Page:Thomson1881.djvu/18

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

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

\therefore the integral

=\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 \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',

\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,