Section 12: Applications of Integration © 2004 by Karl Hahn

12.2 The Area of a Circle

You usually read that calculus began in the 17^th century with Newton and Leibniz. But in the 3^rd century BCE, Archimedes of Syracuse (click for biography of Archimedes. Also see this by Keiichi Suzuki) was indeed doing calculus. He didn't call it calculus, and he didn't seem to realize that the methods he was using were part of a whole new domain of mathematics. But his methods effectively employed the concepts of limits and integration, even though he never formalized those concepts.

Back in high school you learned that the area of a circle is  πr².  But did you ever wonder how we know this? Archimedes was the first to figure it out.

Fig 12.2-1

So how did he do it? The figure on the right shows a circle with polygons inscribed. As the animation progresses, the frames show polygons with 4, 8, 16, 32, 64, and 128 sides. You can see that when the polygon has more sides, its area is a bigger fraction of the circle's area. And you can see that in the limit as the number of sides goes to infinity, the polygon occupies all of the circle. You know this because any point you choose inside the circle is eventually included inside some polygon with sufficient number of sides. Based upon this, a process Archimedes called "exhaustion", he concluded that the area of the circle was the limit of the area of the inscribed polygon as the number of sides goes to infinity. To compute the area of the polygon, he divided it into triangles (as is also shown in the figure), one triangle for each of the polygon's sides.

In this figure, you can see just one triangle of the polygon.

Fig. 12.2-2 Area of One Triangle of the Polygon

The angle it makes at the circle's center is Δθ. The triangle is bisected to make it easier to do the trigonometry that we need to find the base and the height. Simplifying the expression in the figure, we find that the shaded area of the triangle, ΔA, is  r² cos(Δθ/2) sin(Δθ/2). 

But we are interested in the limit as the number of triangles goes to infinity. That means we are interested in the limit as Δθ goes to zero. And if Δθ goes to zero, then so does Δθ/2. Recall that the limit as x goes to zero of  cos(x)  is 1. Recall also that

lim x → 0

sin(x) = 1 eq. 12.2-1 x

That means in the area equation as we go to the limit, we can replace  cos(Δθ/2)  with 1, and we can replace  sin(Δθ/2)  with  Δθ/2.  And to indicate that we are taking it to the limit, we replace  Δθ/2. with  dθ/2  and  ΔA with  dA. We also put the integral sign in front of both sides of the equation to indicate we are adding up all the pieces:

dA =

1 2

r2 dθ eq. 12.2-2

There are 2π radians around the circle. So to get the area of the entire circle, on the right we integrate around the full circle, which goes from 0 to 2π radians. On the left, we know we have to end up with the area, A, and we choose the limits accordingly. Hence the definite integral is

A 0

dA =

1 2

2π 0

r2 dθ eq. 12.2-3

Both definite integrals are trivial to do (remember that r² is a constant), and you should be able to finish this yourself, arriving at the familiar formula for the area of a circle of radius r.

Archimedes did not use integrals in the formal way we did above. Nor did he use limit formulas for trig functions the way we did. Instead he observed that if the polygon has n sides, then each triangle accounts for  1/n  of the circumference of the circle. If you choose any point on the segment,  PQ,  then you can find an n great enough to put that point inside the triangle (his principle of exhaustion again). That means that for sufficiently large n, the radius, r, becomes as close to the length of  OP  as you'd like. That is equivalent to applying the cosine limit we used. Archimedes also observed that as n gets bigger and bigger, the length of line segment,  CD,  gets closer and closer to the length of the circular arc,  CD.  That is equivalent to applying the sine limit that we used. The length of that circular arc is, as we already observed,  1/n  of the circumference of the circle. If you use  2πr  as the circumference, then you have for the length of arc,  CD = 2πr/n ≈ CD.  Hence for the area of the triangle, which is  (1/2)(CD)(OP),  you can use  (1/2)(2πr/n)(r) = πr²/n.  Multiply that by n triangles, then as n grows without limit, you have the area of the circle.

