Dans la question présente, nous avons
![{\displaystyle \varepsilon =-2lc,\qquad r=0,\qquad f=6l\qquad {\text{et}}\qquad p=6l\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80b150c2bdf6d56f4e28a43aad2f0b833f4199ac)
partant,
![{\displaystyle {\begin{aligned}\mathrm {C} \Delta =&\left[b\sin(\varphi -\varpi )-a\cos(\varphi -\varpi )\right]{\frac {24}{5}}\pi \Delta l\cos \theta \\&+\left[b'\sin(2\varphi -2\varpi )-a'\cos(2\varphi -2\varpi )\right]{\frac {48}{5}}\pi \Delta l\sin \theta \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38eb48b6633df01fad5cdf1bf14ced4446d14a61)
ou
![{\displaystyle \mathrm {C} \Delta ={\frac {24}{5}}\pi \Delta l\left(\cos \theta {\frac {\partial z}{\partial \varpi }}+\sin \theta {\frac {\partial x}{\partial \varpi }}\right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4eb6fd60f564780e5afdc514e46c557f0aed0c1)
donc
![{\displaystyle \alpha \mathrm {C} \Delta ={\frac {24}{5}}\pi \alpha \Delta lv\sin \theta \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/928583e7357016e6830e5171b77c6d91d48dc616)
on aura pareillement
![{\displaystyle \mathrm {B} \Delta ={\frac {24}{5}}\pi \Delta lc\sin 2\theta +\int _{0}^{\pi }{\frac {\partial .\mathrm {C} \sin \theta }{\partial \theta }}d\varpi \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99c421e2e75b1d0d7c658284141c1eb83360a8f1)
d’où l’on tirera facilement
![{\displaystyle \alpha \mathrm {B} \Delta ={\frac {24}{5}}\alpha \pi \Delta lu\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dda3b77db2d20d990ceb2315c2249a79b500a06a)
les équations (12) et (13) deviendront ainsi
![{\displaystyle {\begin{aligned}{\frac {d^{2}u}{dt^{2}}}=&lg{\frac {\partial ^{2}u}{\partial \theta ^{2}}}+lg{\frac {\partial ^{2}v}{\partial \varpi \partial \theta }}+lg{\frac {\partial {\cfrac {u\cos \theta }{\sin \theta }}}{\partial \theta }}+{\frac {24}{5}}\pi \Delta lu\\&+\mathrm {K} \left\{\sin 2\theta \left[{\frac {1}{2}}\sin ^{2}\nu -\cos ^{2}\nu +{\frac {1}{2}}\sin ^{2}\nu \cos(2\varphi -2\varpi )\right]\right.\\&\qquad \qquad \qquad \qquad \qquad \qquad \left.+2\cos 2\theta \sin \nu \cos \nu \cos(\varphi -\varpi ){\begin{aligned}\\\\\end{aligned}}\right\}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6076f2d55510efdb078af025e23d40e4a868f075)
![{\displaystyle {\begin{aligned}{\frac {d^{2}v}{dt^{2}}}\sin ^{2}\theta =&lg{\frac {\partial ^{2}v}{\partial \varpi ^{2}}}+lg{\frac {\partial ^{2}u}{\partial \varpi \partial \theta }}+lg{\frac {\cos \theta }{\sin \theta }}{\frac {\partial u}{\partial \varpi }}\\&\qquad +{\frac {24}{5}}\pi \Delta lv\sin ^{2}\theta +\mathrm {K} \sin \nu \cos \nu \sin 2\theta \sin(\varphi -\varpi )\\&\qquad \qquad \qquad +\mathrm {K} \sin ^{2}\nu \sin ^{2}\theta \sin(2\varphi -2\varpi )\,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be16d5949f78fdc1634747774b974e4242ff449c)
en intégrant ces équations par la méthode de l’article VII, on parviendra aux deux suivantes :
(19)
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(20)
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équations qui ne diffèrent des équations (16) et (17) qu’en ce que
se