Page:Thomson1881.djvu/17

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or transforming to polars,

9\sigma\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\frac{\sin^{2}\theta\cos^{2}\theta\cos^{2}\phi}{r^{2}}\sin\theta\ d\phi\ d\theta\ dr

=\frac{36\pi\sigma}{15R}.

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

\frac{d^{2}}{dx\ dy}\left(r^{4}Q_{4}\right)=-12xy;

making this substitution, the integral becomes

-\frac{\sigma}{R^{5}}\int\int\int\frac{36x^{2}y^{2}}{r^{5}}dx\ dy\ dz

=-\frac{\sigma}{R^{5}}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}36r^{3}\sin^{3}\theta\cos^{2}\theta\cos^{2}\phi\ d\phi\ d\theta\ dr

=-\frac{36\pi\sigma}{15R}.

Hence, adding this to the part previously obtained for values of r > R, we see that the coefficient of vv' from F\tfrac{df}{dt} is zero, and, similarly, the coefficient of ww' from this part of the integral vanishes.

Let us now take the terms arising from \int\int\int G\tfrac{dg}{dt}dx dy dz, and take, as before, the part arising from the product of that part of G due to e with the part of \tfrac{dg}{dt} due to e'. The coefficient of uu' in this part will be the same as the coefficient of vv' in the former part, and so will vanish.

The coefficient of vv'

=\sigma\int\int\int\left(r^{2}\frac{d^{2}}{dy^{2}}\frac{1}{r}+\frac{4}{r}\right)\frac{d^{2}}{dx^{2}}\frac{1}{r'}dx\ dy\ dz.

Now for values of r > R we may, as before, substitute \tfrac{d^{2}}{dy^{2}}\tfrac{1}{r} for \tfrac{d^{2}}{dy^{2}}\tfrac{1}{r'}; and it becomes

=\sigma\int\int\int r^{2}\left(\frac{3y^{2}-r}{r}\right)^{2}dx\ dy\ dz.

By transforming to polars, as before, this may be shown to be