how to prove that the volume of the cone is 1/3S*H without using any calculus----no intergration!:D
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)