X , X + ∂ X ∂ x . d x + ∂ X ∂ y . d y , X + ∂ X ∂ x . δ x + ∂ X ∂ y . δ y , Y , Y + ∂ Y ∂ x . d x + ∂ Y ∂ y . d y , Y + ∂ Y ∂ x . δ x + ∂ Y ∂ y . δ y , {\displaystyle {\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
2 Z d Σ = − ( ∂ X ∂ x . d x + ∂ X ∂ y . d y ) ( ∂ Y ∂ x . δ x + ∂ Y ∂ y . δ y ) = − ( ∂ X ∂ x . δ x + ∂ X ∂ y . δ y ) ( ∂ Y ∂ x . d x + ∂ Y ∂ y . d y ) = − ( ∂ X ∂ x . ∂ Y ∂ y − ∂ X ∂ y . ∂ Y ∂ x ) ( d x . δ y − d y . δ x ) . {\displaystyle {\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,
= ∂ X ∂ x . ∂ Y ∂ y − ∂ X ∂ y . ∂ Y ∂ x = ω . {\displaystyle ={\frac {\partial X}{\partial x}}.{\frac {\partial Y}{\partial y}}-{\frac {\partial X}{\partial y}}.{\frac {\partial Y}{\partial x}}=\omega .}
Since
X = − t Z , Y = − u Z , ( 1 + t 2 + u 2 ) Z 2 = 1 , {\displaystyle {\begin{array}{c}X=-tZ,\qquad Y=-uZ,\\(1+t^{2}+u^{2})Z^{2}=1,\end{array}}}
we have
d X = − Z 3 ( 1 + u 2 ) d t + Z 3 t u . d u , d Y = + Z 3 t u . d t − Z 3 ( 1 + t 2 ) d u , {\displaystyle {\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
∂ X ∂ x = Z 3 { − ( 1 + u 2 ) T + t u U } , ∂ Y ∂ x = Z 3 { t u T − ( 1 + t 2 ) U } , ∂ X ∂ y = Z 3 { − ( 1 + u 2 ) U + t u V } , ∂ Y ∂ y = Z 3 { t u U − ( 1 + t 2 ) V } , {\displaystyle {\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
ω = Z 6 ( T V − U 2 ) ( ( 1 + t 2 ) ( 1 + u 2 ) − t 2 u 2 ) = Z 6 ( T V − U 2 ) ( 1 + t 2 + u 2 ) = Z 4 ( T V − U 2 ) = T V − U 2 ( 1 + t 2 + u 2 ) 2 , {\displaystyle {\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