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A history of Greek mathematics - Wilbourhall.org

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3<br />

36 ARCHIMEDES<br />

that, if we continually double the number <strong>of</strong> the sides <strong>of</strong> the<br />

regular polygon inscribed in a circle, segments will ultimately be<br />

left which are together less than any assigned area, Archimedes<br />

has to supplement this (Prop. 6) by proving that, if we increase<br />

the number <strong>of</strong> the sides <strong>of</strong> a circumscribed regular polygon<br />

sufficiently, we can make the excess <strong>of</strong> the area <strong>of</strong> the polygon<br />

over that <strong>of</strong> the circle less than any given area. Archimedes<br />

then addresses himself to the problems <strong>of</strong> finding the surface <strong>of</strong><br />

any right cone or cylinder, problems finally solved in Props. 1<br />

(the cylinder) and 14 (the cone).<br />

Circumscribing and inscribing<br />

regular polygons to the bases <strong>of</strong> the cone and cylinder, he<br />

erects pyramids and prisms respectively on the polygons as<br />

bases and circumscribed or inscribed to the cone and cylinder<br />

respectively. In Props. 7 and 8 he finds the surface <strong>of</strong> the<br />

pyramids inscribed and circumscribed to the cone, and in<br />

Props. 9 and 10 he proves that the surfaces <strong>of</strong> the inscribed<br />

and circumscribed pyramids respectively (excluding the base)<br />

are less and greater than the surface <strong>of</strong> the cone (excluding<br />

the base). Props. 11 and 12 prove the same thing <strong>of</strong> the<br />

prisms inscribed and circumscribed to the cylinder, and finally<br />

Props. 13 and 14 prove, by the method <strong>of</strong> exhaustion, that the<br />

surface <strong>of</strong> the cone or cylinder (excluding the bases) is equal<br />

to the circle the radius <strong>of</strong> which is a mean proportional<br />

between the ' side ' (i. e. generator) <strong>of</strong> the cone or cylinder and<br />

the radius or diameter <strong>of</strong> the base (i.e. is equal to wrs in the<br />

case <strong>of</strong> the cone and 2irrs in the case <strong>of</strong> the cylinder, where<br />

r is the radius <strong>of</strong> the base and s a generator). As Archimedes<br />

here applies the method <strong>of</strong> exhaustion for the first time, we<br />

will illustrate by the case <strong>of</strong> the cone (Prop. 14). .<br />

Let A be the base <strong>of</strong><br />

the cone, C a straight line equal to its<br />

c<br />

E<br />

radius, D a line equal to a generator <strong>of</strong> the cone, E a mean<br />

proportional to G, D, and B a circle with radius equal to E.

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