As most people reading this blog probably know, the area of a circle is given by
where 
The area under a curve
The area under any curve between the x-axis, the curve, and the lines 
where 
The area under any curve 
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
Therefore, to find the area of the circle, all we need to do is integrate between 
To find the area of a circle, in principle all we need to do is sum the rectangles 
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.
The point 
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
To find the area of the circle, we add a series of slices (the shaded region), each with an infinitesimally small angle 
One of the reasons the method of using polar coordinates is easier is that we now have only one variable,
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 
As 
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!
![\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 }](https://s0.wp.com/latex.php?latex=%5Cint_%7B0%7D%5E%7B%5Cpi%2F2%7D+%5Cfrac%7B1%7D%7B2%7D+r+%5Ctimes+r+%5C%3B+d%5Ctheta+%5C%3B+%3D+%5C%3B+%5Cint_%7B0%7D%5E%7B%5Cpi%2F2%7D+%5Cfrac%7B1%7D%7B2%7D+r%5E%7B2%7D+%5C%3B+d%5Ctheta+%3D+%5Cfrac%7B+r%5E%7B2%7D+%7D%7B+2+%7D+%5Cint_%7B0%7D%5E%7B%5Cpi%2F2%7D+d%5Ctheta+%3D+%5Cfrac%7B+r%5E%7B2%7D+%7D%7B+2+%7D+%5B+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+-+0+%5D+%3D+%5Cfrac%7B+%5Cpi+r%5E%7B2%7D+%7D%7B+4+%7D&bg=ffffff&fg=333333&s=0 "\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 }")
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.
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!
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!
[…] 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. […]
[…] 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 […]