Archimedes' Method for Computing Areas and Volumes - Solutions to Exercises

Exercise on Proposition 4:

a) Each cross-section of the cylinder, left where it is, balances the cross-section of the paraboloid, moved to H.

b) Thus the cylinder, left where it is, will balance the paraboloid moved so that its center of gravity is H. By the Law of the Lever we have

\[(\mbox{volume of paraboloid}) \times |AH|=(\mbox{volume of cylinder})\times\frac12|AD|\]

Since \(AD=AH\),

\[\mbox{volume of paraboloid} =\frac12(\mbox{volume of cylinder})\]

c) If \(r\) is the radius of the cylinder, then the height is \(h = r^2\) (since the paraboloid is generated by the curve \(x= y^2\)). So the volume of the cylinder is \(\pi r^2 h= \pi r^4\), and so the volume of the paraboloid is \(\frac12\pi r^4\).

Exercise on Proposition 6:

a) Each cross-section of the hemisphere together with the cross-section of the cone, left where they are, balance the cross-section of the cone moved to H.

b) Thus the hemisphere and cone, left where they are, balance a cone placed so that its center of gravity is at H. Let X be the location of the center of gravity of the hemisphere. By the Law of the Lever,

\[(\mbox{Volume of hemisphere})\times |AX|+(\mbox{Volume of cone})\times\frac34|AG|=(\mbox{Volume of cone})\times |AH|\]

Now we know that the volume of the hemisphere is \(\frac23\pi r^3\), the volume of the cone is \(\frac13\pi r^3\), and \(|AH|=2|AG|\).

Putting this information into the above equation and solving for \(|AX|\) we find that \(|AX|=\frac58|AG|\).

References

For more information, consult:

[1] T. L. Heath, The Works of Archimedes, Dover, New York, 1953.

[2] Sherman Stein, Archimedes: What Did He Do Besides Cry Eureka?, Mathematical Association of America, 1999.