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Re: sphere
Posted:
Jun 26, 1996 7:48 PM
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In article <31D15F98.5E90@dexp.com> Tim Helton <thelton@dexp.com> writes:
>
>I wish someone would explain in real world terms (that I can visualize)
>why a sphere's area is exactly equal to 4 circles of the same radius.
>I'm losing alot of sleep trying to do it myself :)
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This is actually a fascinating question. Even though it's easy enough to
prove this using calculus, that method doesn't provide much insight into WHY
(*) area(sphere) = 4 * area(disk)
where the radii of the sphere and disk are equal.
* * *
Here is a bogus but amusing "proof" which may just possibly be justifiable.
(But probably not.)
Consider the disk D(2r) of radius 2r. Since area magnifies by the square of a
linear magnification, we have
area(D(2r)) = 4 * area(D(r)).
Now notice that if C is the center of D(2r) and P is any point of its boundary
circle, there is just one semicircle (of radius r) whose endpoints are C and P,
and which bulges in the clockwise direction. Let's call this semicircle S(P).
Clearly D(2r) is the union of all the S(P)'s, which are disjoint except for all
sharing the same endpoint C.
If we think of D(2r) as lying in the xy-plane of 3-space, we may now rearrange
all the semicircles so they constitute the latitude lines of a sphere of
radius r. Each one has endpoints = the north and south poles of the sphere of
radius r, and they are otherwise disjoint.
(It may be relevant that this rearrangement can be accomplished without moving
any points on the circle of radius r in the xy-plane.)
IF we could justify that this disassembly and reassembly should preserve
area -- THEN this would be a genuine proof of (*).
--Dan Asimov
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