# Page:Thomson1881.djvu/16

the surface of a sphere vanishes, we may substitute in the integral $\tfrac{d^{2}}{dx^{2}}\tfrac{1}{r}$ for $\tfrac{d^{2}}{dx^{2}}\tfrac{1}{r'}$; then, transforming to polars, the integral

 :$=\sigma\int_{0}^{2\pi}\int_{0}^{\pi}\int_{R}^{\infty}4Q_{2}^{2}\frac{1}{r^{2}}\sin\theta d\phi\ dv\ dr$ $=\frac{16\pi\sigma}{5R}$;

for values of r < R,

$\frac{1}{r'}=\frac{1}{R}+\frac{rQ_{1}}{R^{2}}+\frac{rQ_{2}}{R^{3}}+\dots$.

Now $r^{n}Q_{n}$ is a solid harmonic of the nth order; hence $\tfrac{d^{2}}{dx^{2}}\left(r^{n}Q_{n}\right)$ is a solid harmonic of the (n-2)th order; and in particular $\tfrac{d^{2}}{dx^{2}}\left(r^{4}Q_{4}\right)$ is a solid harmonic of the second order; and, by the same reasoning as before, we may substitute in the integral $\tfrac{1}{R^{5}}\tfrac{d^{2}}{dx^{2}}\left(r^{4}Q_{4}\right)$ for $\tfrac{d^{2}}{dx^{2}}\tfrac{1}{r'}$. Now

 $r^{4}Q_{4}=\frac{35x^{4}-30x^{2}\left(x^{2}+y^{2}+z^{2}\right)+3\left(x^{2}+y^{2}+z^{2}\right)^{2}}{8}$; $\therefore\frac{d^{2}}{dx^{2}}\left(r^{4}Q_{4}\right)=12x^{2}-6\left(y^{2}+z^{2}\right)=12r^{2}Q_{2}$.

So for values of r < R the integral becomes

 $\frac{\sigma}{R^{5}}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}24Q_{2}^{2}r^{3}\sin\theta\ d\phi\ d\theta\ dr$ $=\frac{24\sigma\pi}{5R}$.

Adding this to the part of the integral for r > R, we get for the coefficient of uu' ,$\tfrac{8\sigma\pi}{R}$. The coefficients of uv' and uw' vanish by inspection.

The coefficient of vv'

$=\sigma\int\int\int r^{2}\frac{d^{2}}{dx\ dy}\frac{1}{r}\frac{d^{2}}{dx\ dy}\frac{1}{r'}dx\ dy\ dz$.

Now when r > R we may, by the same reasoning as before, substitute $\tfrac{d^{2}}{dx\ dy}\tfrac{1}{r}$ for $\tfrac{d^{2}}{dx\ dy}\tfrac{1}{r'}$, in the integral, and it becomes

$\sigma\int\int\int\frac{9r^{2}x^{2}y^{2}}{r^{10}}dx\ dy\ dz$,