THE METHOD 
29 
I to Eratosthenes). 
3S mostly taken 
ma which forms 
nedes obtains by 
The area of any 
(parabola) is f of 
; (Prop. 1). The 
Mieroid of revolu- 
,xis of revolution 
here or spheroid 
nd the volume of 
ig'les to the axis, 
l, of revolution), 
(hyperboloid) in 
the segment and 
3 uses his method 
>f a paraboloid of 
lectively, Props, 
cubature of two 
ith a square base 
ilar bases of the 
uares which are 
3 drawn through 
l that diameter of 
:allel to the said 
y two planes and 
ir (a solid shaped 
12-15 prove that 
prism. (2) Sup- 
bhe circular bases 
opposite faces of 
milarly inscribed 
The two cylinders 
lade up of eight 
proves that the 
the cube. Archi- 
■kable fact about 
tion of Heiberg (in 
i (Bibliotheca Mathe- 
these solids respectively is that each of them is equal to a 
solid enclosed by planes, whereas the volume of curvilinear 
solids (spheres, spheroids, &c.) is generally only expressible in 
terms of other curvilinear solids (cones and cylinders). In 
accordance with his dictum that the results obtained by the 
mechanical method are merely indicated, but not actually 
proved, unless confirmed by the rigorous methods of pure 
geometry, Archimedes proved the facts about the two last- 
named solids by the orthodox method of exhaustion as 
regularly used by him in his other geometrical treatises; the 
proofs, partly lost, were given in Props. 15 and 16. 
We will first illustrate the method by giving the argument 
of Prop. 1 about the area of a parabolic segment. 
Let ABC be the segment, BD its diameter, CF the tangent 
at C. Let P be any point on the segment, and let AKF, 
OPNM be drawn parallel to BD. Join CB and produce it to 
meet MO in N and FA in K, and let KH be made equal to 
KC. 
Now, by a proposition ‘ proved in a lemma ’ (cf. Quadrature 
of the Parabola, Prop. 5) 
MO-.OP = CA :AO 
= CK:KN 
= HK.KN. 
ft 
Also, by the property of the parabola, EB = BD, so that 
MN = NO and FK — KA. 
It follows that, if HC be regarded as the bar of a balance, 
a line TO equal to PO and placed with its middle point at H 
balances, about K, the straight line MO placed where it is, 
i.e. with its middle point at N. 
Similarly with all lines, as MO, PO, in the triangle CFA 
and the segment CBA respectively. 
And there are the same number of these lines. Therefore, 
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