Energy in Dielectric Material

The total work done to build up the free charge from zero to the final configuration in the presence of a dielectric material is: $$W = frac{1}{2}int vec{D} cdot vec{E} dtau.$$ We got this by supposing the dielectric material is fixed in position and then we bring in the free charge $Delta rho_f$ a bit at a time. Thus the work done on the incremental free charge is given by:
$$Delta W = int (Delta rho_f)V d tau.$$

Since $nabla cdot vec{D} = rho_f$, $Delta rho_f = nabla cdot (Delta vec{D})$, where $Delta vec{D}$ is the resulting change in $vec{D}$, so
$$Delta W = int [nabla cdot (Delta vec{D})]V d tau.$$ and hence by integration by parts
$$ Delta W = int nabla cdot [(Delta vec{D})V]d tau + int(Delta vec{D}) cdot vec{E} d tau.$$ The divergence theorem turns the first term into a surface integral, which vanishes if we integrate over all space. Therefore , the work done is equal to $$Delta W = int (Delta vec{D}) cdot vec{E} d tau.$$

Questions:
Why does $Delta vec{D}V$ have to go to zero faster that $frac{1}{r^2}$ for the surface integral (from the divergence theorem) $int nabla cdot [(Delta vec{D})V]d tau = oint V Delta vec{D} cdot da = 0$?

In this derivation (a full explanation is here) are we accounting for the fact that as we bring in free charge the dielectric material becomes polarised and hence the electric field (and therefore the potential at any point $V(r)$) changes due to the addition of $Delta rho_f$? If so, where in the derivation?

Thanks.