Page:SearleEllipsoid.djvu/11

Now in fig. 4 let the ellipsoid PQ be determined by $h$, and the ellipsoid RS by $h+dh$.

Let the angular coordinate of P and S be $\phi$, and let that of Q and R be $\phi +d\phi$. Then the area PQRS

${\begin{aligned}&{=}{\frac {d(x,\rho )}{d(h,\phi )}}dh\ d\phi {=}\left({\frac {dx}{dh}}{\frac {d\rho }{d\phi }}-{\frac {dx}{d\phi }}{\frac {d\rho }{dh}}\right)dh\ d\phi \\&{=}{\frac {h^{2}-l^{2}\cos ^{2}\phi }{{\sqrt {\alpha }}{\sqrt {h^{2}-l^{2}}}}}dh\ d\phi \end{aligned}}$.

Now if the area PQRS revolve about the axis O$x$ the volume of the ring traced out is

$2\pi \rho {\frac {d(x,\rho )}{d(h,\phi )}}dh\ d\phi {=}{\frac {2\pi (h^{2}-l^{2}\cos ^{2}\phi )\sin \phi }{\alpha }}dh\ d\phi$.

Thus for the magnetic part of the energy we have

${\begin{aligned}\mathrm {\mathrm {T} } &{=}\int \int {\frac {\mu \mathrm {H} ^{2}}{8\pi }}2\pi \rho {\frac {d(x,\rho )}{d(h,\phi )}}dh\ d\phi \\&{=}{\frac {\mu q^{2}u^{2}}{4}}\int \int {\frac {h^{2}\sin ^{3}\phi \ dh\ d\phi }{(h^{2}-l^{2})(h^{2}-l^{2}\cos ^{2}\phi )}}\end{aligned}}$

Since $\lambda$ goes from 0 to $\infty$, $h$ goes from $a$ to $\infty$. The limits of $\phi$ are 0 and $\pi$.

Now

${\begin{aligned}\int _{0}^{\pi }{\frac {\sin ^{2}\phi \ d\phi }{h^{2}-l^{2}\cos ^{2}\phi }}&{=}-{\frac {1}{l^{2}}}\left[\cos \phi -{\frac {h^{2}-l^{2}}{2hl}}\log {\frac {h+l\ \cos \phi }{h-l\ \cos \phi }}\right]_{0}^{\pi }\\&{=}{\frac {1}{l^{2}}}\left\{2-{\frac {h^{2}-l^{2}}{hl}}\log {\frac {h+l}{h-l}}\right\}\end{aligned}}$.

Hence

${\begin{aligned}\mathrm {T} &{=}{\frac {\mu q^{2}u^{2}}{4l^{2}}}\int _{a}^{\infty }\left({\frac {2h^{2}}{h^{2}-l^{2}}}-{\frac {h}{l}}\log {\frac {h+l}{h-l}}\right)dh\\&{=}{\frac {\mu q^{2}u^{2}}{4l^{2}}}\left[h-{\frac {h^{2}-l^{2}}{2l}}\log {\frac {h+l}{h-l}}\right]\end{aligned}}$