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Your whole derivation is correct. Even in the presence of two different dieletric materials, the $\mathbf{D}$ field will not be affected, but for the free charge density that you already dealt with. So the field will be $$\mathbf{D}=\begin{cases}\dfrac{1}{5}\beta r^3\hat{\mathbf{r}},&\text{if r<a},\\ \dfrac{1}{5}\beta ... 1 We start with the integral$$\oint_{|\vec{r}|=R}\mathrm{d}\Omega\frac{\vec{r}}{|\vec{r}-\vec{r}'|}.$$Since we are integrating over \vec{r}, we can without loss of generality, arrange for \vec{r}'to lie along the +Z axis, so that \vec{r}'=r'\hat{z}. Then the angle between \vec{r} and \vec{r}' is the standard angle (in spherical coordinates) ... 0 To solve the integral, I came across the multipole expansion$$\frac{1}{\left|\vec{r}-\vec{r}_{0}\right|}=\sum_{l=0}^{\infty}\frac{r_{<}^{l}}{r_{>}^{l+1}}P_{l}\left(\cos\sphericalangle\left(\vec{r},\vec{r}_{0}\right)\right),\;\; r_{<}=\min\left(r,r_{0}\right),\; r_{>}=\max\left(r,r_{0}\right)$$With \vec{r}_0=\vec{r}^{\prime} one has ... 1 The covariant formulation of EM is precisely this. The formulation as a gauge theory also does this. (c = 1 in the following) Given the E- and B-fields as spatial three-vectors in some frame, we construct the antisymmetric field strength tensor as (roman indices are spatial indices, summation over repeated indices implied)$$ F^{0i} := E^i \; ...