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.
[…] 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 . […]
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 . If you can see why this is, the length of the other side being follows the same logic.
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.
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).
How can I used line integral to over a circular surface to find the surface area for 3-D sphere equation?