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)`...

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.