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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.

If you are wondering what is interesting about this ?

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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.

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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

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.

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