Flux density via Gauss' Law inside sphere cavity There are many similar questions on here but most seem to deal conceptually with electric field, not the mathematics behind it.
The setup is this: there is some sphere of radius b which has a volume charge density of ρv. However, there is a cavity inside the sphere, centered at the origin, of radius a, inside which ρv = 0. Also, ρv = 0 outside of the sphere of radius b. What we want to determine is the flux density D at r < a, a < r < b, and b < r.
Conceptually, it makes sense to me that the electric field inside the cavity is 0 since all charges on the inner surface are symmetrically distributed. However, there should be some D because flux would be transmitted along unit vector -ar. Same goes for outside the sphere: there should be flux radiating out along ar.
I know I have an error in thinking somewhere because when I use the divergence theorem to relate the forms of Gauss' Law (i.e. the surface integral of flux density is the volume integral of charge density), and I have charge density = 0 inside r < a, I get D = 0. The same happens for when I look at r > b; the integral equates to 0 so, regardless of limits of integration, D = 0.
Between a and b the integrals are obvious and straightforward, so I get a D = 4πρv( $\frac{b^3 - a^3}{3}$). What is not obvious about the cavity and outside the sphere?
 A: Since $\textbf{D}=\epsilon\textbf{E}$ and there is no field, there will be no flux anywhere.  Flux is not transmitted along vectors: it is the sum of $\textbf{D}\cdot d\textbf{S}$ evaluated on each of the surface elements $d\textbf{S}$, 
Now it is true that it’s possible to have $0$ net flux through a closed surface but non-zero field in the region: if the closed surface contains no source charges, then every field line of $\textbf{D}$ that enters the closed surface will also exit the surface.  This would happen for instance if you plunged a cube in the constant field between two infinite plates.  
However, in your geometry, the flux must be in the $\hat{\textbf{r}}$ direction by symmetry, and the vector $\textbf{D}$ must also be along $\hat{\textbf{r}}$ by symmetry, so the contribution from all the charges on the inner surface of your hollow sphere will cancel out when summed over the whole surface.
This is qualitatively different from the situation where you want the flux through a surface containing the entire arrangement: the field outside the arrangement is identical to the field of a single charge of apppropriate magnitude located at the origin; the field lines outside the sphere will point along $+\hat{\textbf{r}}$ at every point of your surface, so that every contribution $\textbf{D}\cdot d\textbf{S}$ will be positive, resulting in a non-zero net flux.
Geometrically, the best way to understand this is by examining the figure below:

Of course with the situation on the left, where you look for the field at the center, the field is $0$ by symmetry.  
For the situation on the right, where we want to field for a field off-center: the charges in angular opening on the right are closer but the area intercepted by the opening is small, whereas the charges in the larger opening opposite are farther but the area intercepted is larger.  The effects exactly cancel out because the field goes like $1/r^2$ but the area intercepted grows like $r^2$.  The result is that the net field
at the off-centered point is still $0$.
