Determine the area "S" which the triangle "R" Project vertically upon the hemisphere. Check the attachment

Hello!
If a surface is given as an image of a scalar function y=f(x,z), defined on some region D on (x,z) plane, then the corresponding surface area is
int int_D sqrt(1+((del f)/(del x))^2+((del f)/(del z))^2) dx dz.
We have f(x,z) = sqrt(4-x^2-z^2), so (del f)/(del x) = -x/(sqrt(4-x^2-z^2)) and (del f)/(del z) = -z/(sqrt(4-x^2-z^2)). The expression under integral therefore is
sqrt(1+x^2/(4-x^2-z^2)+z^2/(4-x^2-z^2)) = 2/sqrt(4-x^2-z^2).
The problem is to express the double integral int int_D (2 dx dz)/sqrt(4-x^2-z^2) as a sequential one-dimensional integral. Because of the symmetry we can integrate only by a half of the triangle and then multiply by 2. The integration region is from -1/2 to 1 by z and from 0 to (1-z)/sqrt(3) by x.
A = 2 int_(-1/2)^1 (int_0^((1-z)/sqrt(3)) 2/sqrt(4-x^2-z^2)dx) dz.
This integral isn't so simple (inner integral is relatively simple), but at least we can compute it approximately. The answer is about 1.343.