60 
ARCHIMEDES 
by planes obliquely inclined to the axis. The base of the 
segment is an ellipse in which BE' 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 BE' 
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’ 1 or PN 2 , i.e. proportional to 
AA r . 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 off by the same planes. Thus 
(first frustum in BF): (first frustum in inscribed figure) 
= BD 2 : PN 2 
= AD: AN 
= BD: TN. 
Similarly 
(second frustum in BF) : (second in inscribed figure) 
= HN: SM, • 
and so on. The last frustum in the cylinder BF has none to