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Planes and sphere

Projective Geometry, Solid Geometry
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Tolaso J Kos
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Planes and sphere


Post by Tolaso J Kos » Wed Dec 16, 2015 12:38 pm

Consider two parallel planes that have a distance $d \leq 2r$ and intersect a sphere of radius $r$. (both planes intersect the sphere.) Prove that the area of the surface of the sphere enclosed by both planes is only dependant by $r, d$ and not by the position of the two planes with respect to the sphere.
Historical Note
This result was first proved by Archimedes in "On the Sphere and Cylinder". His proof is quite amazing since he is using -what we call today - Riemann Sums.
Imagination is much more important than knowledge.
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Grigorios Kostakos
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Re: Planes and sphere


Post by Grigorios Kostakos » Fri Jul 22, 2016 8:20 am

WLOG we assume that the two planes which intersect the sphere of radius $r$ are perpedicular to $x$-axis with equations $x=a$ and $x=b$ respectively.
(In the scheme the points $A$ and $B$ are points of intersection of a meridian cycle with the planes $x=a$ and $x=b$ respectively and the points $A'(a,0,0)$ and $B'(b,0,0)$ are the projections of the points $A$ and $B$ respectively, on $x$-axis.)
The area in question is the area of the surface by revolution of the meridian's arc $\overset{\frown}{AB}$ which lies between the planes, rotating it about the $x$-axis. The arc $\overset{\frown}{AB}$ is the graph of the function $y(x)=\sqrt{r^2-x^2}\,,\; x\in[a,b]$, and the area in question is
A&=2\pi\int_{a}^{b}y(x)\,\sqrt{1+(y'(x))^2}\;dx \\
&=2\pi\int_{a}^{b}\sqrt{r^2-x^2}\sqrt{1+\frac{x^2}{r^2-x^2}}\;dx \\
&=2\pi\int_{a}^{b}\sqrt{r^2-x^2}\sqrt{\frac{r^2}{r^2-x^2}}\;dx \\
&=2\pi\,r\int_{a}^{b}dx \\
Grigorios Kostakos
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