb{R}}^3 : x_1^2+x_2^2+x_3^2=1, x_1^2+x_2^2\leq \frac{2}{3}, x_1^2\leq \frac{1}{3} \}.$ We want to compute the surface area $a(E_2)$ of $E_2.$ That is, $$a(E_2)=\iint_{E_2} d\sigma = 2\iint_{{\mathbb{B}}_2} \frac{dx_1 dx_2}{\sqrt{1-x_1^2-x_2^2}}, \text{where} \ {\mathbb{B}}_2=\{x_1^2+x_2^2\leq 2/3, x_1^2\leq 1/3\}.$$ We tried Monte Carlo, i.e. we uniformly generate enough points on the sphere $x_1^2+x_2^2+x_3^2=1$ and we count how many times are in the set ${\mathbb{B}}_2.$ If I am correct, the probability is 1/3, so $a(E_2)$ must be equal to $V_2(1) = 4\pi/3.$ I am looking for a rigorous proof. One idea to compute the previous integral explicit is to separate ${\mathbb{B}}_2$ to the two triangles \begin{equation*} \begin{split} T_1=\{(x,y):0\leq x\leq 1/\sqrt{3}, -x\leq y\leq x\} \\ T_2=\{(x,y):-1/\sqrt{3}\leq x\leq 0, x\leq y\leq -x\} \end{split} \end{equation*} and two (symmetric) polar regions, \begin{equation*} \begin{split} P_1=\{(r,\theta):0\leq r \leq \sqrt{2/3},\ \pi/4 \leq \theta\leq 3\pi/4\},\\ P_2=\{(r,\theta):0\leq r \leq \sqrt{2/3},\ 5\pi/4 \leq \theta\leq 7\pi/4\}. \end{split} \end{equation*} In fact I want to find $a(E_{n-1})$ where $E_{n-1}$ is the surface, $$E_{n-1} = \big\{{\bf x}=(x_i)_i\in{\mathbb{R}}^n:x_1^2+x_2^2+\cdots +x_n^2=1,\\ x_1^2+\cdots +x_{n-1}^2\leq (n-1)/n,...,\\ x_1^2+x_2^2\leq 2/n,\\ x_1^2\leq 1/n \big\}. $$ By inspection we get $a(E_{n-1})=V_{n}$ where $V_n:$ the volume of $n-$dimensional unit sphere.
Any hint for the rigorous proof either for the case $n=3$ or for the general case?