Integrate the given function over the accompanying surface in Exercises 27–34.
$f(x,y,z)=e^{\sqrt{8y^2-4z+1}}{(4z+8x^2+1)}^{-\frac{1}{2}}$, where S is the portion of the saddle surface with equation $z=y^2-x^2$ that lies above the region in the xy-plane bounded by the line with equation y = x in the first quadrant, the line with equation y = −x in the second quadrant, and the unit circle.

The function:

$$\begin{gathered} f(x,y,z)=e^{\sqrt{8y^2-4z+1}}{(4z+8x^2+1)}^{-\frac{1}{2}} \\ z=y^2-x^2 \end{gathered}$$

The formula for finding the integral over the smooth surface is given by,

${\int }_{S}dS=\int {\int }_{D}f\left(x\right(u,v),y(u,v),z(u,v\left)\right)||r_u\times r_v||dudv$

The surface is parametrized as follows:

$$\begin{gathered} r(u,v)=(u,v,v^2-u^2) \\ {r}_u=\frac{\partial r}{\partial u} \\ =(1,0,-2u) \\ {r}_v=\frac{\partial r}{\partial v} \\ =(0,1,2v) \end{gathered}$$

$$\begin{gathered} r_u\times r_v=(1,0,-2u)\times (0,1,2v) \\ =2ui-2vj+1k \\ ||r_u\times r_v||=\sqrt{{\left(2u\right)}^2+{(-2v)}^2+1^1} \\ =\sqrt{4(u^2+v^2)+1} \end{gathered}$$

$$\begin{gathered} f(x,y,z)=e^{\sqrt{8y^2-4z+1}}{(4z+8x^2+1)}^{-\frac{1}{2}} \\ f\left(x\right(u,v),y(u,v),z(u,v\left)\right)=e^{\sqrt{8v^2-4(v^2-u^2)+1}}(4{(v^2-u^2)+8u^2+1)}^{-\frac{1}{2}} \\ =e^{\sqrt{4v^2+4u^2+1}}{(4v^2+4u^2+1)}^{-\frac{1}{2}} \\ =\frac{e^{\sqrt{4(u^2+v^2)+1}}}{\sqrt{4(u^2+v^2)+1}} \end{gathered}$$

The region of integration is given by,

$$\begin{gathered} D=\left\{\right(r,\theta \left)\overline{)0\le r\le 1,\frac{\mathrm{\pi }}{4}\le \theta \le \frac{3\mathrm{\pi }}{4}}\right\} \\ {\int }_{S}dS=\int {\int }_{D}f\left(x\right(u,v),z(u,v\left)\right)||r_u\times r_v||dudv \\ =\int {\int }_D\frac{e^{\sqrt{4(u^2+v^2)+1}}}{\sqrt{4(u^2+v^2)+1}}\sqrt{4(u^2+v^2)+1}dudv \\ =\int {\int }_D{e}^{\sqrt{4(u^2+v^2)+1}}dA \end{gathered}$$

Evaluate the integral with polar coordinates.

$$\begin{gathered} u^2+v^2=r^2 \\ dA=rdrd\theta \end{gathered}$$

$$\begin{gathered} {\int }_{S}dS=\int {\int }_D{e}^{\sqrt{4(u^2+v^2)+1}}dA \\ ={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}{\int }_0^1{e}^{\sqrt{4r^2+1}}rdrd\theta \\ ={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}{\left[\frac{1}{4}\right(\sqrt{4r^2+1}-1\left]\right)e^{\sqrt{4r^2+1}}]}_0^{1}d\theta \\ ={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}\frac{1}{4}(\sqrt{5}-1)e^{\sqrt{5}}d\theta \\ =\frac{1}{4}(\sqrt{5}-1)e^{\sqrt{5}}{\left[\theta \right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}} \\ =\frac{1}{4}(\sqrt{5}-1)e^{\sqrt{5}}\left[\frac{\mathrm{\pi }}{2}\right] \\ =\frac{\mathrm{\pi }}{8}(\sqrt{5}-1)e^{\sqrt{5}} \end{gathered}$$