Doubt regarding volume element in Spherical Coordinate

[Luca_Mantani]

Homework Statement

Hi everyone. Here's my problem. I know that the volume element in spherical coordinate is ##dV=r^2\sin{\theta}drd\theta d\phi##. The problem is that when i have to compute an integral, sometimes is useful to write it like this:
$$r^2d(-\cos{\theta})dr d\phi$$
because ##d(-\cos{\theta})=sin{\theta}d\theta##, but i often find in textbooks and in the internet the following expression:
$$r^2d(\cos{\theta})dr d\phi$$
without the minus sign. In both cases the integration extreme are -1 and +1. Why? I'm trying to understand and it may be even stupid but i can't figure it out.
Thanks for the help!

Homework Equations

The Attempt at a Solution

 

[STEMucator]

Hi everyone. Here's my problem. I know that the volume element in spherical coordinate is ##dV=r^2\sin{\theta}drd\theta d\phi##.

When speaking in a math sense (not a physics sense), the volume element should read ##\text{dV} = \rho^2 \sin(\phi) \space \text{d} \rho \text{d} \theta \text{d} \phi##. Usually we use ##\rho## instead of ##r## to represent radial distance. Also, ##\theta## is the azimuth angle in the plane, and ##\phi## is the polar angle. Most mathematicians use this convention to remain consistent with the polar and cylindrical co-ordinate systems.

See this link for a visual: https://en.wikipedia.org/wiki/Spherical_coordinate_system#/media/File:3D_Spherical_2.svg

The problem is that when i have to compute an integral, sometimes is useful to write it like this:
$$r^2d(-\cos{\theta})dr d\phi$$

This will now read: ##\text{dV} = \rho^2 \space \text{d} \rho \text{d} \theta \text{d}(- \cos(\phi))##.

Why would you choose to write it like that? Could you provide an actual problem so we can assist you?

 

[Luca_Mantani]

If I'm trying to integrate a function that has a ##\phi## dependence only in the form of ##\cos{\phi}##, i mean ##f(\rho,\theta,\cos{\phi})##, it is useful because then i can make the substitution ##y=-\cos{\phi}## making the integral easier to compute.

 

[STEMucator]

In both cases the integration extreme are -1 and +1. Why?

These are the limits for ##\rho## I assume? The limits for ##\rho## would not change I think.

If I'm trying to integrate a function that has a ##\phi## dependence only in the form of ##\cos{\phi}##, i mean ##f(\rho,\theta,\cos{\phi})##, it is useful because then i can make the substitution ##y=-\cos{\phi}## making the integral easier to compute.

If you provide us with the exact problem statement, perhaps it would be more clear what is required.

 

[RUber]

They are equivalent...it just depends on how you look at the integral.
##\int_0^\pi \sin\phi \, d\phi = \int_\pi ^0 - \sin\phi \, d\phi = \int_{\cos \pi } ^ { \cos 0} 1 \, d( \cos \phi ) = \int_{-1 } ^ { 1} 1 \, d( \cos \phi )##
##\int_0^\pi \sin\phi \, d\phi = \int_{-\cos 0 } ^ { -\cos \pi} 1 \, d( -\cos \phi ) = \int_{-1 } ^ { 1} 1 \, d( -\cos \phi )##

 

[Luca_Mantani]

They are equivalent...it just depends on how you look at the integral.
##\int_0^\pi \sin\phi \, d\phi = \int_\pi ^0 - \sin\phi \, d\phi = \int_{\cos \pi } ^ { \cos 0} 1 \, d( \cos \phi ) = \int_{-1 } ^ { 1} 1 \, d( \cos \phi )##
##\int_0^\pi \sin\phi \, d\phi = \int_{-\cos 0 } ^ { -\cos \pi} 1 \, d( -\cos \phi ) = \int_{-1 } ^ { 1} 1 \, d( -\cos \phi )##

Great, thank you!