# Showing that the surface area of a zone of a sphere that lies between two parallel planes is $2\pi Rh$

Show that the surface area of a zone of a sphere that lies between two parallel planes is $2\pi Rh$,

Where $R$ is the radius of the sphere and $h$ is the distance between the planes.

The fact that the surface area depends only on distance between the planes, and not where they cut the sphere.

I am looking to understand a calculus based solution.

• – Macavity Oct 13 '14 at 6:13
• i looked at it, its complicated, is there an easy way to do it, may be by using parameters – Fusion2 Oct 13 '14 at 6:29
• @KittenButcher The presented solution is using "parameters" and uses some basic calculus and integration. You should instead try to work through the solution and ask a question about a particular part of the presented proof. – AlexR Oct 13 '14 at 6:32
• You can use calculus to find the surface area of a spherical cap and then subtract. – Michael Burr Aug 24 '17 at 11:43

Consider an "infinitesimal latitude annulus" $A$ of geographical width $\Delta\theta$, positioned at the geographical latitude $\theta\in\left]-{\pi\over2},{\pi\over2}\right[$. Its area is given by $${\rm area}(A)=2\pi R\cos\theta\cdot R\Delta\theta\ .$$ Now a glance at the "infinitesimal right triangle" with hypotenuse $R\Delta\theta$ reveals that the increment of the $z$-coordinate across $A$ amounts to $\Delta z=\cos\theta\> R\Delta\theta$. It follows that in fact $${\rm area}(A)=2\pi R\>\Delta z\ .$$ You can put it this way: The projection of $S^2$ along horizontal rays onto the cylinder touching $S^2$ along the equator is area-preserving.

A sphere is generated by rotating the right portion of a circle centered at the origin and with radius R about the y-axis, that is, by rotating the curve x=√R^2-y^2 about the y-axis, so the surface area of the required zone would be ∫ (from -c to c) 2π*√R^2-y^2*√R^2/R^2-y^2 dy, now we know that 2c=h( assume that c>0), we got h*∫ 2πR dy=2πRh