A history of Greek mathematics - Wilbourhall.org

42 ARCHIMEDES
ttt 2 .2 sin -^-
~2n
respectively are (when n is indefinitely increased)
precisely what we should represent by the integrals
47rr 2 .-! sin 6 dO, or 47rr 2 ,
and 7rr w . 2sin0cZ0, or 2irr 2 (l — cos a).
Book II contains six problems and three theorems. Of the
theorems Prop. 2 completes the investigation of the volume of
any segment of a sphere, Prop. 44 of Book I having only
brought us to the volume of the corresponding sector. If
ABB' be a segment of a sphere cut off by a plane at right
angles to AA', we learnt in I. 44 that the volume of the sector
OBAB' is equal to the cone with base equal to the surface
of the segment and height equal to the radius, i.e. \n . AB 2 .r,
where r is the radius. The volume of the segment is therefore
iTr.ABKr-iTr.BMKOM.
Archimedes wishes to express this as a cone with base the
same as that of the segment. Let AM, the height of the segment,
= h.
Now AB BM 2 2 : = A'A : A'M = 2r:(2r-h).
Therefore
in(AB 2 .r-BMKOM)=lw.BM 2 ^^-j i
-(r-h)l
= i7r.BM 2 .h(^h.
3 v
2r—V
That is, the segment is equal to the cone with the same
base as that of the segment and height h(3r—h)/(2r — h).