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*.

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

We then increase the angle by a small amount , and increase the angle by a small amount . As the figure shows, the small surface element which is thus created is just multiplied by , so .

.

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 and we vary . We can go from a (the negative to the positive ), and from a (one complete rotation about the z-axis on the x-y plane), so we have

so, the total surface area of a sphere is

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

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on 28/10/2014 at 07:42 |Derivation of the volume of a sphere | thecuriousastronomer[…] 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 . […]

on 31/07/2016 at 07:26 |Jordan BrownHow did you work out the lengths of the sides of the small area to be deltaTheta * r and deltaPhi * rcos(theta)

on 31/07/2016 at 07:39 |RhEvansLook 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 . If you can see why this is, the length of the other side being follows the same logic.

on 11/05/2017 at 14:06 |Jeremy T.It isn’t a cos in this formula r²cos(Teta)d(Teta)d(Phi). It is a sin.

on 11/05/2017 at 14:27 |RhEvansIf 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 14:41Jeremy T.Then your Theta, when you’re at (0,0, Zmax) is equal to Pi/2 ?

on 11/05/2017 at 14:44RhEvansTheta, 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).