Why Does My Integration Over a Sphere Give Incorrect Results?

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Addez123
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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?
 
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##.
 
Orodruin said:
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.