The Area of a Disk

If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius $ R$ is $ \pi R^2$ .


The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any other circular lamina. Radius for a disk is always the same, irrespective of the location of the point at circumference to which you are joining the center of the disk. The area of disk is defined as the 'measure of surface' surrounded by the round edge (circumference) of the disk.


The area of a disk can be derived by breaking it into a number of identical parts of disk as units -- calculating their areas and summing them up till disk is reformed. There are many ways to imagine a unit of disk. We can imagine the disk to be made up of several concentric very thin rings increasing in radius from zero to the radius of disc. In this method we can take an arbitrary ring, calculate its area and then in similar manner, induce areas of other rings -- sum them till whole disk is obtained.

Circle sections
Rings and Sections

Mathematically, we can imagine a ring of with radius $ x$ and thickness $ dx$ , anywhere in the disk having the same center as disk, calculate its area and then sum up (integrate) it from $ x=0$ to $ x=R$ . Area of a thin ring is since $ \pi x dx$ . And after integrating we get, area of disk $ A=2 \int_0^R \pi x dx$ or $ A=\pi R^2$ .

Another derivation

There is another approach to achieving the area of a disk, A.

An inscribed Triangle

Imagine a disk is made up of a number equal sections or arcs. If there are $ n$ number of arcs then interior angle of an arc is exactly $ \frac{2\pi}{n}$ , since $ 2 \pi$ is the total angle at the center of disk and we are dividing this angle into $ n$ equal parts. If we join two ends of each section --we can get $ n$ identical triangles in which an angle with vertex O is $ \frac{2 \pi}{n}$ . Now, if we can calculate the area of one such section, we can approach the area of the disk intuitively. This approach is called the method of exhaustion.

Let, we draw two lines joining center O of the disk and points A & B at circumference. It is clear that OB and OA are the radius of the disk. We joined points A and B in order to form a triangle OAB. Now consider that the disk is made up of n-number of such triangles. We see that there is some area remaining outside the line AB and inside the circumference. If we had this triangle thinner, the remaining area must be lesser.

Area remaining after the Triangle

If we increase the number of triangles in disk ----we decrease the remaining areas. We can achieve to a point where we can accurately calculate the area of disk when there are infinitely many such triangles or in other words area of one such triangle is very small. So our plan is to find the area of one triangle ---sum it up to n --- make $ n$ tending to infinity to get the area of disk. It is clear that the sum of areas of all identical triangles like OAB must be either less than or equal to area of the disk. We can call triangles like OAB as inscribed triangles.

Now, if we draw a radius-line OT', perpendicular to AB at point T and intersecting the circumference at point T', we can easily draw another triangle OA'B' as shown in figure. AOB and A'OB' are inscribed and superscribed triangles of disk with same angle at vertex O. So, it is clear that the angle A'OB' is equal to the angle AOB. Triangle A'OB' is larger than the circular arc OAB and circular arc OAB is larger than the inscribed triangle AOB. Also, the sum of areas of triangles identical to OA'B' is either greater than or equal to area of the disk.

area of disk area exceeding after superscibed scribed triangle

Let disk be divided into n- such inscribed and (thus) superscribed triangles. Since, total angle at point O is 360° or 2π ----the angles AOB and A'OB' are of $ 2 \pi/n$ . And also since OT and OT' are normals at chord AB and line A'B' respectively, then they must divide the angles AOB and A'OB' in two equal parts, each of $ \pi/n$ radians.

In triangle AOB, the area of triangle AOB is the sum of the area of triangles AOT and BOT. But since both are equal to each other in area, area of AOB must be twice of the area of triangle AOB (or BOT). Our next target is to find, the area of AOT in order to find the area of AOB.


From figure, area of $ \bigtriangleup{AOT}= \frac{1}{2} \times AT \times OT$ ....(1)

$ OA=R$

And, $ \angle{AOT}= \frac{\pi}{n}$ .

Thus, $ \frac{AT}{OA}=\sin {\frac{\pi}{n}}$

or, $ AT=OA \sin {(\pi/n)}$

or, $ AT=R \sin {(\pi/n)}$ .....(2)

Similarly, $ OT=R \cos {(\pi/n)}$

Therefore, area of $ \bigtriangleup {AOT}=\frac{1}{2} \times R \sin {(\pi/n)} \times R \cos {(\pi/n)}=\frac{1}{2} R^2 \sin{(\pi/n)} \cos {(\pi/n)}$

And thus, area of $ \bigtriangleup{AOB}=2 \times \frac{1}{2} R^2 \sin{(\pi/n)} \cos{(\pi/n)}=R^2 \sin{(\pi/n)} \cos{(\pi/n)}$

Since there are $ n$ such triangles: sum of areas of such triangles

$ S_1=n \times R^2 \sin{(\pi/n)} \cos{(\pi/n)}$ .

In triangle A'OB', the total area of triangle A'OB' is the sum of areas two identical triangles A'OT' and B'OT'. Therefore, area of $ \bigtriangleup{A'OB'}=2 \times \text{area of} \bigtriangleup{A'OT'}$ .

And area of $ \bigtriangleup{A'OT'}=\frac{1}{2} \times AT' \times OT'$ .

We have $ OT'=R$

and angle A'OT'=$ \frac{\pi}{n}$

Thus, A'T'/OT'= $ \tan{\frac{\pi}{n}}$

or, $ A'T'=OT' \tan{\frac{\pi}{n}} =R \tan{\frac{\pi}{n}}$ .

Therefore, area of triangle A'OT'= $ \frac{1}{2} \times R \times R \tan{\frac{\pi}{n}}=\frac{1}{2} R^2 \tan{\frac{\pi}{n}}$ .

Hence, area of $ \bigtriangleup A'OB'=2 \times \frac{1}{2}R^2 \tan{\frac{\pi}{n}}$

As, there are $ n$ such triangles: sum of areas of those triangles $ S_2=n \times R^2 \tan{\frac{\pi}{n}}$ .

As it is clear that Sum of areas of triangles like $ \bigtriangleup AOB$ is an approximation for the area of disk from below, i.e., $ S_1 \le A$ when $ n \to \infty$ or,

$ \displaystyle{\lim_{n \to \infty}} n \times R^2 \sin{(\pi/n)} \cos{(\pi/n)} \le A$

$ \displaystyle{\lim_{n \to \infty}} n \times R^2 \frac{\pi}{n} \dfrac{\sin{(\pi/n)}}{\pi/n} \cos{(\pi/n)} \le A$

$ \pi R^2 \le A \ldots (I)$

Similarly, $ \displaystyle{\lim_{n\to \infty}} n \times R^2 \tan{\frac{\pi}{n}} \ge A$

or, $ \pi R^2 \ge A \ldots (II)$

From (I) and (II), we have $ A=\pi R^2$ .

So the area of a disk is $ \pi R^2$ .