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

where is the ratio of the diameter of a circle to its circumference, and 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 (see the figure below) is given by

where is the function (the curve), and is an infinitesimally small width. Effectively, what we are doing is summing a series of rectangles of area from the lower limit to the upper limit .

## 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 the equation which describes this circle is just

Therefore, to find the area of the circle, all we need to do is integrate between , and then multiply the answer by 4 (as we have only found the area of quarter of the circle).

The integration to do this is

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.

We write the coordinates in terms of two new variables , where is the radius of the circle and 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 and .

Then, to determine the area of the quadrant were are positive, we can instead integrate using . To see how we do this, consider the figure below.

One of the reasons the method of using polar coordinates is easier is that we now have only one variable, , as the radius is a constant. When we were using as our variables, moving along the circle involved *both* variables changing, but with polar coordinates only one variable changes, .

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 , an infinitesimally small angle, and the radius of each slice is . This is shown by the shaded region in the figure above. We then sum these slices between an angle of and . 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 becomes infinitesimally small, the slice becomes a triangle, and the area of a triangle is given by . The height is, of course, just the radius , 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 .

The integral we wish to do is therefore

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

just as the famous formula states!

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

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.

on 15/07/2014 at 10:36 |Boxing PythagorasInteresting! 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!

on 15/07/2014 at 13:05 |Feng WuWow, this is interesting. Mind if I use this for a bit of inspiration for one of my posts?

on 15/07/2014 at 13:07 |RhEvansThank you and yes, go ahead!

on 21/10/2014 at 06:31 |Derivation of the surface area of a sphere | thecuriousastronomer[…] 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. […]

on 25/11/2014 at 07:30 |Derivation of the volume of a sphere – method 2 | thecuriousastronomer[…] 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 […]