Page:Joseph Louis de Lagrange - Œuvres, Tome 12.djvu/292

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MÉCANIQUE ANALYTIQUE.

ces valeurs, substituées dans l’expression de $\operatorname {D} z,$ donneront

$\operatorname {D} z={\cfrac {{\cfrac {\partial z}{\partial a}}\left({\cfrac {\partial x}{\partial b}}{\cfrac {\partial y}{\partial c}}-{\cfrac {\partial x}{\partial c}}{\cfrac {\partial y}{\partial b}}\right)+{\cfrac {\partial z}{\partial b}}\left({\cfrac {\partial x}{\partial c}}{\cfrac {\partial y}{\partial a}}-{\cfrac {\partial x}{\partial a}}{\cfrac {\partial y}{\partial c}}\right)+{\cfrac {\partial z}{\partial c}}\left({\cfrac {\partial x}{\partial a}}{\cfrac {\partial y}{\partial b}}-{\cfrac {\partial y}{\partial a}}{\cfrac {\partial x}{\partial b}}\right)}{{\cfrac {\partial x}{\partial a}}{\cfrac {\partial y}{\partial b}}-{\cfrac {\partial y}{\partial a}}{\cfrac {\partial x}{\partial b}}}}dc.$

Pour avoir de même la valeur de $\operatorname {D} y,$ on supposera $\operatorname {D} x=0$ et $\operatorname {D} z=0,$ ce qui donne

$dc=0,\qquad {\frac {\partial x}{\partial a}}da+{\frac {\partial x}{\partial b}}db=0\,;$

d’où l’on tire

$da=-{\frac {\cfrac {\partial x}{\partial b}}{\cfrac {\partial x}{\partial a}}}db,$

et cette valeur ainsi que celle de $dc$ égale à $0,$ étant substituées dans l’expression de $\operatorname {D} y,$ donneront

$\operatorname {D} y={\cfrac {{\cfrac {\partial x}{\partial a}}{\cfrac {\partial y}{\partial b}}-{\cfrac {\partial x}{\partial b}}{\cfrac {\partial y}{\partial a}}}{\cfrac {\partial x}{\partial a}}}db.$

Enfin, pour avoir la valeur de $\operatorname {D} x,$ on fera $\operatorname {D} y=0,\ \operatorname {D} z=0,$ ce qui donne

$db=0,\quad dc=0,$

et par conséquent,

$\operatorname {D} x={\frac {\partial x}{\partial a}}da.$

Multipliant ensemble ces valeurs de $\operatorname {D} x,\,\operatorname {D} y,\,\operatorname {D} z,$ on aura

$\operatorname {D} x\operatorname {D} y\operatorname {D} z=$

$\left[{\frac {\partial z}{\partial a}}\left({\frac {\partial x}{\partial b}}{\frac {\partial y}{\partial c}}-{\frac {\partial x}{\partial c}}{\frac {\partial y}{\partial b}}\right)+{\frac {\partial z}{\partial b}}\left({\frac {\partial x}{\partial c}}{\frac {\partial y}{\partial a}}-{\frac {\partial x}{\partial a}}{\frac {\partial y}{\partial c}}\right)+{\frac {\partial z}{\partial c}}\left({\frac {\partial x}{\partial a}}{\frac {\partial y}{\partial b}}-{\frac {\partial y}{\partial a}}{\frac {\partial x}{\partial b}}\right)\right]da\,db\,dc.$

Faisant donc $\operatorname {D} x\operatorname {D} y\operatorname {D} z=dadbdc,$ on aura tout de suite l’équation (E).

Il est bon de remarquer que cette valeur de $\operatorname {D} x\operatorname {D} y\operatorname {D} z$ est celle qu’on doit employer dans les intégrales triples relatives à $x,\,y,\,z,$ lorsqu’on