Derivation of the divergence in the spherical coordinate

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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
 

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The procedure is described here: https://en.wikipedia.org/wiki/List_of_common_coordinate_transformations#From_spherical_coordinates

If we change coordinates in an integral we have to adjust the volume elements by the determinant of the transformation matrix, and this determinant is ##r^2 \sin \vartheta##. It is the transformation theorem for integrals. In words: you must not neglect the radius, as it is part of the formulas for volume. Without it you would set ##r=1## which makes no sense if we want to consider an infinitesimal volume element,
 

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