Homework Help: Doubt regarding volume element in Spherical Coordinate

1. Aug 20, 2015

Luca_Mantani

1. The problem statement, all variables and given/known data
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 text books 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!
2. Relevant equations

3. The attempt at a solution

2. Aug 20, 2015

Zondrina

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

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?

3. Aug 20, 2015

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.

4. Aug 20, 2015

Zondrina

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

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

5. Aug 20, 2015

RUber

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

6. Aug 21, 2015

Luca_Mantani

Great, thank you!