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## Derivation of the surface area of a sphere

In this blog, I used polar coordinates to derive the well-known expression for the area of a circle, $A =\pi r^{2}$. 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 $A = 4 \pi r^{2}$. To do this, we need to use the 3-dimensional equivalent of polar coordinates, which are called spherical polar coordinates.

Let us imagine drawing a line in 3-D space of length $r$ into the positive part of the $x,y,z$ coordinate system. We will draw this line at an angle $\theta$ above the x-y (horizontal) plane, and at an angle $\phi$ to the y-z (vertical) plane (see the figure below). When we drop a vertical line from our point $(x,y,z)$ onto the x-y plane it has a length $r \cos \theta$, as shown in the figure below.

We then increase the angle $\theta$ by a small amount $d\theta$, and increase the angle $\phi$ by a small amount $d \phi$. As the figure shows, the small surface element $dA$ which is thus created is just $r d\theta$ multiplied by $r \; \cos \theta \; d\phi$, so $dA = r^{2} \; \cos \theta \; d\theta d\phi$.

Using spherical polar coordinates, the area element $dA$ on the surface of a sphere is given by $r^{2} \cos \theta d \theta d\phi$.

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To find the surface area of the sphere, we need to integrate this area element over the entire surface of the sphere. Therefore, we keep $r \text{ constant}$ and we vary $\theta \text{ and } \phi$. We can go from a $\theta \text{ of } -\pi/2 \text{ to } +\pi/2$ (the negative $z-direction$ to the positive $z-direction$), and from a $\phi \text{ of } 0 \text{ to } 2\pi$ (one complete rotation about the z-axis on the x-y plane), so we have

$\text{ total surface area } = \int dA = r^{2} \int_{-\pi/2}^{+\pi/2} \cos \theta d \theta \int_{0}^{2\pi} d\phi$

$\text{ total surface area } = r^{2} \left[ \sin \theta \right]_{-\pi/2}^{+\pi/2} \cdot \left[ \phi \right]_{0}^{2\pi} = r^{2} \cdot (1-(-1)) \cdot (2\pi) = r^{2} \cdot (2) \cdot (2\pi)$

so, the total surface area of a sphere is

$\boxed{ \text{surface area } = 4\pi r^{2} }$

as required. In a future blog I will use spherical polar coordinates to derive the volume of a sphere, where $r$ will no longer be constant as it is here.

## 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.