- #1

elgen

- 58

- 5

Hi,

I am reviewing some vector calculus and have a problem on the derivation of the divergence in the spherical coordinate.

Assume there is a small volume located at [tex]r_0, \theta_0, \phi_0[/tex] with a volume of [tex]r_0^2\sin\theta_0 \Delta r \Delta \theta \Delta \phi[/tex].

My question is that why there is an [tex]r^2[/tex] sitting behind the partial derivative operator? The problem is attached.

My feeling is that as I take the limit of the volume going to 0, I can cannot cancel [tex]r_0^2[/tex]. However, I cannot give a good justification to that.

Would anyone knowledgeable with vector calculus give an explanation? Thank you.

elgen

I am reviewing some vector calculus and have a problem on the derivation of the divergence in the spherical coordinate.

Assume there is a small volume located at [tex]r_0, \theta_0, \phi_0[/tex] with a volume of [tex]r_0^2\sin\theta_0 \Delta r \Delta \theta \Delta \phi[/tex].

My question is that why there is an [tex]r^2[/tex] sitting behind the partial derivative operator? The problem is attached.

My feeling is that as I take the limit of the volume going to 0, I can cannot cancel [tex]r_0^2[/tex]. However, I cannot give a good justification to that.

Would anyone knowledgeable with vector calculus give an explanation? Thank you.

elgen