Greetings,

Note: I will be using Z for integration

The spherical coordinate system, which is widely used in physics when there is a spherical symmetry. For our purpose, it will be useful in solving hydrogen atom problem later

A hydrogen atom is composed of a proton and an electron, and the Coulomb potential of the electron due to the presence of proton only depends on the distance to the proton and not on the direction to the proton.

It is given by “−ke2/r,” where r is the distance. Therefore, there is a spherical symmetry (i.e. independence from the direction) for the potential energy of electron. In this article, we will use the spherical coordinate system to obtain the formulas for the surface area of sphere and the volume of ball, which most high school students learn, but without their derivation. We have denoted spherical coordinate system. Instead of x, y, and z to denote a position in 3-dimensional space, we have r, θ and φ. Notice that the distance from the origin is r. Notice also that the right triangle, which is denoted by dotted line, has a height of r cos θ and a base of r sin θ.

This suggests the following relation between x,y,z and r,θ,φ.

x = r sin θ cos φ

y = r sin θ sin φ

z = r cos θ

where r ≥ 0, 0 ≤ θ ≤ π and 0 ≤ φ < 2π. Notice again that the Coulomb potential an Given this, let’s apply spherical coordinate to obtain the surface area of a sphere with radius r.The shaded region is an infinitesimal part of the surface of the sphere. It is a rectangle and the area is given by (r sin θdφ)·(rdθ). If we add all these up for different θ and φ, we will get the surface area of sphere. Let’s do this: A = Z π θ=0 Z 2π φ=0 (r sin θdφ) · (rdθ) = Z π θ=0 r 2 sin θdθ2π = 2πr2 (− cos π + cos 0) = 4πr2 . So, this is the surface area of the sphere. Now, what is the enclosed volume of the sphere? The inner region of a sphere is called a “ball.” So what is the volume of ball with radius R?

We see that the infinitesimal volume area is given as follows:

dV = (r sin θdφ) · (rdθ) · dr

Integrating, we get: V = Z r=R r=0 Z π θ=0 Z 2π φ=0 dV = Z r=R r=0 Z π θ=0 dr(r 2 sin θdθ2π) (6) = Z r=R r=0 4πr2 dr = 4 3 πR3