This is briefly the finding of areas (or volumes) of curved regions by successive approximations using inscribed and exscribed polygons (or polyhedra) and other shapes whose areas (or volumes) are already known. For example, finding the area of a circle using approximations by regular polygons with increasing numbers of sides. A more complicated example by Archimedes follows:
We require the shaded area A enclosed by the spiral and the x-axis.
Draw radii with angles in A.P.:
r = a, r = 2a, r = 3a, ....etc. Where n = 2.
This divides the area into n sectors (the ith sector is shown)
Enclose the arc between circular sectors i and c, and a third sector s with radius 2a, for all i.
So s = s = ........ = s = (2a).
Let S = s C = c I = i
For the ith sector we have the proportions:
I < < C - - - - - - - - - - - - - - - -(1)
At this stage the modern argument would be something like:
C - I = (c - i) = c ( since c = i and i = 0 )
and the area c can be made as small as we like by taking n large enough,
i.e. C - I 0 as n .
Hence (by (1)) C and I as n .
i.e. the area A = , where S = (2a)2
so A = a
Archimedes' argument was:
Either A < , A > or A = .
If A < , choose n so that C - A < - A - - - - - - - - (2)
This can be done because C - A < C - I = c , which can be made as small as we like by taking n large enough.
Now from (2) : C < and from (1) : Cn > .
- a contradiction, so A < is false.
A similar argument (leading to I > and I < ) shows A > is also false
- so we must have A = . Note that Archimedes' argument does not use the limit concept, and is completely rigorous.
Since the Greeks did not use algebraic notation, the version above would not be recognisable by Archimedes, but the argument is his.
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