When you polarize a neutral dielectric, the charge moves a bit, but the total remains zero.
This fact should be reflected in the bound charges $\sigma _b $ and $\rho _b $.
Prove from Eqs. 4.11 and 4.12 that the total bound charge vanishes.
Let's start by recalling that the bound charges are given by (Given some polarization $\mathbf{P} $)
\[\sigma _b=\mathbf{P}\cdot \hat{\mathbf{n}} \]And
\[\rho _{b}=-\nabla \cdot \mathbf{P} \]Where the first denotes surface (bound) charge (density) and the latter volume (bound) charge (density).
Now that we have the (bound) densities, the total bound charge is given by integrals of the aforementioned densities:
\[Q_{\text{total,bound} }=\oint _{\mathcal{S}} \sigma _bda+\int _{\mathcal{V}} \rho _bd\tau \]But we know what $\sigma _b $ and $\rho _b $ are, so let's replace them in the above to get:
\[Q_{\text{total,bound} }=\oint _{\mathcal{S}} \mathbf{P}\cdot \hat{\mathbf{n}}-\int _{\mathcal{V}} \nabla \cdot \mathbf{P}d\tau \]But the divergence theorem tells us that closed surface integrals are related to volume integrals through a divergence:
\[ \int _{\mathcal{V}}\nabla \cdot \mathbf{P}d\tau =\oint _{\mathcal{S}}\mathbf{P}\cdot d\mathbf{a} \]Therefore, the above expression vanishes and we conclude that
\[Q_{\text{total,bound} }=0 \]