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2006-12-02 16:42:07 · 4 answers · asked by MANDY 1 in Science & Mathematics Mathematics

4 answers

What's the question? I'll assume you're not referring to a SOLID SPHERE like a billiard ball or bee-bee (sp?) pellet, but rather a sphere made out of some material having its own thickness, i.e. a sphere with a greater or lesser cavity inside it.

A real world inflated "truly spherical ball" (extremely hard to make!) would have a constant thickness, e.g the thickness of the rubber, if the inner and outer surfaces were perfect concentric spheres. A synonym for a "sphere having thickness" would also be "a hollow sphere" (for which, generally, constant finite or infinitesimal thickness would be understood depending on a real world or idealized mathematical context, respectively.)

A tennis ball is an approximation to a (constant finite thickness) "hollow sphere." This is because of the fuzziness of the surface as well as the slightly depressed "curly figure of 8 pattern" of glue-like substance that generally joins the two halves. Similarly basketballs, close approximations except for their distinctive surface panel patterns, the slightly "bubbly" surface and the place where air is pumped in. A table tennis ball is a closer approximation, though even here there tends to be a thicker equatorial belt joining the two thin halves.

If you're interested in the volume of stuff making up the material thickness of an idealized hollow sphere, that volume is of course:

4 pi {(R_2)^3 - (R_1)^3}/3, where R_2 and R_1 are the outer and inner radii, respectively.

In the limit as these two radii tend to one another, this "skin volume" tends to:

4 pi R^2 (delta R), where delta R is their difference, and R is an intermediate value of radius lying somewhere between R_2 and R_1.

In the ultimate limit used in calculus, this becomes, of course:

4 pi r^2 dr, for dr ---> 0 and r the radius of the idealized sphere (now of zero thickness). You can think of this as "the surface area" times the "infinitesimal thickness," just as you'd estimate it if the surface were able to lie flat. (The reason you can't write it that way when the inner and outer radii have a finite difference is a direct consequence of 3-D curvature of the inner and outer surfaces. Cut out any part of a kind of ball we've been discussing --- you can't lay it flat without buckling or differentially stretching it somehow!)

I hope this helps.

Live long and prosper.

2006-12-02 17:20:35 · answer #1 · answered by Dr Spock 6 · 1 0

A sphere is a 3 dimensional object. So how about a pool ball?

2006-12-03 00:47:23 · answer #2 · answered by wilkes_in_london 3 · 0 0

Any spherical shape has a thickness which is equal to it's diameter because it has three dimensions.

2006-12-03 01:10:21 · answer #3 · answered by Jason S 1 · 0 0

Do you get any anxiety..Then i can suggest that you ask someone that does..They should be able to answer your questions.

2006-12-03 00:52:28 · answer #4 · answered by Danielle 1 · 0 0

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