Why does Gauss' theorem use ∂V instead of S for the surface?

[Wrichik Basu]

I was reading the book "Mathematical Methods for Physicists", and in the first chapter, under Gauss's Theorem, the statement given was:

The surface integral of a vector over a closed surface equals the volume integral of the divergence of the vector over the entire closed surface.

But the in the mathematical form, $\partial V$ was used instead of $S$ to denote the surface integral.

I could understand that $\partial V$ is the same as $S$. Can anyone explain how?

 

[Orodruin]

I was reading the book "Mathematical Methods for Physicists"

Always quote the author(s) as well as the book title when you quote something. There are many books with this title and similar and without the authors we cannot know which.

The surface integral of a vector over a closed surface equals the volume integral of the divergence of the vector over the entire closed surface.

This is not correct. The integral of the divergence should be over the enclosed volume, not the surface.

For any volume $V$, $\partial V$ denotes its boundary.

 

[Wrichik Basu]

Always quote the author(s) as well as the book title when you quote something. There are many books with this title and similar and without the authors we cannot know which.

Authors are Arfken, Weber and Harris.

This is not correct. The integral of the divergence should be over the enclosed volume, not the surface

My mistake. In the book, it was written over the entire volume.

For any volume V, ∂V denotes its boundary.

Understood, thanks.