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

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

So for values of r < R the integral becomes

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

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