# Cylinder and a ball weighing

When I was questor in Sicily I managed to track down his <Archimedes> grave. The Syracusians knew nothing about it, and indeed denied that any such thing existed. But there it was, completely surrounded and hidden by bushes of brambles and thorns.

I remembered having heard of some simple lines of verse which had been inscribed on his tomb, referring to a sphere and cylinder modelled in stone on top of the grave.

And so I took a good look round all the numerous tombs that stand beside the Agrigentine Gate. Finally I noted a little column just visible above the scrub: it was surmounted by a sphere and a cylinder. I immediately said to the Syracusans, some of whose leading citizens were with me at the time, that I believed this was the very object I had been looking for.

Cicero (106—43 BC), Tusculan Disputations, Book V, Sections 64—66.

Translation by Michael Grant in Cicero — On the Good Life, Penguin Books, New York, 1971, Pages 86—87.

The ball and the cylinder. The cylinder base radius is equal to the ball radius and the cylinder height is equal to the ball diameter. With these dimensions, it is possible to fit the ball into the cylinder.

What is the relationship between the volumes of the cylinder and the ball? How to place these objects on the seesaw scale arms to get them in balance?

It can be verified that relationship of scale arms (arm is the distance from the fulcrum to the point where the weighted object is placed) in equilibrium will be $2:3$. Thus, the ball volume is equal to two thirds of the cylinder volume. Interestingly, the areas of objects’ surfaces are in the same relationship.

It is believed that out of his discoveries, Archimedes most appreciated the finding of the relationship between the volumes of the ball and the cylinder, thereby determining the ball volume.

The formula for a ball volume can be derived from this relationship.

Let’s use a formula for a cylinder volume — product of the cylinder base area by its height. The base area is $\pi \cdot R^2$, cylinder height is $2 \cdot R$, where $R$ is the ball radius. Thus, the cylinder volume is $(\pi \cdot R^2) \cdot (2 \cdot R) = 2 \cdot \pi \cdot R^2$.

Multiplying by the coefficient of $2/3$, we get the formula for the ball volume — $4/3 \cdot π \cdot R^3$.

It is worth considering the possibility of dividing the cylinder into two parts equal in height. If we balance the ball with the cylinder having the height equal to the ball radius, the scale’s arms relation will be exactly $4:3$