CodalSearch this book — or all of Codal…⌘K
nydus/ArchimedesPublic
Page 44 of 71
Table of Contents

Archimedes

revolution are inscribed and circumscribed to a segment less than a hemisphere, and the theorem about the surface of the segment is finally proved in Prop. 42. Prop. 43 deduces the surface of a segment greater than a hemisphere. Prop. 44 gives the volume of the sector of the sphere which includes any segment.

Book II begins with the problem of finding a sphere equal in volume to a given cone or cylinder; this requires the solution of the problem of the two mean proportionals, which is accordingly assumed. Prop. 2 deduces, by means of 1., 44, an expression for the volume of a segment of a sphere, and Props. 3, 4 solve the important problems of cutting a given sphere by a plane so that (a) the surfaces, (b) the volumes, of the segments may have to one another a given ratio. The solution of the second problem (Prop. 4) is difficult. Archimedes reduces it to the problem of dividing a straight line AB into two parts at a point M such that

MB : (a given length) = (a given area) : AM².

The solution of this problem with a determination of the limits of possibility are given in a fragment by Archimedes, discovered and preserved for us by Eutocius in his commentary on the book; they are effected by means of the points of intersection of two conics, a parabola and a rectangular hyperbola. Three problems of construction follow, the first two of which are to construct a segment of a sphere similar to one given segment, and having (a) its volume, (b) its surface, equal to that of another given segment of a sphere. The last two propositions are interesting. Prop. 8 proves that, if V, V′ be the volumes, and S, S′ the surfaces, of two segments into which a sphere is divided by a plane, V and S belonging to the greater segment, then

S² : S′ ² > V : V′ > S3/2 : S′ 3/2.

Prop. 9 proves that, of all segments of spheres which have equal surfaces, the hemisphere is the greatest in volume.

44