If $\oint \vec D \cdot d\vec a = 0$ everywhere then $\vec D = \vec 0$? I'm reading a book on Electricity and Magnetism.  In an example using the electric displacement they write that, since $\oint \vec D \cdot d\vec a = 0$ in the example, we can infer that $\vec D = \vec 0$.  I would buy that argument if the integral were any path integral, but here I believe that $\oint$ refers to an oriented closed-path integral, yes?  
If it helps, here's as much as I know about the objects involved: $\vec D = \epsilon_0\vec E + \vec P$ where $\epsilon_0$ is a physical constant, $\vec E$ is the net electric field, and $\vec P$ is the polarization, which mathematically is just any (presumably continuous, I would think) vector field.  Only $\vec E$ is guaranteed to be conservative.
So how do we know that $\vec D = \vec 0$ rather than merely every closed path integral being zero, so that in effect this only shows that $\vec D$ is conservative?  Is there some further feature of the physical system that would imply a property of $\vec D$ which allows us to make this inference? 

To give context, in case it helps, the example comes from Griffith's Electrodynamics 3rd Ed.  It's problem 4.4.15 where $\vec P = \frac k r \hat r$ inside a thick spherical shell of inner radius a and outer radius b, where k is some arbitrary constant. 
 A: 
We shall assume that we are analyzing a problem in electrostatics.  

Let $\displaystyle V$ denote the open domain $\displaystyle V=\{\vec r|0<a<|\vec r|<b\}$ and $\displaystyle \partial V$ denote its boundary.  We shall denote the magnitude of the position vector by $r$.  That is, $r=|\vec r|$.

Suppose that the Polarization density $\displaystyle \vec P(\vec r)$ is given by 
$$\vec P(\vec r)=
\begin{cases}
k\,\frac{\vec r}{r^2}&, \vec r\in \bar V\\\\ 0&,\vec r \,\,\text{otherwise}
\end{cases}$$
Then the volume polarization charge density $\displaystyle \rho_P(\vec r)-\nabla \cdot \vec P(\vec r)$ is 
$$  \rho_P(\vec r)=-\frac{k}{r^2}$$
and the polarization surface charge density $\sigma_P(\vec r)=\left. \hat n \cdot \vec P(\vec r)\right|_{\vec r \in \partial V}$ is 
$$\sigma_P(\vec r)=\begin{cases}
\frac{k}{b}&, r=b\\\\
-\frac{k}{a}&,r=a
\end{cases}$$

Suppose further that the free charge density $\rho(\vec r)=0$ for $\vec r \in \mathbb{R}^3$.   Then by the spherical symmetry of the problem we see that $\vec P=\hat r P_r(r)$, $\vec E=\hat r E_r(r)$, and $\vec D=\hat r D_r(r)$. That is all vector fields are radial and depend on $r$ only.

Since there is no free charge density, $\nabla \cdot \vec D=0$.  From the divergence theorem we see that for all closed surfaces $S$, $\oint_{S}\vec D(\vec r)\cdot \hat n\,dS=0$.  Take $S$ to be a sphere of raidus $r$.  Then,
$$\oint_{S}\vec D(\vec r)\cdot \hat n\,dS=4\pi r^2 D_r(r)=0$$
and we conclude that $\vec D=0$.
