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Topic: Crossed Cylinders (Read 7333 times) 

SWF
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Posts: 879


Crossed Cylinders
« on: Jan 22^{nd}, 2003, 7:40pm » 
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From William Wu's Widdles page: Two cylinders of equal radius are intersected at right angles as shown at left. Find the volume of the intersection between the two cylinders, without using calculus! A 3D picture of the intersection is shown on the riddle page. Hint (medium hint  exactly which high school formulae you need):1) Area of circle = pi * radius2, and 2) Volume of sphere = (4/3) * pi * radius3 Note: Solved by the mathematician Archimedes (287 B.C.  212 B.C.), waaay before calculus came around!


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SWF
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Posts: 879


Re: Crossed Cylinders
« Reply #1 on: Jan 22^{nd}, 2003, 7:42pm » 
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Based on the hints I think the solution he is looking for is: Imagine a sphere of radius R next to the shape. Slices through this shape are squares that vary such that their area is 4/pi times the area of the corresponding section of the sphere. So volume is 4/pi times that of the sphere or 16R^{3}/3. There is a solution which I think is better: I will just give a hint. There is no need to use the formula for volume of a sphere or any shape other than a cube. The rest can be figured out from basic geometry and no calculus. Actually, I'd say that both of these solutions are not too far from being calculus.


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Harry Dewulf
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tentative answer:the volume of a cube of edge length d (equal to the diameter of the cylinder) minus half the difference between the volume of the same cube and a cylinder whose length is equal to its diameter (d) No maths at all, just visualisation.


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Matt
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the intersection of the 2 cylinders is very similar to a sphere, one made up of 4 "orange slices" if you will. the difference in these slices and 1/4 of a sphere being that the outside edge of the slice is flat rather than curved. By finding the ratio in the area of a triangle to the area of a semi circle, you can then take this ratio and multiply it by the volume of the sphere.


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rloginunix
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Posts: 1026


Re: Crossed Cylinders
« Reply #4 on: Oct 29^{th}, 2013, 7:27pm » 
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Known nowadays as Steinmetz solids this classical volume and its surface area was calculated by Arcimedes more than two thousand years ago. The answer for the volume is 16/3 R*R*R  sixteen thirds R cubed. My sample solution is the demonstration of the Division basic approach  break the whole into parts so simple that each one can be solved by itself, then find a way to combine the solutions of the parts into a solution of the whole. In this particular case the elementary volume is a square tile of some known easily obtained volume  square area of the base times the height, etc. Sum all these elementary volumes. Thus obtained volume will be an approximation. To find the exact volume apply the concept of a limit to it. Calculations only seem laborious but are actually quite simple. This is not calculus but something rather close to it as is anything that divides the whole into parts. Read the whole article here: http://romanyandronov.elementfx.com/bas/division/ryabassteinmetz.html At the bottom I also calculate the volume of the intersection of two elliptic cylinders.


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