A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

THE COLLECTION. BOOK IV 383
a certain ratio shown in the second figure where ABC is
a quadrant of a circle equal to a great circle in the sphere,
namely the ratio of the segment ABC to the sector DABC.
384 PAPPUS OF ALEXANDRIA
»
Now (sector HPN on sphere) :
(sector HKL on sphere)
= (chord HP) 2 : (chord HL) 2
(a consequence of Archimedes, On Sphere and Cylinder, I. 42).
And HP* : HL = 2 CB 2 : CA
2
= CB 2 :CE 2 .
Therefore
(sector HPN) :
(sector HKL) = (sector CBG) :
(sector CEF).
Draw the tangent CF to the quadrant at C. With C as
centre and radius CA draw the circle AEF meeting CF in F.
Then the sector CAF is equal to the sector ADC (since
CA 2 = 2 AD 2 ,
while Z ACF = \ Z ADC).
It is
required, therefore, to prove that, if S be the area cut
off by the spiral as above described,
S: (surface of hemisphere) = (segmt. ABC) :
(sector CAF).
Let KL be a (small) fraction, say I /nth, of the circumference
of the circle KLM, and let HPL be the quadrant of the
great circle through H, L meeting the spiral in P. Then, by
the property of the spiral,
(arc HP) :
(arc HL) = (arc KL) :
= l:n.
(circumf . of KLM
Let the small circle NPQ passing through P be described
about the pole H.
Next let FE be the same fraction, \/nth, of the arc FA
that KL is of the circumference of the circle KLM, and join EC
meeting the arc ABC in B. With C as centre and CB as
radius describe the arc BG meeting CF in G.
Then the arc CB is the same fraction, 1/^th, of the arc
CBA that the arc FE is of FA (for it is easily seen that
IFCE = \LBDCy while Z FCA = \LCDA). Therefore, since
(arc CBA) = (arc HPL), (arc CB) = (arc HP), and chord CB
= chord HP.
Similarly, if
the arc L1J be taken equal to the arc KL and
the great circle through H, II cuts the spiral in P',. and a small
circle described about H and through P / meets the arc HPL
in j) ; and if likewise the arc BB r is made equal to the arc BO,
and CB' is produced to meet AF in E' , while again a circular
arc with C as centre and CB' as radius meets CE in b,
(sector HP']) on sphere) :
And so on.
(sector HLU on sphere)
= (sector CB'b) : (sector (7^^).
Ultimately then we shall get a figure consisting of sectors
on the sphere circumscribed about the area S of the spiral and
a figure consisting of sectors of circles circumscribed about the
segment CBA
;
and in like .manner we shall have inscribed
figures in each case similarly made up.
The method of exhaustion will then give
$: (surface of hemisphere) = (segmt. ABC) :
= (segmt. ABC) :
(sector CAF)
(sector DAC).
[We may, as an illustration, give the analytical equivalent
of this proposition. If p, a> be the spherical coordinates of P
with reference to H as pole and the arc HNK as polar axis,
the equation of Pappus's curve is obviously co = 4 p.
If now the radius of the sphere is taken as unity, we have as
the element of area
dA. — dec (1 —cos/)) — 4dp(l — cos/o).
rh-
Therefore A = idp (1 —cos/)) = 2 7T — 4.