The Archimedes Problem
Archimedes' greatest achievement was his anticipation of some of the fundamental ideas of calculus. The problem given here is a classic example of a problem that most mathematician today would regard unsolvable without a knowledge of calculus (indeed it is found in many calculus textbooks) but that yielded readily to Archimedes ingenious method.
Two circular cylinders intersect at right angles. If each cylinder has a radius of one unit, what is the volume of solid figure that is common to both the cylinders ?
No Surviving record was found how Archimedes solved this problem. there is, however, a startling simple way to obtain the answer; indeed, one need to know a little more than formula for the area of the circle(Pi times square of the radius) and the formula for the volume of the sphere (Four-thirds Pi times the cube of the radius). It may have been the method Archimedes used. In any case, it has become a famous illustration of how calculus often can be sidestepped by finding a single approach to the problem.
Two circular cylinders intersect at right angles. If each cylinder has a radius of one unit, what is the volume of solid figure that is common to both the cylinders ?
No Surviving record was found how Archimedes solved this problem. there is, however, a startling simple way to obtain the answer; indeed, one need to know a little more than formula for the area of the circle(Pi times square of the radius) and the formula for the volume of the sphere (Four-thirds Pi times the cube of the radius). It may have been the method Archimedes used. In any case, it has become a famous illustration of how calculus often can be sidestepped by finding a single approach to the problem.
Labels: friday special, mathemagic, puzzle
posted by Rajesh Lal at
9:41 PM
7 Comments:
sooooo.....what's the question?
Hint/spoiler (http://www.retards.org/projects/rot13/):
Vzntvar n fcurer vafpevorq va gur vagrefrpgvba bs gur plyvaqref naq n pebff-frpgvba bs gur vagrefrpgvba ol n cynar cnenyyry gb obgu nkrf bs gur plyvaqref. Ng rnpu cbvag gur pebff frpgvba bs gur vagrefrpgvba vf n fdhner naq gur pebff frpgvba bs gur fcurer vf n pvepyr vafpevorq va gung fdhner.
Pbafvqre gur engvb bs nernf bs gurfr pvepyrf naq fdhnerf.
I agree kind of with surge, but other than that good job rajesh this was a real toughie
Is the question what is the volume of the cylinders put together, or the common area both cylinders occupy?
common to both
it would be the area of what the cylinders are at so, about 90 units of area and 120 units cubed
Imagine a sphere inscribed in the intersection of the cylinders and a cross-section of the intersection by a plane parallel to both axes of the cylinders. At each point the cross section of the intersection is a square and the cross section of the sphere is a circle inscribed in that square.
Consider the ratio of areas of these circles and squares. If radius of one of these circles is r, then its area is pi*r^2 and the corresponding square's area is 4r^2, making the ratio of the square's to circle's area 4/pi.
Since this holds true at every cross section, the ratio of the intersection volume to volume of the inscribed sphere is also 4/pi. Since the inscribed sphere's volume is 4/3 pi (because the radius is unit), volume of the intersection is 16/3 or 5 1/3.
Illustrated here.
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