proof-sphere in a cylinder(3D)

There is a classic Greek proof, which does not explicitly use calculus. [Calculus as we know it was created much later.]

The secret is this:
Take a hemisphere. Surround it by a cycliner of the same radius as the hemisphere, and the height of the hemisphere. We assume you know the volume of this cylinder.

(pi R2) (R) = Area of base x height.

Now, take an inverted right circular cone in the cylinder. with the 'base' of the cone at the top of the cyliner, and the point at the bottom (at the center of the hemisphere). We assume you know the volume of a cone:

(1/3) base x height = (1/3)(pi R2)(R)
Proposition:
On any horizontal slice of this configuration, the area of the cross section of the hemisphere
= the area of the cross section of the cylinder - the area of the cross section of the inverted cone.

Proof:
This just involves writing out the areas of the three different circles and seeing that the way the radius of the circular cross section of the hemisphere changes and the way the cross section of the cone changes match up.

Conclusion (General principle):
If two solids have cross sections of equal area for all horizontal slices, then the have the same volume.

Therefore the volume of the hemisphere
= volume of cylinder - volume of cone
= (pi R3) - (1/3) (pi R3)
= (2/3) (pi R3)

The volume of the sphere is twice that = (4/3)(pi R3).

One neet feature of this approach is that it DOES lead to a general method which WILL be used later in calculus - the so called 'slab method' for finding volumes.


This non-calculus proof is copied from:
http://mathcentral.uregina.ca/QQ/database/QQ.09.01/rahul1.html

You may go to the Physics Forum.
Many people had discussed about it and posted some calculus proofs there as well:
http://www.physicsforums.com/showthread.php?s=d1349578fcc86fcc592b990b90b14262&t=63654