{"id":1652,"date":"2008-12-14T09:25:19","date_gmt":"2008-12-14T01:25:19","guid":{"rendered":"http:\/\/www.intmath.com\/blog\/?p=1652"},"modified":"2019-03-30T09:25:13","modified_gmt":"2019-03-30T01:25:13","slug":"archimedes-and-the-area-of-a-parabolic-segment","status":"publish","type":"post","link":"https:\/\/www.intmath.com\/blog\/mathematics\/archimedes-and-the-area-of-a-parabolic-segment-1652","title":{"rendered":"Archimedes and the area of a parabolic segment"},"content":{"rendered":"
\"180px-Archimedes_greece_1983\"
\nArchimedes
\n[Image source<\/a>]<\/div>\n

Archimedes was a Greek mathematician who lived 2300 years ago. <\/p>\n

Many of his inventions and mathematical discoveries were years ahead of their time. It's amazing how inventive you become when you are constantly under attack from your enemies. <\/p>\n

Finding the area under a curve had been a problem for many years. Fair trade depended (in part) on being able to work out volumes of cylinders and spheres. Archimedes worked out good approximations for the area of a circle and the value of π<\/span> (pi).<\/p>\n

The most remarkable thing about the following example of Archimedes' thinking is that it pre-dates the work of Isaac Newton and Gottfried Leibniz (the 17th century mathematicians who developed differential calculus) by around 2000 years.<\/p>\n

I have a special interest in this kind of thing because the name of this blog, squareCircleZ, is derived from the historical attempt to construct a square with the same area as a given circle (called \"squaring the circle\").<\/p>\n

The following discussion follows the technique as outlined by Archimedes in his Quadrature of the Parabola<\/em>, using the \"method of exhaustion\", a pre-cursor to infinitesimal calculus. The concept came via Antiphon (5th century BCE), and Eudoxus of Cnidus (4th century BCE).<\/p>\n

The area of a parabolic segment<\/h2>\n

A parabolic segment<\/strong> is a region bounded by a parabola and a line, as indicated by the light blue region below:<\/p>\n

\"parabola-1\" <\/p>\n

[See Parabola<\/a> for some background on this interesting shape.]<\/p>\n

I have used the simple parabola y = x<\/em>2<\/sup><\/span> and chosen the end points of the line as A<\/em> (−1, 1)<\/span> and B<\/em> (2, 4).<\/span> The theorem will work for any parabola and any line passing through that parabola, intersecting in 2 points. [Of course, Archimedes did not use the x-y<\/i> co-ordinate system, since it was not invented by Descartes until the 17th century.]<\/p>\n

Archimedes then located point C such that the x<\/em>-value of C<\/em> is half-way between the x<\/em>-values of points A<\/em> and B<\/em>. He then constructed triangle ABC, as follows:<\/p>\n

\"parabola-2\"<\/p>\n

 <\/p>\n

In my example, C<\/em> has x<\/em>-value 0.5, which is 1.5 units from both x<\/em> = −1<\/span> and x<\/em> = 2.<\/span> <\/p>\n

Archimedes showed that the area of the (light blue) parabolic segment is 4\/3 of the area of the triangle ABC<\/em>.<\/p>\n

The way Archimedes achieved this result was to use the Method of Exhaustion, which involves finding the area of a curved shape by inscribing successively smaller polygons until the shape is filled. We can find the area of those polygons and hence the area of the curved shape. <\/p>\n

We now construct another triangle by choosing point D on the parabola such that its x<\/em>-value is half-way between the x<\/em>-values of A<\/em> and C<\/em>, similar to what we did before.<\/p>\n

Let's zoom in and see the result.<\/p>\n

\"parabola-3\" <\/p>\n

We do the same thing on line BC<\/em>, by locating point E<\/em> such that its x<\/em>-value is half-way between C<\/em> and B<\/em>.<\/p>\n

\"parabola-4\" <\/p>\n

We can see already that if we add the areas of triangles ABC, ACD<\/em> and BCE<\/em>, we have a reasonable approximation for the area of the parabolic segment, but we can do better.<\/p>\n

