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## Derivation of the area of a circle

As most people reading this blog probably know, the area of a circle is given by

$\boxed{ \text{ area of a circle} = \pi r^{2} }$

where $\pi$ is the ratio of the diameter of a circle to its circumference, and $r$ is the radius of the circle. But, how would we go about proving this well known formula? To do so, we can use something called integration, which is a branch of calculus.

## The area under a curve

The area under any curve between the x-axis, the curve, and the lines $x_{1} \text{ and } x_{2}$ (see the figure below) is given by

$\int_{ x_{1} }^{ x_{2} } f(x) \; dx$

where $f(x)$ is the function (the curve), and $dx$ is an infinitesimally small width. Effectively, what we are doing is summing a series of rectangles of area $ydx$ from the lower limit $x_{1}$ to the upper limit $x_{2}$.

The area under any curve $y=f(x)$ can be found by summing infinitesimally small rectangles, each of height $y$ and of width $dx$ between the lower x-limit $x_{1}$ and the upper x-limit $x_{2}$

## The equation of a circle

To do this for a circle, we need to write the equation for a circle. To make things easier we will centre the circle at the origin. In this case, if the circle has a radius $r$ the equation which describes this circle is just

$x^{2} + y^{2} = r^{2}$

Therefore, to find the area of the circle, all we need to do is integrate between $x=0 \text{ and } x=r$, and then multiply the answer by 4 (as we have only found the area of quarter of the circle).

To find the area of a circle, in principle all we need to do is sum the rectangles $ydx$ from $x=0$ to $x=r$ between the x-axis and the circle, then multiply the answer by 4.

The integration to do this is

$\int_{0}^{r} y \; dx \; = \; \int_{0}^{r} \sqrt{ r^{2} - x^{2} } \; dx$

This is not an integral which we can do, so we appear to be stuck!

## Changing variables to use “polar coordinates”

Luckily for us, there is a way around this problem. Rather than using x-y co-ordinates (more correctly known as Cartesian co-ordinates), we can change the co-ordinate system to something called polar co-ordinates, and when we do this we get an integral that we can do.

To see how to go from Cartesian to polar co-ordinates, consider the figure below.

The point $(x,y)$ on the circle can be written in terms of the radius $r$ and the angle $\theta$. We can write that $x = r \cos \theta$ and $y = r \sin \theta$

We write the $x \text{ and } y$ coordinates in terms of two new variables $r \text{ and } \theta$, where $r$ is the radius of the circle and $\theta$ is the angle between the line from the centre of the circle to the point and the x-axis. When we do this, we can write that $x = r \cos \theta$ and $y = r \sin \theta$.

Then, to determine the area of the quadrant were $x \text{ and } y$ are positive, we can instead integrate using $r \text{ and } \theta$. To see how we do this, consider the figure below.

To find the area of the circle, we add a series of slices (the shaded region), each with an infinitesimally small angle $d\theta$ and radius $r$ between an angle of $\theta = 0$ and $\theta = \pi/2$, then multiply the answer by 4

One of the reasons the method of using polar coordinates is easier is that we now have only one variable, $\theta$, as the radius $r$ is a constant. When we were using $x \text{ and } y$ as our variables, moving along the circle involved both variables changing, but with polar coordinates only one variable changes, $\theta$.

Instead of finding the area of a rectangle and summing those, we instead consider a slice of the circle, where the angle in the slice is $d\theta$, an infinitesimally small angle, and the radius of each slice is $r$. This is shown by the shaded region in the figure above. We then sum these slices between an angle of $\theta =0$ and $\theta = 90^{\circ}$. But, when using calculus, we do not use degrees, but rather we have to use radians, which as I explained in this blog, are a more natural unit for measuring an angle.

As $d\theta$ becomes infinitesimally small, the slice becomes a triangle, and the area of a triangle is given by $\text{ half the base } \times \text{ the height}$. The height is, of course, just the radius $r$, but what about the base? The base is the length of the arc, which you will recall from the definition of a radian is just $\text{ base } = r \; d\theta$.

The integral we wish to do is therefore

$\int_{0}^{\pi/2} \frac{1}{2} r \times r \; d\theta \; = \; \int_{0}^{\pi/2} \frac{1}{2} r^{2} \; d\theta = \frac{ r^{2} }{ 2 } \int_{0}^{\pi/2} d\theta = \frac{ r^{2} }{ 2 } [ \frac{ \pi }{2} - 0 ] = \frac{ \pi r^{2} }{ 4 }$

But, remember, this is the area of just the positive quadrant, so the area of the whole circle is going to be 4 times this, or

$\boxed { 4 \times \frac{ \pi r^{2} }{ 4 } = \pi r^{2} }$

just as the famous formula states!

When I derived the acceleration of an object moving in a circle (the centripetal acceleration)

$\vec{a} = \frac{ v^{2} }{ |\,\vec{r}\,| } \hat{r}$

one of the comments stated that I could have derived the same formula using polar coordinates. Now that I have introduced polar coordinates, I will in a future blog re-derive the formula using them.

### 5 Responses

1. Interesting! I actually did a post on my own blog, about a month ago, which also derived the formula for the Area of the Circle; however, I used a method from Geometry. Seeing a method from Calculus is one of those lovely ways we can see that mathematics always remains the same despite the fact that there are multiple methods in which to describe it!

2. Wow, this is interesting. Mind if I use this for a bit of inspiration for one of my posts?

• Thank you and yes, go ahead!

3. […] In this blog, I used polar coordinates to derive the well-known expression for the area of a circle, . In today’s blog, I will go from 2 to 3-dimensions to derive the expression for the surface area of a sphere, which is . To do this, we need to use the 3-dimensional equivalent of polar coordinates, which are called spherical polar coordinates. […]

4. […] a circle with radius which is centred on the origin has the equation , but as I pointed out in my blog on deriving the area of a circle, because we can’t integrate we are stuck in trying to use conventional Cartesian […]