Derivation of bound surface and volume charge density

Question

I have read two different derivations of the bound surface and volume charge densities and I'm not sure how to reconcile between those two.

$$V(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int_{\textit{V}} \frac{\textbf{P(r')}\cdot\hat{\mathfrak{r}}}{\mathfrak{r}^2}d\tau'\tag{1}$$

is transformed into

$$\frac{1}{4\pi\epsilon_0}\oint_S\frac{1}{\mathfrak{r}}\textbf{P}\cdot d\textbf{a'} -\frac{1}{4\pi\epsilon_0}\int_{V}\frac{1}{\mathfrak{r}}(\nabla'\cdot\textbf{P})d\tau'\tag{2}$$

Both approaches calls the second term the volume bound charge and defines $\rho_b=-\nabla\bullet\textbf{P}$

and this is where the two approaches differ:

First approach (Griffiths) calls the first term surface charge and defines $$\sigma_b=\textbf{P}\cdot\hat{\textbf{n}}\tag{2}$$

Second approach (http://physics.unl.edu/tsymbal/teaching/EM-913/section4-Electrostatics.pdf page 4) claims the first term is zero, and derives the surface term using the volume bound charge as a delta function(page 5).

What I'm confused about is that when you first look at (2) how would you know whether you should try the delta function method or surface charge method? I know it doesn't matter which one you choose as long as you only do one of these two, but I'm uncomfortable that if you do take into account the surface charge you'd have to ignore the surface in the second integral, or vice versa with the delta function; surely the two terms are independent and what you do to the limit of one shouldn't affect the other integral?

I started thinking about this when I was reading about Gauss's Law in dielectrics and the fact that you can 'ignore' surface charges, so if you could refer to that in your answer I'd be grateful.

Answer 1

The two claims are both alright. The first takes the closed surface for integration to be inside the material body (below the actual boundary surface of the body), so $\mathbf P$ is non-zero on the integration surface and there is some surface term; divergence of $\mathbf P$ inside the surface may not be zero, but usually it is zero, if the body has spatially uniform dielectric constant. The second claim uses slightly bigger surface that contains the whole body in vacuum, so $\mathbf P$ is zero on the integration surface and the surface term vanishes and only the divergence term contributes, near the real surface of the body.

Either way, one gets the same polarization charge near the real surface of the body.

Answer 2

I think there is no problem with the two ways of seeing the surface bound charge density. On page (4) they say that $(P\cdot \hat{n})$ becomes zero because the density of bound charge is being measured on a closed surface (finite volume) since the total amount of charge that crosses the surface is in opposite directions this flow is canceled and the net gives zero.

we take a macroscopic volume, it will contain equal amount of positive and negative charges and the net charge will be zero.

on page (5) there is a discontinuity of $P$, we must consider the boundary condition and here the surface is infinitesimal and $\sigma_b$ will not become zero .