If we continue the process, we form 4 more triangles as follows:<\/p>\n

\"parabola-5\"<\/p>\n

There is very little 'white space' left and if we were to add the areas of the 7 triangles, we would have an even better approximation for the area of the parabolic segment.<\/p>\n

If we kept going and added the areas of an infinite<\/em> number of such triangles, we would have an exact<\/em> area for the parabolic segment.<\/p>\n

Now, the area of each light green triangle is 1\/8<\/span> the area of the pink triangle. This is because the green triangle has 1\/2<\/span> the width of the pink triangle (we made it so) and 1\/4<\/span> of the height (which can be shown using parametric equations<\/a>.)<\/p>\n

Then, considering the red triangles, they will be 1\/8<\/span> the area of the light green triangles.<\/p>\n

So if we call the area of the large pink triangle X<\/em>, then the area of all of the triangles is given by:<\/p>\n

X + <\/em>2(X<\/em>\/8) + 4(X<\/em>\/64)<\/span> + 8(X<\/em>\/512) + ...<\/p>\n

= X + X<\/em>\/4 + X<\/em>\/16 + X<\/em>\/64 + ...<\/p>\n

= X<\/em>(1 + <\/em>1\/4 + 1\/16 + 1\/64 + ...)<\/p>\n

We recognize that the expression in brackets is a Geometric Series<\/a>, with common ratio r<\/em> = 1\/4<\/span> and first term a<\/em> = 1.<\/span> The sum of this series is given by:<\/p>\n

Sum = a<\/em>\/(1 − r<\/em>) = 1\/(1 − 1\/4)<\/span> = 1\/(3\/4) = 4\/3<\/span><\/p>\n

[Archimedes used a geometric proof for this portion.]<\/p>\n

So the total area of the triangles (which gives us the area of our light blue parabolic segment) is 4X<\/em>\/3,<\/span> which is 4\/3<\/span> the area of the pink triangle, as Archimedes claimed.<\/p>\n

The thinking behind this solution is very similar to the ideas behind the development of calculus.<\/p>\n

<\/ins><\/div>\n

Using Calculus<\/h2>\n

Let's use integral calculus to check the answer we obtained using Archimedes' approach.<\/p>\n

In my particular example, with y = x<\/em>2<\/sup><\/span> and the line y = x<\/em> + 2<\/span> intersecting the parabola at (−1, 1)<\/span> and (2, 4),<\/span> the pink triangle has width AC = 1.68<\/span> units and perpendicular height 4.02<\/span> units, so the area is:<\/p>\n

Area ΔABC = 0.5 × 1.68 × 4.02 = 3.38<\/span> unit2<\/sup><\/p>\n

So according to Archimedes, the area of the (light blue) parabolic segment will be:<\/p>\n

Area segment = 4\/3 × 3.38 = 4.5<\/span> unit2<\/sup><\/p>\n

Now, let's compare this result using calculus. The required area is an area between 2 curves. The upper curve is the line y<\/em>2<\/sub> = x<\/em> + 2<\/span> and the lower curve is y<\/em>1<\/sub> = x<\/em>2<\/sup><\/span>. The limits of integration are x<\/em> = −1<\/span> and x<\/em> = 2.<\/span><\/p>\n

\"parabola-6\"<\/p>\n

So we get the same answer, 4.5<\/span> unit2<\/sup>.<\/p>\n

As I said at the beginning, Archimedes' solution for finding the area of a parabolic segment is a remarkable result. Almost 2000 years before Newton and Leibniz formalized differential and integral calculus, Archimedes already had a good handle on the basics.<\/p>\n

It's interesting (and dangerous) how knowledge can be 'forgotten' for thousands of years.<\/p>\n

Go here to see an interesting English translation of Archimedes' Quadrature of the Parabola<\/a>.<\/p>\n

The graphs in this article were produced using Geogebra, the free computer algebra system. See my review: GeoGebra math software - a review<\/a>.<\/p>\n

See the 26 Comments<\/a> below.<\/p>\n","protected":false},"excerpt":{"rendered":"

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