# Page:Thomson1881.djvu/18

${\displaystyle {\tfrac {16\sigma \pi }{5}}}$. For values of r<R we may, as before, substitute ${\displaystyle {\tfrac {1}{R^{5}}}{\tfrac {d^{2}}{dy^{2}}}\left(r^{4}Q_{4}\right)}$ for ${\displaystyle {\tfrac {d^{2}}{dy^{2}}}{\tfrac {1}{r'}}}$ in the integral. Now

 ${\displaystyle {\frac {d^{2}}{dy^{2}}}\left(r^{4}Q_{4}\right)={\frac {36y^{2}+12z^{2}-48x^{2}}{8}}}$

${\displaystyle \therefore }$ the integral

${\displaystyle ={\frac {\sigma }{R^{5}}}\int \int \int {\frac {r^{2}\left(3y^{2}-r^{2}\right)\left(36y^{2}+12z^{2}-48x^{2}\right)}{8r^{5}}}dx\ dy\ dz}$.

By transforming to polars, this may be shown to be ${\displaystyle {\tfrac {9\pi \sigma }{5R}}}$. Adding this to the part of the integral due to values of r > R, we get for the coefficient of vv',

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

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

The coefficient of ww'

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

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

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

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

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

On making this substitution, the integral

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

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

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

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