${\tfrac {16\sigma \pi }{5}}$. 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,