
\item{\bf Cartesian quadrupole moment of an ellipsoid.} 
\medskip

\noindent (a) We have
\begin{eqnarray} \nonumber
  \left(\frac{x^1}{c^1}\right)^2 + \left(\frac{x^2}{c^2}\right)^2 +
  \left(\frac{x^3}{c^3}\right)^2 &=& 1~~~{\rm with}~~~
  c^1=c^2=a\,,~~c_3=b \\ \nonumber
  {\rm and\ charge\ density}~~ \rho(\vec{x}) &=& 
  \begin{cases} \rho = {\rm const}~~{\rm within\ the\
            ellipsoid,}\\ 0~~~{\rm otherwise.} \end{cases} 
\end{eqnarray}
The total charge is 
\begin{eqnarray} \nonumber
  q&=&\int_{\sum_{i=1}^3\left(\frac{x^i}{c^i}\right)^2\le 1}d^3x\,\rho
  \,=\, c^1\,c^2\,c^3\,\rho \int_{\sum_{i=1}^3 (x'^i)^2 \le 1} d^3x'\\
  \nonumber &=& \frac{4\pi}{3}\, c^1\,c^2\,c^3\,\rho 
  \,=\, \frac{4\pi}{3}\, a^2\,b\,\rho 
\end{eqnarray}
We write the quadrupole moments of the ellipsoid as
\begin{eqnarray} \nonumber
  Q^{ij} &=& Q^{ij}_S - Q^{ij}_D~~{\rm with}~~
  Q^{ij}_S = 3\,\rho\int_{\sum_{i=1}^3\left(\frac{x^i}{c^i}\right)^2\le 1}
          d^3x\,x^i\,x^j\,~~
  Q^{ij}_D = \rho\int_{\sum_{i=1}^3\left(\frac{x^i}{c^i}\right)^2\le 1}
          d^3x\,r^2\,\delta^{ij}\,.
\end{eqnarray}
These moments transform into
\begin{eqnarray} \nonumber
  Q^{ij}_S &=& c^1\,c^2\,c^3\,\rho \int_{\sum_{i=1}^3 (x'^i)^2 \le 1} 
  d^3x'\,c^i\,c^j\,x'^i\,x'^j
          \\ \nonumber
  Q^{ij}_D &=& 
   c^1\,c^2\,c^3\,\rho \int_{\sum_{i=1}^3 (x'^i)^2 \le 1} d^3x' 
   \delta^{ij}\sum_{k=1}^3 (c^k\,x'^k)^2\ .
\end{eqnarray}
We have $Q^{ij}_D=0$ for $i\ne j$ and the relevant integrals for
$Q^{ij}_S$ are of the form
$$ \int_{\sum_{i=1}^3 (x^i)^2 \le 1} d^3x\,x^i\,x^j = \delta^{ij} 
   \int_{\sum_{i=1}^3 (x^i)^2 \le 1} d^3x\,(x^i)^2\ 
   \Rightarrow\  Q^{ij} = 0~~{\rm for}~~i\ne j $$
and
$$ \int_{\sum_{i=1}^3 (x^i)^2 \le 1} d^3x\,(x^i)^2 = 
   \int_{\vec{x} \le 1} d^3x\,(x^3)^2 = 2\pi\int_0^1 r^4\,dr\,
   \int_{-1}^{+1} (\cos\theta)^2\,d\cos\theta = \frac{2\pi}{5}\,
   \frac{2}{3} = \frac{4\pi}{15} $$
$$ \Rightarrow~~ Q^{ii}_S = \frac{q}{5}\,3\,(c^i)^2\,,~~
   Q^{ii}_D = \frac{q}{5}\,\sum_{k=1}^3 (c^k)^2\,.          $$
With $c^1=c^2=a$ and $c_3=b$:
$$ Q^{11}_S=Q^{22}_S=\frac{3\,q}{5}\,a^2\,,~~ Q^{33}_S=
   \frac{3\,q}{5}\,b^2\,,~~Q^{11}_D=Q^{22}_D=Q^{33}_D=
   \frac{q}{5}\,(2\,b^2+a^2)\,                              $$
resulting into the traceless tensor given by
$$ Q^{11}=Q^{22}=\frac{q}{5}\,\left(a^2-b^2\right),\qquad
   Q^{33}=\frac{2\,q}{5}\,\left(b^2-a^2\right)\,.  $$
(b) The energy is
$$ W = q\,\Phi(0) - \vec{p}\cdot\vec{E}(0) - \frac{1}{6}\sum_{i=1}^3
   \sum_{j=1}^3 Q^{ij}_S\,\frac{\partial E^j}{\partial x^i}\ . $$
With the normalization $\Phi(0)=0$ and
$$ \vec{p} = \rho \int_{\sum_{i=1}^3\left(\frac{x^i}{c^i}\right)^2\le 1}
   d^3x\,\vec{x} = 0 $$
the result for an external electric field $\vec{E}=E(z)\,\hat z$ becomes
$$ W = -\frac{1}{6}\,Q^{33}_S\,\frac{d\,E}{d\,z} = -
    \frac{q\,b^2}{10}\,\frac{d\,E}{d\,z}\ .  $$
\bigskip \eject

