Pagina:Werkecarlf03gausrich.djvu/365

15.

Nullo negotio perspicitur, integralia

${\begin{aligned}&\int {\frac {\cos T\operatorname {d} T}{(G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\\&\int {\frac {\sin T\operatorname {d} T}{(G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\\&\int {\frac {\cos T\sin T\operatorname {d} T}{(G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\end{aligned}}$

a $T=180^{\circ }$ usque ad $T=360^{\circ }$ extensa obtinere valores aequales iis, quos nanciscantur, si a $T=0$ usque ad $T=180^{\circ }$ extendantur, sed signis oppositis affectos; quapropter ista integralia a $T=0$ usque ad $T=360^{\circ }$ extensa manifesto fiunt $=0.$ Hinc colligimus, esse

${\begin{aligned}&\xi =\int {\frac {((\gamma -e\alpha )\alpha G-(\gamma ^{\prime }-e\alpha ^{\prime })\alpha ^{\prime }G^{\prime }{\cos T}^{2}-(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\alpha ^{\prime \prime }G^{\prime \prime }{\sin T}^{2})\operatorname {d} T}{2\pi a(G+G^{\prime }{cosT}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\\&\eta =\int {\frac {((\gamma -e\alpha )\beta G-(\gamma ^{\prime }-e\alpha ^{\prime })\beta ^{\prime }G^{\prime }{\cos T}^{2}-(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\beta ^{\prime \prime }G^{\prime \prime }{\sin T}^{2})\operatorname {d} T}{2\pi b(G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\\&\zeta =\int {\frac {((\gamma -e\alpha )\gamma +(\gamma ^{\prime }-e\alpha ^{\prime })\gamma ^{\prime }{\cos T}^{2}+(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\gamma ^{\prime \prime }{\sin T}^{2})C\operatorname {d} T}{2\pi (G+G^{\prime }{\cos T}^{2}+G^{\prime \prime }{\sin T}^{2})^{\frac {3}{2}}}}\end{aligned}}$

integralibus a $T=0$ usque ad $T=360^{\circ }$ extensis. Quodsi itaque valores integralium, eadem extensione acceptorum,

${\begin{aligned}&\int {\frac {{\cos T}^{2}\operatorname {d} T}{2\pi ((G+G^{\prime }){\cos T}^{2}+(G+G^{\prime \prime }){\sin T}^{2})^{\frac {3}{2}}}}\\&\int {\frac {{\sin T}^{2}\operatorname {d} T}{2\pi ((G+G^{\prime }){\cos T}^{2}+(G+G^{\prime \prime }){\sin T}^{2})^{\frac {3}{2}}}}\end{aligned}}$

per $P,$ $Q$ denotamus, erit

${\begin{alignedat}{8}a\xi &=((\gamma -e\alpha )\alpha G&&-(\gamma ^{\prime }-e\alpha ^{\prime })\alpha ^{\prime }G^{\prime })P&&+((\gamma -e\alpha )\alpha G&&-(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\alpha ^{\prime \prime }G^{\prime \prime })Q\\b\eta &=((\gamma -e\alpha )\beta G&&-(\gamma ^{\prime }-e\alpha ^{\prime })\beta ^{\prime }G^{\prime })P&&+((\gamma -e\alpha )\beta G&&-(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\beta ^{\prime \prime }G^{\prime \prime })Q\\\zeta &=((\gamma -e\alpha )\gamma &&+(\gamma ^{\prime }-e\alpha ^{\prime })\gamma ^{\prime })CP&&+((\gamma -e\alpha )\gamma &&+(\gamma ^{\prime \prime }-e\alpha ^{\prime \prime })\gamma ^{\prime \prime })CQ\end{alignedat}}$

quo pacto problema nostrum complete solutum est.