91影视

The function:

$$\begin{gathered} f(x,y,z)=x-z+y^2 \\ r(u,v)=(u+v)i+2\sqrt{u^2+v^2}j+(u-v)k \end{gathered}$$

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

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

The surface is parametrized as follows:

$$\begin{gathered} r(u,v)=(u+v)i+2\sqrt{u^2+v^2}j+(u-v)k \\ {r}_u=\frac{鈭�r}{鈭�u} \\ =(1,\frac{2u}{\sqrt{u^2+v^2}},1) \\ {r}_v=\frac{鈭�r}{鈭�v} \\ =(1,\frac{2v}{\sqrt{u^2+v^2}},-1) \end{gathered}$$

$$\begin{gathered} r_u脳r_v=(1,\frac{2u}{\sqrt{u^2+v^2}},1)脳(1,\frac{2v}{\sqrt{u^2+v^2}},-1) \\ =-\frac{2(u+v)}{\sqrt{u^2+v^2}}i+2j+\frac{2(v-u)}{\sqrt{u^2+v^2}}k \\ ||r_u脳r_v||=\sqrt{{(-\frac{2(u+v)}{\sqrt{u^2+v^2}})}^2+2^2+{\left(\frac{2(v-u)}{\sqrt{u^2+v^2}}\right)}^2} \\ =\sqrt{4(\frac{{(u+v)}^2}{u^2+v^2}+1+\frac{{(v-u)}^2}{u^2+v^2})} \\ =\sqrt{4\left(\frac{3(u^2+v^2)}{u^2+v^2}\right)} \\ =\sqrt{12} \\ =2\sqrt{3} \end{gathered}$$

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

The region of integration is given by,

$$\begin{gathered} D=\left\{\right(u,v\left)\overline{)0鈮�u鈮�1,u^2鈮�v鈮�u}\right\} \\ {鈭�}_{S}dS=鈭�{鈭�}_{D}f\left(x\right(u,v),z(u,v\left)\right)||r_u脳r_v||dudv \\ ={鈭�}_0^1{鈭�}_{u^2}^u(4u^2+4v^2+2v)2\sqrt{3}dvdu \\ =2\sqrt{3}{鈭�}_0^1{[4u^{2}v+\frac{4v^3}{3}+v^2]}_{u^2}^{u}du \\ =2\sqrt{3}{鈭�}_0^1[-\frac{4}{3}{u}^6-5u^4+\frac{16}{3}{u}^3+u^2]du \\ =2\sqrt{3}{[-\frac{4}{21}{u}^7-u^5+\frac{4u^4}{3}+\frac{u^3}{3}]}_0^1 \\ =2\sqrt{3}\left[\frac{10}{21}\right] \\ =\frac{20\sqrt{3}}{21} \end{gathered}$$