# Stoke's formula for a sphere

I have a question, raised form kinetic theory (pure mathematical). Imagine, that $\Psi (\overrightarrow{r},\overrightarrow{p},t)$ - sufficiently smooth function, where $\overrightarrow{r}$ - radius vector, $\overrightarrow{p}$ - unit vector of orientation, $t$ - time.

Can someone explain me, how to proove this identities (I think, Stoke's theorem must be used here):

$$\int_{S} \overrightarrow{p} \left(\nabla_p \cdot \left( \frac {d\overrightarrow{p}} {dt} \Psi\right) \right)d\overrightarrow{p}=-\int_{S} \frac {d\overrightarrow{p}} {dt} \Psi d\overrightarrow{p}$$

$$\int_{S} \nabla_p \cdot \left( \frac {d\overrightarrow{p}} {dt} \Psi\right) d\overrightarrow{p}=0$$

Here $S$ is the surface of the unit sphere, so we integrate over a unit sphere.

This formula are used in physical article. I am sure, that you don't need more information for this identities.

Edit: If this is simple Integration by parts, I am interested, what boundary term is, and why it is zero

• hint: integration by parts Commented Sep 3, 2015 at 19:39
• Thanks, I figured that. Can you explain me, why boundary term disappears? I don't know, what boundary is here Commented Sep 3, 2015 at 19:44
• I have an explanation that avoids differential geometry in the appendix of this article: faculty.missouri.edu/~stephen/preprints/pde-sphere.html Commented Sep 3, 2015 at 19:48
• But if you want to use Stoke's Theorem, the clever thing is that on the surface of the sphere, the boundary is the empty set. Commented Sep 3, 2015 at 19:49
• You said that $\vec p$ is a unit vector. Is that correct? And, what is the relevance or $\vec r$ then? Commented Sep 3, 2015 at 19:51