Full text: From Aristarchus to Diophantus (Volume 2)

♦ 
ON CONOIDS AND SPHEROIDS 
61 
base of the 
i its plane is 
isses through 
Hyperbola, or 
is cut into a 
s are drawn 
3, cutting the 
, QQ', &c., are 
J and passing 
pe draw the 
these frusta, 
ribed frustum 
second to the 
ibed frustum 
iircumscribed 
}f which BB' 
i be made as 
drcum scribed 
assigned solid 
applying tlie 
correspond to it in the inscribed figure, and we should write 
the ratio as (BD : zero). 
Archimedes concludes, by means of a lemma in proportions 
forming Prop. 1, that 
(frustum BF) : (inscribed figure) 
= {BD + HN +...): {TNP SM+ ... + XO) 
= 7i"k \ (Ji 2 k 3 k ti — 1 k), 
where XO = k, so that BD = 7hk. 
In like manner, he concludes that 
(frustum BF): (circumscribed figure) 
— 71"k (1c -i- 2k 3k ... -p Tik). 
But, by the Lemma preceding Prop. 1, 
k 2 k 3 k 7i — 1 k < -g- 7i" k <c k -P 2k -P 3 k -P ... 4* 7ik, 
whence 
(frustum BF): (inscr. fig.) > 2 > (frustum BF): (circumscr. fig.). 
This indicates the desired result, which is then confirmed by 
,raboloid, the 
the method of exhaustion, namely that 
(frustum BF) = 2 (segment of paraboloid), 
which FP' is 
oportional to ’ 
into n equal 
the inscribed 
diole cylinder 
or, if Fbe the volume of the ‘segment of a cone’, with vertex 
A and base the same as that of the segment, 
(volume of segment) = §V. 
Archimedes, it will be seen, proves in effect that, if k be 
indefinitely diminished, and n indefinitely increased, while 7ik 
remains equal to c, then 
>ed figure) 
limit of k {k -p 2 k -t- 3 k -P .., -P (a — 1) k} — ^ c", 
that is, in our notation, 
f xdx = -|c 2 . 
Jo 
d figure) 
t 
Prop, 23 proves that the volume is constant for a given 
length of axis AD, whether the segment is cut off by a plane 
perpendicular or not perpendicular to the axis, and Prop. 24 
shows that the volumes of two segments are as the squares on 
? lias none to 
their axes.
	        
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