Once Archimedes had added up the triangles using his method of exhaustion, he knew everything he needed to for the area of a circle except for one detail. That is the value of π. Until Archimedes, nobody knew how to compute the value of this fundamental constant. The estimates that had existed up to his time were all derived empirically using methods such as wrapping a cord around a cylindrical object or rolling a circular object along a ruler. They ranged anywhere from 3 to 3.2, depending upon which ancient culture you asked. But Archimedes observed that he could calculate the value of π to as fine a precision as he liked by adding up the base-lengths of the triangles for one of the polygons shown in figure 12.2-1. The higher the number of sides of the polygon, the more accurate the approximation for π would be.

In the diagram to the right, you can see

Fig 12.2-3 Bisecting the Triangle

how Archimedes did it. If you know the length of CD, then you can apply the Pythagorean formula to find the length of CQ. If you start with a regular polygon having n sides, each of length CD, then a regular polygon with 2n sides will have each side of length CQ. The formula at the lower left of the diagram shows how you would turn half the length of CD (that is, the length, a) into the length of CQ (which is the length, b). If you set the radius to  r = 1,  then n times the length of CD is an approximation for 2π. But 2n times the length of  CQ is an even better approximation for 2π. And if you keep doubling the number of sides again and again, the approximations get better and better. In the limit, the number of sides times the length of each side goes to 2π.

Let x₁ be the length of side of a two-sided polygon (that is, simply the diameter line). When the radius of the circle is  r = 1, then  x₁ = 2.  Multiplying that by the number of sides, 2, we have 4 as our first approximation for 2π. With a little algebra on the formula given in the diagram, you find that to proceed from one polygon to the next, the relationship between the side lengths is (remembering that x_n is twice the length, a, that is shown in the diagram)

   xn+1  =

                                     eq. 12.2-4

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

2.00000000000000
1.41421356237310
0.765366864730180
0.390180644032257
0.196034280659121
0.0981353486548360
0.0490824570458246
0.0245430765714399
0.0122717692983090
0.00613591352593195
0.00306796037256953
0.00153398063748541
0.000766990375142791
0.000383495194621407
0.000191747598191954
0.0000958737992061338
0.0000479368996168364
0.0000239684498101394
0.0000119842249052848
0.00000599211245266932
0.00000299605622633802
0.00000149802811316943
0.000000749014056584768
0.000000374507028292390
0.000000187253514146196

    2
    4
    8
   16
   32
   64
  128
  256
  512
 1024
 2048
 4096
 8192
16384
32768
65536
131072
262144
524288
1048576
2097152
4194304
8388608
16777216
33554432

4.00000000000000
5.65685424949238
6.12293491784144
6.24289030451610
6.27309698109188
6.28066231390951
6.28255450186555
6.28302760228860
6.28314588073418
6.28317545055432
6.28318284302240
6.28318469114024
6.28318515316975
6.28318526867713
6.28318529755397
6.28318530477318
6.28318530657799
6.28318530702919
6.28318530714199
6.28318530717019
6.28318530717724
6.28318530717900
6.28318530717944
6.28318530717955
6.28318530717958

Because of the accumulation of roundoff errors, you need to carry out the Archimedean calculations for the numbers shown in the table to at least twice the desired precision. So to get 15 digit accuracy for the numbers in the table, I had to use a calculator that has at least 30 digits of precision.

Archimedes had one more hurdle to get over in order to make a table like the one you see here. Until Archimedes' time, nobody knew how to calculate square roots accurately. Would it surprise you to find out that Archimedes was the first to establish an algorithm for calculating square roots to any desired degree of accuracy?

Isaac Newton is known to have said in 1676, "If I have seen farther than most men, it is because I stood on the shoulders of giants." Archimedes stands tall among Newton's giants.

Click to see an important coached exercise showing an alternative method for getting the area of a circle using concentric rings.

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