Full text: From Aristarchus to Diophantus (Volume 2)

ON CONOIDS AND SPHEROIDS 
63 
l let the axis 
let AA'—a. 
i, cylinder on 
i axis to the 
frustum BF 
so that 
hS n = ah (h+ 2h+ ... +nh) + h {h 2 + {2h) 2 + ... +{nh) 2 }. 
The limit of this latter expression is what we should write 
r>b 
{ax + x 2 ) dx = b 2 {%a +1 h), 
-0 
') 
and Archimedes’s procedure is the equivalent of this integration. 
III. In the case of the spheroid (Props. 29, 30) we take 
a segment less than half the spheroid. 
As in the case of the hyperboloid, 
b 
(frustum in BF): (frustum on base QQ') 
= BD 2 : QM 2 
rh + (rh) 2 }, 
= AD . A'D : AM. A'M; 
but, in order to reduce the summation to the same as that in 
%. rh + {rh) 2 }. 
Prop. 2, Archimedes expresses AM. A'M in a different form 
equivalent to the following. 
Let AD (=b) be divided into n equal parts of length h, 
and suppose that AA'= a, CD = \c. 
: {t a + ^nh) 
a. rh + {rh) 2 }. 
Then AD . A'D = ^a 2 — ^c 2 , 
and AM. A'M = i a 2 - {%c + rh) 2 {DM = rh) 
= AD . A'D— {c . rh + {rh) 2 } 
:{%a + ^nh), 
B') 
{AD+ 2CA)] 
= cb + b 2 — {c . rh + {rh) 2 }. 
Thus in this case we have 
(frustum BF): (inscribed figure) 
m. 
nt to proving 
is indefinitely 
i 
= n {cb + b 2 ) :[n {cb + b 2 ) — {c .rh-f- {rh) 2 } ] 
and 
(frustum BF): (circumscribed figure) 
= n{cb + b 2 ): \n{cb + h 2 ) — '2A l ~ x {c . rh + {rh) 2 ]'\. 
And, since h = nh, we have, by means of Prop. 2, 
n{cb + b 2 ) : \n{cb + b 2 ) — . rh + {rh) 2 }] 
> {c+b): {c + b — (1 c + § b)} 
2 +... + {nh) 2 ] 
> n{ch + b 2 ): \n{cb + b 2 ) —. rh + {rh) 2 }\
	        
Waiting...

Note to user

Dear user,

In response to current developments in the web technology used by the Goobi viewer, the software no longer supports your browser.

Please use one of the following browsers to display this page correctly.

Thank you.