La valeur de
est facile à déterminer lorsqu’on connaît le rayon
du sphéroïde, car on a, par l’article I,
![{\displaystyle {\begin{aligned}\mathrm {A} =&\int _{0}^{\pi }\int _{0}^{\pi }\left[2\sin ^{3}p\sin ^{2}q(1+\alpha y)+\alpha (y'-y)\sin p\right]dpdq\\=&{\frac {4}{3}}\pi -{\frac {2}{3}}\alpha \pi y+\alpha \int _{0}^{\pi }\int _{0}^{\pi }y'\sin pdpdq,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68370b609c74ef2c98bdd894b47533cf67c976b9)
ce qui donne
![{\displaystyle \mathrm {D} =2\alpha \int _{0}^{\pi }\int _{0}^{\pi }y'\sin pdpdq.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0e64c3631c6b1b49d1cd08ef54ffe5bb0b7c6ff)
XXIII.
Si l’on suppose
et
les équations (6), (7) et (9) de l’article précédent, deviendront
![{\displaystyle {\begin{aligned}y\sin \theta =&-l{\frac {\partial .u\sin \theta }{\partial \theta }}-l{\frac {\partial v}{\partial \varpi }}\sin \theta ,\\{\frac {d^{2}u}{dt^{2}}}=&-g{\frac {\partial y}{\partial \theta }}+{\frac {\partial \mathrm {D} }{\partial \theta }}\Delta +{\frac {\partial \mathrm {R} }{\partial \theta }},\\{\frac {d^{2}v}{dt^{2}}}\sin ^{2}\theta =&-g{\frac {\partial y}{\partial \varpi }}+{\frac {\partial \mathrm {D} }{\partial \varpi }}\Delta +{\frac {\partial \mathrm {R} }{\partial \varpi }}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f68d3e62408d4df5c150ee010f81f86b45543909)
La seconde de ces équations donne
![{\displaystyle {\begin{aligned}-l{\frac {d^{2}}{dt^{2}}}{\frac {\partial .u\sin \theta }{\partial \theta }}=&lg{\frac {\partial ^{2}y}{\partial \theta ^{2}}}\sin \theta +lg{\frac {\partial y}{\partial \theta }}\cos \theta \\&-l{\frac {\partial ^{2}\mathrm {D} }{\partial \theta ^{2}}}\Delta \sin \theta -l{\frac {\partial \mathrm {D} }{\partial \theta }}\Delta \cos \theta \\&-l{\frac {\partial ^{2}\mathrm {R} }{\partial \theta ^{2}}}\sin \theta -l{\frac {\partial \mathrm {R} }{\partial \theta }}\cos \theta .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8471bee48b669afadf177f6f6420dd4bc45b2bd6)
La troisième donne
![{\displaystyle -l{\frac {d^{2}}{dt^{2}}}{\frac {\partial v}{\partial \varpi }}\sin \theta ={\frac {lg}{\sin \theta }}{\frac {\partial ^{2}y}{\partial \varpi ^{2}}}-{\frac {l}{\sin \theta }}{\frac {\partial ^{2}\mathrm {D} }{\partial \varpi ^{2}}}\Delta -{\frac {l}{\sin \theta }}{\frac {\partial ^{2}\mathrm {R} }{\partial \varpi ^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c9b18e66016ab542480ba4aaf8a6dc64d32fa62)