Why Does My Integration Over a Sphere Give Incorrect Results?

[Addez123]

Homework Statement

Integrate 3 over a sphere with radius R.

Relevant Equations

Polar coordinates

$\iiint 3 dr d\rho d\phi$
The volume of a sphere is $4\pi /3 r^3$ so naturally the answer is $4 \pi R^3$
But when I integrate I do:

$3 \iint r |_0^R d\rho d\phi$

$3R \int \rho |_0^{2\pi} d\phi$

$6R\pi * \phi |_0^\pi = 6R\pi^2$

What am I doing wrong?

 

[Orodruin]

What is the actual homework statement? Please reproduce the full statement exactly as given.

You have also used the wrong volume element. The volume element in spherical coordinates is $dV = r^2 \sin(\theta) dr \, d\theta\, d\phi$.

 

[Addez123]

What is the actual homework statement? Please reproduce the full statement exactly as given.

You have also used the wrong volume element. The volume element in spherical coordinates is $dV = r^2 \sin(\theta) dr \, d\theta\, d\phi$.

it wasn't a homework but I keep being forced to post it in homework section by mods.

Eitherway thanks, I forgot to add $r^2sin(\theta)$ when converting from normal coordinates.