Page:Thomson1881.djvu/18

 $$\tfrac{16\sigma\pi}{5}$$. For values of r R, we get for the coefficient of vv',

$$\frac{5\sigma\pi}{R}$$.

As before, the coefficients of uv', vu', uw', &c. disappear by inspection.

The coefficient of ww' 

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

substituting, for values of r > R, as before $$\tfrac{d^{2}}{dy\ dz}\tfrac{1}{r}$$ for $$\tfrac{d^{2}}{dy\ dz}\tfrac{1}{r'}$$ for in the integral, it becomes

$$\sigma\int\int\int\frac{9y^{2}z^{2}}{r^{8}}dx\ dy\ dz$$,

which, by transforming to polars, may be shown to be $$\tfrac{12\sigma\pi}{5R}$$. For values of r < R we may, as before, substitute $$\tfrac{1}{R^{5}}\tfrac{d^{2}}{dy\ dz}\left(r^{4}Q_{4}\right)$$ for $$\tfrac{d^{2}}{dy\ dz}\tfrac{1}{r'}$$ in the integral. Now

$$\frac{d^{2}}{dy\ dz}\left(r^{4}Q_{4}\right)=3yz$$.

On making this substitution, the integral

$$=\frac{\sigma}{R^{5}}\int\int\int\frac{9y^{2}z^{2}}{r^{3}}dx\ dy\ dz=\frac{3\sigma\pi}{5R}$$.

Adding this to the part obtained before, we get for the coefficient of ww',

$$\frac{12\sigma\pi}{5R}+\frac{3\sigma\pi}{5R}$$, or $$3\sigma\pi$$.

From the part of $$\int\int\int H\tfrac{dh}{dt}dx\ dy\ dz$$ which arises from that part of H due to e and that part of $$\tfrac{dh}{dt}$$ due to e', we can see,