Page:Mémoires de l’Académie des sciences, Tome 10.djvu/582

$\int _{0}^{\pi }\int _{0}^{2\pi }\operatorname {F} \left(\cos .\theta '\cos .\theta +\sin .\theta '\sin .\theta \cos .(\omega '-\omega )\right)\sin .\theta 'a'\theta 'd\omega '=$

$2\pi \int _{0}^{\pi }\operatorname {F} (\cos .u)\sin .udu,$

quelle que soit la fonction $\operatorname {F} .$ Il en résulte qu’on aura

${\begin{aligned}\int _{0}^{\pi }&\int _{0}^{2\pi }\cos .\left[\rho r'\left(\cos .\theta '\cos .\theta +\sin .\theta '\sin .\theta \cos .(\omega -\omega ')\right)\right.\\&\left.+\rho (x-2h)\cos .\theta -\rho y\sin .\theta \sin .\omega -\rho z\sin .\theta \cos .\omega \right]\sin .\theta 'd\theta 'd\omega \\&={\frac {4\pi }{\rho r'}}\sin .\rho r'\cos .\rho ((x-2h)\cos .\theta -y\sin .\theta \sin .\omega -z\sin .\theta \cos .\omega ,\\\int _{0}^{\pi }&\int _{0}^{2\pi }\cos .\left[\rho r'\left(\cos .\theta '\cos .\theta +\sin .\theta '\sin .\theta \cos .(\omega -\omega ')\right)\right.\\&-{\frac {\rho }{a'}}(x-h){\sqrt {a^{2}-a'^{2}\sin .^{2}\theta }}-\rho h\cos .\theta -\rho y\sin .\theta \sin .\omega \\-\rho &\left.z\sin .\theta \cos .\omega \right]\sin .\theta 'd\theta 'd\omega '={\frac {4\pi }{\rho r'}}\sin .\rho r'\cos .\rho ({\frac {x-h}{a'}}{\sqrt {a^{2}-a'^{2}\sin .^{2}\theta }}\\&\qquad \qquad \qquad \qquad \left.+h\cos .\theta +y\sin .\theta \sin .\omega +z\sin .\theta \cos .\omega \right).\end{aligned}}$

Cela étant, si nous faisons, pour abréger,

${\begin{aligned}&(2h-x)\cos .\theta +y\sin .\theta \sin .\omega +z\sin .\theta \cos .\omega \pm at=\varpi ,\\&{\frac {x-h}{a'}}{\sqrt {a^{2}-a'^{2}\sin .^{2}\theta }}+h\cos .\theta +y\sin .\theta \sin .\omega +z\sin .\theta \cos .\omega \pm at=\varpi ',\end{aligned}}$

les formules (b) deviendront

$\left.{\begin{alignedat}{2}\Pi \ =&{\frac {1}{2\pi ^{2}}}\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{{\frac {1}{2}}\pi }\int _{0}^{2\pi }&&{\frac {\left(a^{2}\cos .\theta -a'{\sqrt {a^{2}-a'^{2}\sin .^{2}\theta }}\right)\sin .\theta }{a^{2}\cos .\theta +a'{\sqrt {a^{2}-a'^{2}\sin .^{2}\theta }}}}\\&&&\times \rho r'\sin .\rho r'\cos .\rho \varpi \,fr'dr'd\rho \,d\theta \,d\omega ,\\\Pi '=&{\frac {1}{\pi ^{2}}}\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{{\frac {1}{2}}\pi }\int _{0}^{2\pi }&&{\frac {a'^{2}\cos .\theta \sin .\theta }{a^{2}\cos .\theta +a'{\sqrt {a^{2}-a'^{2}\sin .^{2}\theta }}}}\\&&&\times \rho r'\sin .\rho r'\cos .\rho \varpi \,fr'dr'd\rho \,d\theta \,d\omega .\end{alignedat}}\right\}$

(d)