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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.

### 7 Responses

1. […] In this blog, I derived the expression for the surface area of a sphere, . In today’s blog, I will derive the expression for the volume of a sphere. Actually, once one has understood how to derive the surface area of a sphere using spherical polar coordinates, deriving the volume is pretty straight forward. It only involves one extra step, and that is to create a volume element with the same surface area that we had before, but with a thickness , and to integrate over in addition to integrating over . […]

2. How did you work out the lengths of the sides of the small area to be deltaTheta * r and deltaPhi * rcos(theta)

• Look at the diagram carefully. From the definition of angles expressed in radians, you should be able to see that the length of one side is $\delta \thera * r$. If you can see why this is, the length of the other side being $\delta \phi * r cos( \theta)$ follows the same logic.

3. on 11/05/2017 at 14:06 | Reply Jeremy T.

It isn’t a cos in this formula r²cos(Teta)d(Teta)d(Phi). It is a sin.

• If you look more carefully at my diagram, I think that you’ll find that it is a cosine. I have defined theta as the angle between the horizontal (x-y) plane and the radius vector.

• on 11/05/2017 at 14:41 Jeremy T.

Then your Theta, when you’re at (0,0, Zmax) is equal to Pi/2 ?

• Theta, as I have defined it, would go from -pi/2 to +pi/2 (the -z direction to the +z direction). You can see this in the limits of the integral for the theta part of the integral (we integrate over both theta and phi).