There is a solved exercise in which a charged sphere of radium $r=a$ is inside a dielectric material. The surface bound charge and the volume bound charge are respectively:
$$p_{sp}=\vec{P}\cdot \hat{n}|_{r=a},\quad p_{vp}=-\nabla \cdot \vec{P}$$
where the polarization $P$ is known.
The exercise says that the total bound charge is:
$$Q_{T}=\int_{S}p_{sp}dS+\int_{V}p_{vp}dV=f(\epsilon_{r},q)\neq 0$$
But for the total bound charge it is true that:
$$Q_{T}=\int_{S}p_{sp}dS+\int_{V}p_{vp}dV=\oint_{S} \vec{P}\cdot \hat{n}dS-\int_{V}\nabla \cdot \vec{P}dV=0$$
How can this be when the total bound charge should always be equal to 0?
Best Answer
The integral over the total surface bound charge , inner and outer, of the dielectric is zero. Equivalently, the volume integral of the bound charge is zero. It is not a good idea to mix surface and volume integrals to avoid double counting.