A history of Greek mathematics - Wilbourhall.org

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62 ARCHIMEDES
II.
In the case of the hyperboloid (Props. 25, 26) let the axis
AD be divided into n parts, each of length h, and let AA'=a.
Then the ratio of the volume of the frustum of a cylinder on
the ellipse of which any double ordinate QQ' is an axis to the
volume of the corresponding portion of the whole frustum BF
takes a different form ;
for, if AM = rh, we have
(frustum in BF) :
(frustum on base QQ')
= BD 2 : QM
= AD .
2
A'D :
AM
.
A'M
— [a.nh + (nh) 2 } : {a . rh+ (rh) 2 }.
By means of this relation Archimedes proves that
(frustum BF) :
(inscribed figure)
and
(frustum BF) :
= n {a.nh+ (nh) 2 } : S^-
(circumscribed figure)
1 { a . rh + (rh) 2 }
= n{a.nh+(nh) 2 } : 2
x
n {a.rh-\-(rh)
2
}.
But, by Prop. 2,
n{a.nh + (nh) 2 } :\ n ~ l
{a.rh + (rh) 2 } > (a + nh):(±a + %nh)
From these relations it is inferred that
(frustum BF) :
> n{a.nh + (nh) 2 } :2
{
n {a.rh-\-(rh)
2 }.
(volume of segment) = (a + nh) : (^a + ^nh),
or (volume of segment) : (volume of cone ABB')
= (AD+3CA):(AD + 2CA);
and this is confirmed by the method of exhaustion.
The result obtained by Archimedes is equivalent to proving
that, if h be indefinitely diminished while n- is indefinitely
increased but nh remains always equal to b,
then
limit of n(ab + b 2 )/S n
= (a + b)
'(£« + §&),
or
where
limit of - S n
= b 2 (\a + J
b),
it
S n
= a(h + 2h+3h+...+nh) + {h 2 + (2h) 2 + (3h) 2 +...+(nh) 2 }