We'll, I think there are many ways to calculate the surface area of a sphere. The method that is most recent to me is using surface integrals. Here's how:
In this case you of course integrate 1 over the surface.
If S is the surface:
So we want to solve the following:
Integrate[1 dS, over S]
To find that, we need a parametrization for S. This could easily be:
p(u,v) = r*cos(u)*cos(v)*i + r*sin(u)*cos(v)*j + r*sin(v)*k
0<=u<=2*pi
-pi/2<=v<=pi/2
Now we can say that:
Integrate[1 dS, over S] = Integrate[1 |dp/du x dp/dv| du dv, 0<=u<=2*pi, -pi/2<=v<=pi/2]
|dp/du x dp/dv| is a lot of writing, so I'll just put the answer here:
|dp/du x dp/dv| = r^2*cos(v)
So we integrate:
Integrate[r^2*cos(v) du dv, 0<=u<=2*pi, -pi/2<=v<=pi/2] =
2*pi*Integrate[r^2*cos(v) dv, -pi/2<=v<=pi/2] =
2*pi*(r^2*sin(pi/2) - r^2*sin(-pi/2)) = 4*pi*r^2
Is that enough proof?
Kevil