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