Someone just sent me this one:
A ten-inch long cylindrical hole is bored through the center of a sphere so that the axis of the cylinder coincides with the diameter of the sphere. Find the volume of the resulting object. Note: The length of the cylinder is measured after the hole is bored; i.e., you don't count the heights of the two spherical caps (one at each end) as part of the length of the cylinder. (FYI: The volume of a sphere is 4/3 pi times the cube of the sphere's radius.)
A ten-inch long cylindrical hole is bored through the center of a sphere so that the axis of the cylinder coincides with the diameter of the sphere. Find the volume of the resulting object. Note: The length of the cylinder is measured after the hole is bored; i.e., you don't count the heights of the two spherical caps (one at each end) as part of the length of the cylinder. (FYI: The volume of a sphere is 4/3 pi times the cube of the sphere's radius.)
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Re: drilling through a sphere
Thu, February 23, 2006 - 4:45 AMIf the solution is independent of the radius of the hole drilled,
then the volume must be that of sphere diameter 10 inches, which is the limiting case of the radius of the hole approaching zero.
Karl -
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Re: drilling through a sphere
Mon, February 27, 2006 - 7:05 AMThen the next challage is to prove that the volume is indeed independent of the radius of the hole drilled.
Karl -
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Re: drilling through a sphere
Mon, February 27, 2006 - 9:00 AMI'd like to see that. I started thinking about how to prove it, but then stopped thinking about it after 2-3 seconds. I guess you would basically just need the formulas for determining volumes of cylinders and "spherical caps." Shouldn't be more complicated than that, right? -
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Re: drilling through a sphere
Wed, March 1, 2006 - 7:55 PMSee the cylinder isn't specified as a solid object.
So its volume makes no diference. -
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Re: drilling through a sphere
Sun, March 5, 2006 - 11:48 AMI don't think that's it, this isn't a trick question, but now I have to think back 20 years to when I was in high school and was given a problem much like this one.
let h=cyl height, rc= cyl radius, rs= sphere radius
Sphere vol=(4*pi*rs^3)/3
Cyl vol=h*pi*rc^2
relationship between h and rc, rs in this problem (I have to think about triangles here.... Pythaogrian theorem)
4rs^2 = h^2 + 4rc^2
h=10
so... 4rs^2 = 100 + 4rc^2
and now I want to use calculus to get the volume of the object. Integrate pi*2r*sqrt(4rs^2-4r^2) with respect to r, r=rc .. rs
ah there has to be an easier way... and there is.... found on web at mathforum.org/library/drm...w/55631.html
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