boundary condition of electrical field The boundary condition of electrical field is well known.
$$ E^{\perp} _{above} -E^{\perp} _{below} = \frac{\sigma}{\epsilon_0}$$
Griffiths wrote that where there is no surface charge $E^{\perp}$ is continuous, as for instance at the surface of a uniformly charged solid sphere.
Let's think of electrical field of solid sphere which carries a uniformly volume charge density $\rho$. It's obviously true that $E$ is continuous by Gauss theorem, but I think it has surface charge density $\sigma$ too. Why doesn't it have $\sigma$ in this situation?
 A: Your general analysis is correct except the fact that you expect to exist a surface density just because there is a charge volume density.
One way to see this is the following. In your uniform sphere consider a shell with small thickness. What is the amount of charge in this shell? If this shell has radius $r$ it is $\Delta Q=4\pi r^2 \Delta r \rho$ (volume times density). If you want to calculate the surface density you should divide for the total area of the surface and send the thickness to zero. It gives $\Delta r \rho \,\to\,0$ as $\Delta r \,\to\,0$. Put it simply, you have a volume charge density therefore the total charge of an object goes to zero as the volume goes to zero which is what happens in any surface (by definition it has volume zero).
Suppose now you want to solve another problem in which, besides a charge density sphere, you will also put as a condition of the problem a surface charge $\sigma$  and want to determine the electric field. In this case in applying Gauss theorem there is a discontinuity on the total charge enclosed by a sphere as it crosses this surface: $\Delta Q(R+\delta r)-\Delta Q(R-\delta r) = \sigma 4\pi R^2$. But again, this only happens if surface density is directly imposed and not because you have a volume charge density.
