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Some Applications of the Notion of
an Infinitesimal in Geometry

The Greek mathematician Archimedes long ago demonstrated the power of the concept of infinitesimals. For example, he derived the formula for the area enclosed with in a circle, by considering that area as composed of infinitesimal triangles with an apex at the center and a base on the circle. The area of such a triangle is one half of the height of the triangle, in this case the radius R of the circle, times the length of the base Rdθ. The total length of the bases of all the triangles is just the circumference of the circle, which is the radius times the sum of all the infinitesimal angles, 2π; i.e., 2πR. Thus the area enclosed by a circle of radius R is ½(R)(2πR), which is equal to πR².

Likewise the volume of a ball enclosed within a sphere of radius R is sum of all the infinitesimal pyramids with their apices at the center and their bases on the sphere. The volume of a pyramid is one third of the product of its height times the area of its base. The height of each pyramid is the sphere radius R. The sum of the area of the bases is just the area of the sphere 4πR². Thus the volume of the ball is (1/3)r(4πR²) which is equal to (4/3)πR³.

One construction using infinitesimals that is not valid is the derivation of the area of a sphere. Suppose the surface of a sphere was divided up into a large number of segments. A segment bounded by two longitudinal lines to the equator and the portion of the equator between them might be thought to be equivalent in area to a triangle whole height is equal to the arc distance from the equator to the pole. If the radius of the sphere is R then the arc distance from a pole to the equator is (2πR)/4. The area of such a supposed triangel would be ½(πR/2)ds where ds is the length of the base of the segment. The sum of the bases is 2πR. This would mean the area of both hemispheres of a sphere would be 2(πR/4)(2πR) which is equal to π²R². The correct value is πR².

The correct derivation of the area of sphere is as follows. Let θ be an angle measured from a pole and let r be the radius of a latitude circle.


r = R*sin(θ)

If the width of the segment encompasses a longitudinal angle Δφ then the area of the band at θ with a height Rdθ is

(R*sin(θ)Δφ)(Rdθ) = R²Δφsin(θ)dθ

The area of the segment is then

which is equal to
R²Δφ[−cos(θ)]0π/2 = R²Δφ[0 − (−1)]
= R²Δφ

This is the area of the segment in one hemisphere; in two hemispheres the value is 2R²Δφ.

For the entire sphere Δφ is equal to 2π and thus the area of a sphere is

2R²(2π) = 4πR²

The Area of a Generalized Parallelogram

Consider a figure of constant width such as shown below.

An infinitesimal parallelogram of width b and height dh has an area of bdh. The sum of all such parallelograms is b∫dh which is equal to bh. This applies no matter how sinuous the sides so long as the direction is not reversed.

The Area of a Generalized Prism

Another simple application of infinitesimals is to find the area of the sides of a prism, cylinder or any other figure of constant cross section. Let ds be an infinitesimal length on the perimeter of the base.

A parallelogram with a base of ds and height of h has an area of hds. The area of the sides is equal to h∫ds which is equal to h times the circumberence of the base. This would also apply to a distorted prism or cylinder so long as the the cross section is constant.

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