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

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

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

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

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

Now

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

Hence

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