A history of Greek mathematics - Wilbourhall.org

60 ARCHIMEDES
by planes obliquely inclined to the axis. The base of the
segment is an ellipse in which BB' is an axis, and its plane is
at right angles to the plane of the paper, which passes through
the axis of the solid and cuts it in a parabola, a hyperbola, or
an ellipse respectively. The axis of the segment is cut into a
number of equal parts in each case, and planes are drawn
through each point of section parallel to the base, cutting the
solid in ellipses, similar to the base, in which PP', QQ', &c, are
axes. Describing frusta of cylinders with axis AD and passing
through these elliptical sections respectively, we draw the
circumscribed and inscribed solids consisting of these frusta.
It is evident that, beginning from A, the first inscribed frustum
is equal to the first circumscribed frustum, the second to the
second, and so on, but there is one more circumscribed frustum
than inscribed, and the difference between the circumscribed
and inscribed solids is equal to the last frustum of which BB'
is the base, and ND is the axis. Since ND can be made as
small as we please, the difference between the circumscribed
and inscribed solids can be made less than any assigned solid
whatever. Hence we have the requirements for applying the
method of exhaustion.
Consider now separately the cases of the paraboloid, the
hyperboloid and the spheroid.
I. The 'paraboloid (Props. 20-22).
The frustum the base of which is the ellipse in which PP' is
an axis is proportional to PP' 2 or PN 2 ,
i.e. proportional to
AX. Suppose that the axis AD (= c) is divided into n equal
parts. Archimedes compares each frustum in the inscribed
and circumscribed figure with the frustum of the whole cylinder
BF cut oft' by the same planes. Thus
(first frustum in BF) :
Similarly
(second frustum in BF) :
(first frustum in inscribed figure)
2
= BD 2 : PN
= AD:AN
= BD : TK
(second in inscribed figure)
= HN:3M,
and so on. The last frustum in the cylinder BF has none to