1. Oct 22, 2013

### joshmccraney

hey pf!

i had a subtlety about the divergence theroem/gradient theorem. specifically, the following was presented: $$\iint_s p \vec{dS} = \iiint_v \nabla (p) dv$$

i am familiar with the divergence theorem, but that is for a vector "dotted" with a surface element (flux) related to the divergence (expansion) through the volume. i believe the above is the gradient theorem. can anyone verify the legitimacy of this (and perhaps offer some physical intuition if it holds). i should say $p$ is not a constant.

thanks!

2. Oct 22, 2013

### SteamKing

Staff Emeritus
3. Oct 22, 2013

### arildno

We may write:
$$\vec{dS}=dS(n_{x}\vec{i}+n_{y}\vec{j}+n_{z}\vec{k}) [tex]\int\int_{S}p\vec{dS}=\int\int_{S}pn_{x}dS\vec{i}+\int\int_{S}pn_{y}dS\vec{j}+\int\int_{S}pn_{z}dS\vec{k} (*)$$
Since the Cartesian unit vector can be extracted from the integrand!
--
But, we have:
$$pn_{x}dS=(p\vec{i}\cdot\vec{dS})$$
And the first term in RHS in (*) is then simply, by the divergence theorem:
$$\int\int_{S}pn_{x}dS\vec{i}=+int\int\int_{V}\frac{\partial{p}}{\partial{x}}\vec{i}dV$$

Do the similar thing to the other two terms on RHS in (*), and you get your identity.

4. Oct 22, 2013

### joshmccraney

thank you arildno!!!!! this makes a lot of sense now! youre brilliant!

5. Oct 23, 2013

### joshmccraney

on another note, arildno, did you just figure this out by thinking about what's going on or did you see it in a book before? either way, thanks again

6. Oct 23, 2013

### arildno

I noticed it a long time ago. I don't remember if it was in a book or on my own inspiration.

7. Oct 23, 2013

### vanhees71

I'd apply Gauß's theorem to the vector field
$$\vec{V}(\vec{x})=\vec{a} p(\vec{x})$$
for a constant vector $\vec{a}$. Then you have
$$\vec{\nabla} \cdot \vec{V}=\vec{a} \cdot \vec{\nabla} p.$$
Then Gauß's theorem gives
$$\int_V \mathrm{d}^3 \vec{x} \vec{a} \cdot \vec{\nabla} p = \vec{a} \cdot \int_V \mathrm{d}^3 \vec{x} \vec{\nabla} p = \int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{a} p = \vec{a} \int_{\partial V} \mathrm{d}^2 \vec{F} p.$$
Since this is true for all $\vec{a}$ you have proven your vector identity.

8. Oct 23, 2013

### arildno

Sure. It's matter of taste, really, which one approach is used.
After all, my method is just as general as the one you present.