Solving the Mystery: Why Phi is Limited to 0-Pi in Spherical Coord System

[gikiian]

Whyyyyyy??! Whhhhhy?!?

 

[SteamKing]

It depends. Often, symmetry is used in evaluating definite integrals. If something is symmetrical with respect to the range of phi, then 2*integral|0-pi = integral|0-2pi.

 

[I like Serena]

Spherical coordinates have 2 angles.
It's like a position on earth, which has latitude and longitude.
Longitude goes all the way around (total angle 2π).
And latitude goes from pole to pole (total angle π).

Oh, and welcome to PF!

 

[LCKurtz]

There is a real reason. In triple integration, if you use the standard volume element:

$$dV = \rho^2\sin(\phi)d\rho d\phi d\theta$$

you want to let θ to from 0 to 2π and φ go from 0 to π, otherwise the sin(φ) factor can be negative. If you don't do that you need absolute values around the sine factor, generally causing twice the work, or worse, incorrect calculation by being unaware of that.

 

[I like Serena]

There is a real reason. In triple integration, if you use the standard volume element:

$$dV = \rho^2\sin(\phi)d\rho d\phi d\theta$$

you want to let θ to from 0 to 2π and φ go from 0 to π, otherwise the sin(φ) factor can be negative. If you don't do that you need absolute values around the sine factor, generally causing twice the work, or worse, incorrect calculation by being unaware of that.

Good one!

I never realized that and I have often wondered why spherical coordinates didn't use a latitude-like angle, which for instance wouldn't turn the zero-point into a singular point.

 

[Mute]

It's because you'll double count the contribution of the integrand to the integral if both angles run from 0 to 2pi. Think about integrating over the sphere to find its volume: If you integrate over phi from 0 to pi, you get half of a circle; if you then integrate theta from 0 to 2pi that half-circle sweeps out the volume of the sphere; however, if you integrated phi from from 0 to 2pi, then that gives you a full circle, which if you then integrate theta from 0 to 2pi, the circle sweeps out the volume of the sphere twice. You only need to integrate phi from 0 to pi to sweep out the full volume of the sphere.

 

[gikiian]

Thanks, I got it :)