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how to prove that the volume of the cone is 1/3S*H without using any calculus----no intergration!:D
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This works just like calculus, except you don't actually use an integral.
The strategy is to find the ratio of the volumes of a cone and it's respective cylinder, since once you find that, that remains constant no matter how you scale them.
Let the radius of the base = r, and without loss of generality (you'll see why), assume that r is an integer. Let the height = r also (to spare us of complexity)
We can approximate the volume of the cone by adding the volumes of r discs, where each disk has an integral radius from 1 to r.
(Ex, the first disk has radius 1, the next has 2, the next has 3 etc... the last has radius r)
Let the height of each disk be 1.
The volume is therefore:
summation of k as k ranges from 1 to r of
The volume of a cylinder with the same base and height is
Note that if we had used a height other than r, the height would've canceled out anyway once we divided by the volume of the cylinder.
Obviously, the more disks we have (the bigger r is), the more accurate the ratio will be.
As r gets very very big,
At this point, note that since r is getting very very big, the assumption that it is an integer get's less and less important, because we are taking an infinte number of "samples".
So effectively, the ratio is
, and the volume of the cone is the volume of the cylinder.Volume of a cylinder is Area of base x Height, so the volume of a cone is 1/3 x Area of Base x Height
Again, these kinds of approximations work just like calculus and are the foundation of calculus, but you should be able to understand it without calculus.
PS - what is the latex tag on this forum? I'm sick of writing without latex...
PPS - Thanks John
Last edited by God (2006-01-28 04:14:08)
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T h e L a T e X t a g i s [ m a t h ]Equation here.[ / m a t h ].
igloo myrtilles fourmis
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