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Hannibal lecter

Member

Registered: 2016-02-11

Posts: 392

Archimedes and limit

Archimedes of Syracuse first developed the idea of limits to measure curved figures and the volume of a sphere in the third century b.c. By carving these figures into small pieces that can be approximated, then increasing the number of pieces, the limit of the sum of pieces can give the desired quantity

how was Archimedes  doing that? is there examplew with simple numbers for a simple sphere or any curved figure

what what does " the limit of the sum " mean?

Last edited by Hannibal lecter (2023-06-23 04:44:36)

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Wisdom is a tree which grows in the heart and fruits on the tongue

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Archimedes and limit

I'll have a go at this. I've not written out the proof before so wish me luck

The formulas for the volume of a cylinder and for a cone were already known.  Archimedes used a technique rather like integral calculus to show that there is a relationship between the volumes of a cylinder (height = diameter); a sphere with the same diameter; and a cone (base diameter = height).  He split the solids in half and dealt with the top half only.  You can split a cone into two smaller cones, each half the height with one upside down sittting on top of the other. This relationship is

VOLUME OF SMALL CONE + VOLUME OF HEMISPHERE = VOLUME OF HALF THE CYLINDER

If the formulas for the cone and the cylinder are known then the formula for the sphere can be deduced.

I'll start with the cone:

If you create two smaller cones, each height h/2 then

Hopefully this will explain the following diagram:

I have started with a cylinder and taken a slice through the centre. The cylinder is ABCD.

Inside is  a sphere centre at I.  And two cones, one upside down, with each vertex at I. Thus ABI is the slice through the top cone.

The axis for all three is JIK. Consider a line at an arbitary point E on JIK drawn at right angles to this axis.

Let IE = z and r be the radius of the cylinder, the sphere and the base of the cone.

As ABCD is a square and IK = KB = r then IE = EF = z

Imagine that EH represents an infinitesimally thin slice through all three solids.

The volume of the slice EH for the cylinder = pi r^2  Δz

The volume of the slice EG for the sphere is pi (r^2 - z^2) Δz

The volume of the slice EF for the cone is pi z^2 Δz

You can see that slice EF + slice EG = slice EH

The same is true for the botton half.

volume of small cone + volume of hemisphere = volume of cylinder height h.

As cylinder = pi r^2 h (=r);
volume of cone = 1/3 pi r^2 h/2 = 1/3 pi r^3 we can deduce that the volume of half the sphere is 2/3 pi r^3 and hence for the whole sphere is 4/3 pi r^3

Bob

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