Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/115

Hence we have

$$d\phi ={\frac {\zeta Y\,d\xi -\zeta X\,d\eta -(Y\xi -X\eta )\,d\zeta +\xi \zeta \,dY-\eta \zeta \,dX}{(Y\xi -X\eta )^{2}+\zeta ^{2}}}.$$

The denominator of this expression is

$${\begin{aligned}&=Y^{2}\xi ^{2}-2XY\xi \eta -X^{2}\eta ^{2}+\zeta ^{2}\\&=-(X\xi +Y\eta )^{2}+(X^{2}+Y^{2})(\xi ^{2}+\eta ^{2})+\zeta ^{2}\\&=-Z^{2}\zeta ^{2}+(1-Z^{2})(1-\zeta ^{2})+\zeta ^{2}\\&=1-Z^{2},\end{aligned}}$$

or

$$d\phi ={\frac {\zeta Y\,d\xi -\zeta X\,d\eta +(X\eta -Y\xi )\,d\zeta -\eta \zeta \,dX+\xi \zeta \,dY}{1-Z^{2}}}.$$

We verify readily by expansion the identical equation

$${\begin{array}{c}\eta \zeta (X^{2}+Y^{2}+Z^{2})+YZ(\xi ^{2}+\eta ^{2}+\zeta ^{2})\\=(X\xi +Y\eta +Z\zeta )(Z\eta +Y\zeta )+(X\zeta -Z\xi )(X\eta -Y\xi )\end{array}}$$

and likewise

$${\begin{array}{c}\xi \zeta (X^{2}+Y^{2}+Z^{2})+XZ(\xi ^{2}+\eta ^{2}+\zeta ^{2})\\=(X\xi +Y\eta +Z\zeta )(X\zeta +Z\xi )+(Y\xi -X\eta )(Y\zeta -Z\eta ).\end{array}}$$

We have, therefore,

$${\begin{aligned}\eta \zeta &=-YZ+(X\zeta -Z\xi )(X\eta -Y\xi ),\\\xi \zeta &=-XZ+(Y\xi -X\eta )(Y\zeta -Z\eta ).\end{aligned}}$$

Substituting these values, we obtain

$$d\phi ={\frac {Z}{1-Z^{2}}}(Y\,dX-X\,dY)+{\frac {\zeta Y\,d\xi -\zeta X\,d\eta }{1-Z^{2}}}$$

$$+{\frac {X\eta -Y\xi }{1-Z^{2}}}{\bigl \{}d\zeta -(X\zeta -Z\xi )\,dX-(Y\zeta -Z\eta )\,dY{\bigr \}}.$$

Now

$${\begin{alignedat}{4}&X\,dX&&+Y\,dY&&+Z\,dZ&&=0,\\&\xi \,dX&&+\eta \,dY&&+\zeta \,dZ&&=-X\,d\xi -Y\,d\eta -Z\,d\zeta .\end{alignedat}}$$

On substituting we obtain, instead of what stands in the parenthesis,

$$d\zeta -Z(X\,d\xi +Y\,d\eta +Z\,d\zeta ).$$

Hence

$$d\phi ={\frac {Z}{1-Z^{2}}}(Y\,dX-X\,dY)+{\frac {d\xi }{1-Z^{2}}}\{\zeta Y-\eta X^{2}Z+\xi XYZ\}$$

$${\begin{aligned}&-{\frac {d\eta }{1-Z^{2}}}\{\zeta X+\eta XYZ-\xi Y^{2}Z\}\\&+d\zeta (\eta X-\xi Y).\end{aligned}}$$