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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 ].
This works just like calculus, except you don't actually use an integral.
times the sum of the first r perfect squares
, by the formula for the sum of the first r perfect squares.
The volume of a cylinder with the same base and height is
The ratio of the area of the cone to the cylinder is therefore
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, get's very very small, and get's very very small.
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
how to prove that the volume of the cone is 1/3S*H without using any calculus----no intergration!