In the last article (Dated: 1 July, 2008) we had discussed about the origination of successive approximation technique, discretization process and its relation with interpolation. Another intellectual achievement of archimedes is the invention of integration.
In this article,
- We will discuss about the origination of Integral calculus by explaining his method (geometrical proof) to calculate the area enclosed by a parabola
- We will solve an example to gain more knowledge on the process. In this example, the area enclosed by a parabola is determined using modern integral calculus and the result is compared with Archimedes method of calculation.
- The relationship between geometric series and integration will also be exploited using Archimedes method
In a letter (Quadrature of parabola) to his friend Dositheus, archimedes wrote that the total area under the parabola [Fig 1] is 4/3 of the area of the blue triangle (i.e).
Area under the parabola = 4/3 * Area of the blue triangle
To prove this archimedes first determined that blue triangle has 8 times the area of each green triangle, each green triangle has 8 times the area of each yellow triangle and so on
(i.e) He proved that Δ ACB = 8 * Δ ADC [Refer Fig 2]
Proof:
In Figure 2, M is the midpoint of the chord AB. A property of parabolas in general is that if AM is parallel to DE then
AM^2 / DE^2 = MC/EC ———– (1)
Archimedes refers to the classical work on conics by Euclid and aristaeus for the derivation of this property [ Refer: The Works of Archimedes, Page:235 ]
Further in figure 2, P is the midpoint of AM
Therefore, AM = 2 * DE ———– (2)
Substitute (2) in (1)
MC/EC = AM^2 / DE^2 = (2*DE)^2/DE^2 = 4 * DE^2/DE^2 =4
Which implies MC = 4*EC ————— (3)
We know that, MC = ME+EC
Therefore ME = 3*EC , EC = ME/3 ———- (4)
Substituting (4) in (3) Gives
MC = 4/3 * ME = 4/3 * PD [Since PD = ME]
Δ ADC is similar to Δ DCW, Hence MC = 2 * PW
PW = 2/3 * PD [ Since MC = 4/3 * PD] and PW = 2 * WD.
Therefore Δ ACP = 2 * Δ ADC, Δ ACM = 4 * Δ ADC and Δ ACB = 8 * Δ ADC
=> Δ ADC = (1/8 )* Δ ACB
The triangle formed by the chord CB also has an eight of the area of Δ ACB
Similarly we can inscribe 4 yellow triangles which has an eight of the area of the blue triangle. This process continues indefinetely
Assuming that the blue triangle in Figure 1 has area 1, the total area under the parabola is an infinite sum
1 + 2*( 1/8 ) + 4*(1/8)^2 + 8*(1/8)^3 + ………
Here the first term represents the area of the blue triangle, the second term the area of two green triangles, the third term the area of four yellow triangles and so on. Simplifying the ablove fraction gives
1+1/4+1/16+1/64 + ……
This is a Geometric ratio with common ratio 1/4. The sum of the geometric series is given by 1/(1-r) where ‘r’ is the common ratio
=> 1/1-(1/4) = 4/3
In general we can say that the area under the parabola is 4/3 * area of the blue triangle.
Example:
To gain more knowledge into the proof we will find the area under parabola y = x^2 +1 intersected by a line at y=10
Archimedes method:
We know by archimedes method, Area under the parabola = 4/3 * Area of Δ ABC
From Figure 3, Area of Δ ABC = 1/2 (b * h)
h=9, b= abs(-3) + abs(3) = 6
=(1/2 )* 54 => Area of Δ ABC = 27
Therefore area under the parabola = 4/3*(27) = 36
Modern (or) well known Integral calculus:
Where f(x) = y-(x^2)
as we had expected.
References:
- The Works of Archimedes, Page:235
- Gordon swain and Thomas Dence, Archimedes’ Quadrature of the parabola revisited, Mathematics magazine. Vol. 71. No.2 (Apr.1998), pp.123-130 [This article is not freely downloadable. However it is very intresting and worth to pay
]
Posted by Christober rayappan 
