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Random question about cones and cylinders volume 
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#1
Mar3114, 05:20 PM

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A cone's volume with height ##x## and radius ##y## is ##1/3## of the volume of a cylinder with height ##x## and radius ##y##.I was trying to visualize it in my head and struggled a bit.Take a rectangle triangle with height ##x## and the other side of lenght ##y## which isn't the hypothenuse , then you get the cone if you rotates the triangle such that the two corners forming height ##x## rotates on themselves at the center , and you get the the rest of the cylinder if you rotates (from the same center) the triangle that would form a rectangle with the initial rectangle triangle.This means a rotation of a rectangle triangle where only one corner is rotating on itself at the center covers 2 times the volume of a rectangle triangle rotating such that two corners rotate on themselves at the center.
(Obviously I assume the triangle is filling the volume anywhere it goes during the rotation and I assume the rotation is complete) So at the center of the rotation , in the first exemple constructing the cone the side ##x## is only "filling" the volume at the very center while it's filling all the curved area of the cylinder in the second exemple.I'm not exactly sure what I'm trying to accomplish here but I guess putting more rigor on this phenomenon in my mind would be a good start. Any help? thanks 


#2
Mar3114, 09:31 PM

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BTW: it does not matter what shape the base is. Does it matter of the cone is skewed? (i.e. make a box and cut a pyramid from it so that all the diagonal edges meet at one corner of the box. Compare volume of box to volume of pyramid. A cone base radius r and height h would be the solid of rotation of the line y=rrx/h inside 0<x<h about the line y=0. The corner of the triangle at (x,y)=(0,h) maps onto itself after a rotation of 2pi. The two y=0 corners always map onto themselves. The corresponding cylinder would be the solid of rotation of the line y=r inside 0<x<h. But I think the best way for you to get a proper feel for this it to actually build the various coneshapes and their corresponding rectangular shapes. Make them out of stiff card or paper. Pour flour or fine sand into the cone version, and see how many conefulls fit into the rectangular version. 


#3
Mar3114, 09:52 PM

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If you really prefer to think in terms of algebra, try this exercise:
imagine a rectangular block in the positive octant in Cartesian coordinates: base sides 0<x<a and 0<x<b with height 0<z<h (sketch it out).  the volume of box is V=abh If you slice it in half, diagonally, perpendicular to the xz plane, then you have a wedge. (sketch the wedge)  the volume of the wedge is: V/2 = abh/2 (prove this) Slice the wedge diagonally, perpendicular to the yz plane this time, leaving a skewed pyramid. You can see this does not halve the volume again because the widest part of the bit removed is towards the narrow end of the wedge. So the question is: What is the volume remaining? V/3 perhaps? You could do this by symmetry off the formula for a regular pyramid, but it is probably most informative to do this by integration... 


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