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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

#1

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.
sfaira.JPG
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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

#2

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.
spharea_bwn_2planes.png
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(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
\begin{align*}
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 \\
&=2\pi\,r\,(b-a)\\
&=2\pi\,r\,d\,.
\end{align*}
Grigorios Kostakos
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