Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/110

$${\begin{array}{c}X,\\X+{\frac {\partial X}{\partial x}}.dx+{\frac {\partial X}{\partial y}}.dy,\\X+{\frac {\partial X}{\partial x}}.\delta x+{\frac {\partial X}{\partial y}}.\delta y,\end{array}}\qquad {\begin{array}{c}Y,\\Y+{\frac {\partial Y}{\partial x}}.dx+{\frac {\partial Y}{\partial y}}.dy,\\Y+{\frac {\partial Y}{\partial x}}.\delta x+{\frac {\partial Y}{\partial y}}.\delta y,\end{array}}$$

From this the double area of the element is found to be

$${\begin{aligned}2Z\,d\Sigma &={\phantom {-}}\left({\frac {\partial X}{\partial x}}.dx+{\frac {\partial X}{\partial y}}.dy\right)\left({\frac {\partial Y}{\partial x}}.\delta x+{\frac {\partial Y}{\partial y}}.\delta y\right)\\&{\phantom {={}}}-\left({\frac {\partial X}{\partial x}}.\delta x+{\frac {\partial X}{\partial y}}.\delta y\right)\left({\frac {\partial Y}{\partial x}}.dx+{\frac {\partial Y}{\partial y}}.dy\right)\\&={\phantom {-}}\left({\frac {\partial X}{\partial x}}.{\frac {\partial Y}{\partial y}}-{\frac {\partial X}{\partial y}}.{\frac {\partial Y}{\partial x}}\right)(dx.\delta y-dy.\delta x).\end{aligned}}$$

The measure of curvature is, therefore,

$$={\frac {\partial X}{\partial x}}.{\frac {\partial Y}{\partial y}}-{\frac {\partial X}{\partial y}}.{\frac {\partial Y}{\partial x}}=\omega .$$

Since

$${\begin{array}{c}X=-tZ,\qquad Y=-uZ,\\(1+t^{2}+u^{2})Z^{2}=1,\end{array}}$$

we have

$${\begin{aligned}dX&=-Z^{3}(1+u^{2})\,dt+Z^{3}tu.du,\\dY&=+Z^{3}tu.dt-Z^{3}(1+t^{2})\,du,\end{aligned}}$$

therefore

$${\begin{alignedat}{2}{\frac {\partial X}{\partial x}}&=Z^{3}{\bigl \{}-(1+u^{2})T+tuU{\bigr \}},\qquad &{\frac {\partial Y}{\partial x}}&=Z^{3}{\bigl \{}tuT-(1+t^{2})U{\bigr \}},\\{\frac {\partial X}{\partial y}}&=Z^{3}{\bigl \{}-(1+u^{2})U+tuV{\bigr \}},\qquad &{\frac {\partial Y}{\partial y}}&=Z^{3}{\bigl \{}tuU-(1+t^{2})V{\bigr \}},\end{alignedat}}$$

and

$${\begin{aligned}\omega &=Z^{6}(TV-U^{2}){\bigl (}(1+t^{2})(1+u^{2})-t^{2}u^{2}{\bigr )}\\&=Z^{6}(TV-U^{2})(1+t^{2}+u^{2})\\&=Z^{4}(TV-U^{2})\\&={\frac {TV-U^{2}}{(1+t^{2}+u^{2})^{2}}},\end{aligned}}$$

the very same expression which we have found at the end of the preceding article. Therefore we see that