[ SiS ] 



Solution. Let the rad, AC — r,C:«= z, the meafurie of the 

 L AC R to rad. unity = a, and. let on, be a circular arc. defcribed" 

 with radius C // (z). Becaufe the furface of a zone contained: 

 betvvrqen a great circle and a pa,r^llel circle equals the redangle 

 p. contained by the circumferpncp of the great, circle, apd the per- 



pendicular diflance betwee.n, th?. circles*; it readily fpUows that. 

 the furface of the fphere infiftjng perpendicularly, above and, 

 below P R «. = 2 , P R. x AcIrTZTc"^ = 2 P R >c V;.j —zk'^ 

 This furface is to the furface ftanding perpendiicularly on P»j.«-R, 

 ultimately in the ratio of equality. Therefore, the fluxion of the- 

 furface over AC « R;= 2 ra ^r^.—z^J • Again it eafily follows- 

 from the lemma that the folid infifting perpendicularly, on P R «o = 



— — — X ^ cube of the perp. at « = - —— X j-i — z^Xt, and the 

 2 C R 3 CR \ 



ratio of this folid to the folid infifting on PKfim is ultimately 



the ratio of equality. Hence the fluxion of the folid infifting 



on AC«R = 4.ar2— zAh 



It follows therefore that the value of a muft be fuch that 



the fluents o^ j.-^ " U ■ ^V l maybe both algebraically ex- 



preflTed. This it readily appears can take place by aflTuming 

 an infinite variety of values of a in terms of z , z and conft. 

 quantities. The problem confequently is truly unlimited. 



To 



• Archimedes de Sphatra & Cylindro. 



