Surface area of a sphere segment proof

Question

I have been searching proof for area of a spherical segment $S = 2\pi Rh$ but ı coulnt find a proper explaining one yet

Answer 1

The surface area of a sphere, or more precisely the zone, is $$2\pi Rh \\$$ Proof: Let R denote the radius of the sphere, a and b denote the upper and lower radii as shown in your figure, and h denote the height of the segment. The surface is a surface of revolution about the z-axis: $$S(s)=\int{2\pi x \sqrt{1+{x'}^2}dz}$$ Note that we're working in the xz-plane, and the equation of the zone in the xz-plane, is the equation of a circle: $$x = \sqrt{R^2-z^2} \Rightarrow x'=\frac{-z}{\sqrt{R^2-z^2}} \Rightarrow {x'}^2=\frac{z^2}{R^2-z^2}.$$ Plugging back into our integral, $$\int_{\sqrt{R^2-a^2}}^{\sqrt{R^2-b^2}}{2\pi \sqrt{R^2-z^2}\sqrt{1+\frac{z^2}{R^2-z^2}}dz}$$ $$= \int_{\sqrt{R^2-a^2}}^{\sqrt{R^2-b^2}}{2\pi R \space dz}$$ $$= 2 \pi R(\sqrt{R^2-b^2}-\sqrt{R^2-a^2}) \iff \space 2\pi R h \space \space \space \square$$

Answer 2

S(R(h))=2 $\pi$ R(h) ; R is a function of h
What is 2 $\pi$ R(h) in the first place?
well its the curved surface area of the spherical zone
since R changes with the change of h, in the integration world its bounded by [0,h] //it means the area from 0 to h of our main function
integration = surface =area
I really don't know how to explain it without using calculus
You may see this : https://en.wikipedia.org/wiki/Spherical_segment