Recall the following information about cylinders and cones with radius r and height h:
|
Solid |
Volume |
Center of Gravity |
Cylinder |
\(\pi r^2 h\) |
On the cylinder's axis, half-way between top and bottom |
Cone |
\(\frac13\pi r^2 h\) |
On the cone's axis, three times as far from the vertex as from the base |
Suppose a sphere with radius r is placed inside a cylinder whose height and radius both equal the diameter of the sphere. Also suppose that a cone with the same radius and height also fits inside the cylinder, as shown below.
![Cylinder, sphere, and cone Cylinder, sphere, and cone](/sites/default/files/images/cms_upload/p405139.jpg)
We place the solids on an axis as follows:
![](/sites/default/files/images/cms_upload/p506564.jpg)
For any point S on the diameter AC of the sphere, suppose we look at a cross section of the three solids obtained by slicing the three solids with a plane containing point S and parallel to the base of the cylinder. The cross-sections are all circles with radii SR, SP, and SN, respectively. What Archimedes discovered was that if the cross-sections of the cone and sphere are moved to H (where |HA| = |AC|), then they will exactly balance the cross section of the cylinder, where HC is the line of balance and the fulcrum is placed at A.
![](/sites/default/files/images/cms_upload/p607941.jpg)
This is not hard to show. If the radius of the sphere is \(r\), the origin is at \(A\), and the \(x\) coordinate of \(S\) is \(x\), then the cross-section of the sphere has area \(\pi(r^2-(x-r)^2)=\pi(2r x-x^2)\), the cross-section of the cone has area \(\pi x^2\), and the cross-section of the cylinder has area \(4\pi r^2\). So according to the law of the lever, in order for the above balancing relationship to hold we need to following equation to be true: \[2r\left[\pi x^2+\pi(2r x-x^2)\right]=4\pi r^2 x\] which can easily be verified.