Page:Philosophical Transactions of the Royal Society A - Volume 184.djvu/45

Rh Section I.—Preliminary Analysis.

§ 1. Take a fine ring in the plane of $$x$$, $$y$$, centre at the origin, and consisting of attracting matter of density $$k\cos n\phi^\prime$$, where $$\phi^\prime$$ is the azimuth of any point.

The potential of this ring at a point $$\mathrm{P}$$, whose coordinates are $$r, \theta, \phi$$, is

$\int_0^{2\pi}\frac{k\cos n\phi^\prime\;cd\phi^\prime}{\sqrt\left\{r^2 + c^2 - 2cr\sin\theta\cos\left(\phi - \phi^\prime\right)\right\}}$.|undefined

Put $$\phi^\prime = \phi + \psi$$, then the potential

$= \int_0^{2\pi}\frac{ck\left(\cos n\phi\;\cos n\psi - \sin n\phi\;\sin n\psi\right)\;d\psi}{\sqrt{\left\{r^2 + c^2 - 2cr\sin\theta\;\cos\psi\right\}}}$|undefined

$= \cos n\phi\int_0^{2\pi}\frac{ck\cos n\psi\;d\psi}{\sqrt{\left\{r^2 + c^2 - 2cr\sin\theta\;\cos\psi\right\}}}$|undefined

as the second integral vanishes between the limits.

Therefore

$\cos n\phi\int_0^{\pi}\frac{\cos n\phi\;d\phi}{\sqrt{\left(r^2 + c^2 - 2cr\sin\theta\;\cos\phi\right)}}$|undefined

or

$\cos n\phi\int_0^{\pi}\frac{\cos n\phi\;d\phi}{\sqrt{\left(z^2 + c^2 - 2c\varpi\cos\phi + \varpi^2\right)}}$|undefined

is a solution of 's equation.