91Ó°ÊÓ

Consider the given question,

If a surface S is parametrized by $r\left(u,v\right)$for $\left(u,v\right)∈D$, then the integral of $f\left(x,y,z\right)$ over a smooth surface S is given below,

${∫}_{S}f\left(x,y,z\right)dS=âˆ\neg {}_{D}f\left(x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)\right)||{\mathit{r}}_u×{\mathit{r}}_w||dA......\left(i\right)$

Now, find ${\mathit{r}}_{\mathbf{u}},{\mathit{r}}_v$,

role="math" localid="1650303986567" $$\begin{gathered} {\mathit{r}}_u=\frac{∂}{∂u}\mathit{r}\left(u,v\right) \\ =\frac{∂}{∂u}\left(2u\mathbf{i}+\left(\mathrm{u}+\mathrm{v}\right)\mathbf{j}+\left(2\mathrm{u}-2\mathrm{v}\right)\mathbf{k}\right) \\ =2\mathit{i}+\mathit{j}+2\mathit{k} \\ {r}_v=\frac{∂}{∂v}\mathit{r}\left(u,v\right) \\ =\frac{∂}{∂v}\left(2u\mathbf{i}+\left(\mathrm{u}+\mathrm{v}\right)\mathbf{j}+\left(2\mathrm{u}-2\mathrm{v}\right)\mathbf{k}\right) \\ =0\mathit{i}+\mathit{j}-2\mathit{k} \end{gathered}$$

On solving $||{\mathit{r}}_u×{\mathit{r}}_{\mathbf{v}}||$,

$$\begin{gathered} {\mathit{r}}_u×{\mathit{r}}_{\mathrm{v}}=\left(2\mathit{i}+\mathit{j}+2\mathit{k}\right)×\left(0\mathit{i}+\mathit{j}-2\mathit{k}\right) \\ =\left|\begin{array}{ccc}\mathit{i}& \mathit{j}& \mathit{k}\\ 2& 1& 2\\ 0& 1& -2\end{array}\right| \\ =-4\mathit{i}+4\mathit{j}+2\mathit{k} \end{gathered}$$

Then,$$\begin{gathered} ||{\mathit{r}}_u×{\mathit{r}}_{\mathrm{v}}||=||-4\mathbf{i}+4\mathbf{j}+2\mathbf{k}|| \\ =\sqrt{{\left(-4\right)}^2+4^2+2^2} \\ =\sqrt{36} \\ =6 \end{gathered}$$

Consider the following function, localid="1650304882776" $f\left(x,y,z\right)$,

localid="1650304885354" $$\begin{gathered} f\left(x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)\right)=\frac{1}{2u\ln \left(\left(2u-2v\right)+2\left(u+v\right)\right)} \\ =\frac{1}{2u\ln \left(4u\right)} \end{gathered}$$

Substitute the values in equation (i),

$D=\left\{\left(u,v\right)\overline{)3≤u≤7,2≤u≤10}\right\}$

Take the integral of f over the surface S,

$\begin{array}{r}{∫}_S f(x,y,z)dS={âˆ\neg }_D f\left(x\right(u,v),y(u,v),z(u,v\left)\right)∥{\mathbf{r}}_u×{\mathbf{r}}_v∥dA\\ ={∫}_2^{10} {∫}_3^7 \left[\frac{1}{2u\ln \left(4u\right)}\right)\left(6\right)dudv\\ =3{∫}_2^{10} {∫}_3^7 \frac{1}{u\ln \left(4u\right)}dudv\\ =3{∫}_2^{10} \left[{∫}_3^7 \frac{1}{u\ln \left(4u\right)}du\right]dv\\ =3{∫}_2^{10} \left[\ln \right(\ln \left(4u\right)){]}_3^{7}dv\end{array}$

On solving the above equation,

$\begin{array}{r}{∫}_S f(x,y,z)dS=3{∫}_2^{10} \left[\ln \right(\ln \left(28\right))−\ln (\ln \left(12\right)\left)\right]dv\\ =3{∫}_2^{10} \ln \left(\frac{\ln \left(28\right)}{\ln \left(12\right)}\right)dv\\ =3\ln \left(\frac{\ln \left(28\right)}{\ln \left(12\right)}\right){∫}_2^{10} dv\\ =3\ln \left(\frac{\ln \left(28\right)}{\ln \left(12\right)}\right)[v{]}_2^{10}\\ =3\ln \left(\frac{\ln \left(28\right)}{\ln \left(12\right)}\right)[10−2]\\ =24\ln \left(\frac{\ln \left(28\right)}{\ln \left(12\right)}\right)\end{array}$