91Ó°ÊÓ

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,

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

The surface is parametrized as follows:

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

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

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

The region of integration is given by,

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

Evaluate the integral with polar coordinates.

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

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