Gradient version of divergence theorem?

[Cygnus_A]

So we all know the divergence/Gauss's theorem as
$∫ (\vec∇ ⋅ \vec v) dV = ∫\vec v \cdot d\vec S$

Now I've come across something labeled as Gauss's theorem:
$\int (\vec\nabla p)dV = \oint p d\vec S$
where p is a scalar function.

I was wondering if I could go about proving it in the following way (replacing dot products with implied sums):
With $e_i := \hat e_i$ and $d\vec S := ds_1 \hat x + ds_2 \hat y + ds_3 \hat z = ds_i e_i$,

$\oint p d\vec S = \oint p (e_i ds_i) = \oint (p e_i)(ds_i) = \oint \vec p \cdot d\vec s$ (this p vector has scalar functional dependence still, it's just $(p)(\vec v)$, a scalar times a vector, but still overall a vector in my mind)

then applying divergence theorem and getting
$= \int (\vec \nabla \cdot \vec p) dV = \int \partial_i (p e_i) dV$

and finally applying the product rule and the fact that $e_i$ is a unit vector
$\int (e_i \partial_i p + p \partial_i e_i) dV$.

The second term is zero, since it's a partial of a unit vector, which has no spatial dependence, leaving
$\int (e_i \partial_i p)dV = \int (\vec \nabla p) dV$

Does that make sense? I think it seems to work out, but I'm concerned that it's flawed due to my free conversions between sums and vectors. It seems unnatural that I've said $d\vec S =d\vec s = ds_i$, despite defining them differently. One, I suppose has actual vector components, whereas the other is just a list of components.

 

[TMO]

I believe the gradient version of the divergence theorem would be your typical statement that the integral of the path going through a potential is just the difference in potentials.

 

[Cygnus_A]

I believe the gradient version of the divergence theorem would be your typical statement that the integral of the path going through a potential is just the difference in potentials.

You must be thinking of the fundamental theorem of calculus for gradients -- $\phi (\vec b) - \phi (\vec a) = \int_a^b \vec \nabla \phi \cdot d\vec r$

This was what was presented to me during a proof, and I had never seen it before
$\int (\vec\nabla p)dV = \oint p d\vec S$(oops, and I just realized I forgot to close the surface integral of divergence theorem in my original post :P)

 

[davidmoore63@y]

The usual proof of the scalar version of the divergence theorem involves replacing the vector field v by (p k), where p is a scalar field and k is an arbitrary constant vector. Then div v is equal to k.grad p. Since k is arbitrary it can then be removed from both sides giving the result.

 

[Cygnus_A]

Ah, ok. The steps are essentially the same then. Just slightly different in how you get your constant vector. Thanks!