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Consider the given question,

$$\begin{gathered} \mathbf{F}(x,y,z)=鈭�y\mathbf{i}+x\mathbf{j}鈭�e^{yz}\mathbf{k}\mathbf{,} \\ {x}^2+y^2=9 \end{gathered}$$

If a surface S is parametrized by $\mathit{r}\left(u,v\right)$for $\left(u,v\right)鈭�D$, then the Flux of $\mathit{F}\left(x,y,z\right)$ through S is given below,

${鈭�}_S鈥�\mathbf{F}(x,y,z)鈰�\mathbf{n}dS={鈭�}_D鈥�\left(\mathbf{F}\right(x,y,z)鈰�\mathbf{n})鈭�{\mathbf{r}}_u脳{\mathbf{r}}_v鈭�dA......\left(i\right)$

Here, the surface S is the cylinder, so this surface parametrized by as follows, $\mathbf{r}(u,v)=鉄�3\cos 鈦�u,3\sin 鈦�u,v鉄�$.

Where, $D=\left\{\right(u,v)鈭�0鈮�u鈮�2蟺,2鈮�v鈮�4\}$.

Now, find $r_u,r_v$,

$\begin{array}{r}{\mathbf{r}}_u=\frac{\mathrm{鈭�}}{\mathrm{鈭�}u}\mathbf{r}(u,v)\\ =\frac{\mathrm{鈭�}}{\mathrm{鈭�}u}鉄�3\cos u,3\sin u,v鉄�\\ =鉄�鈭�3\sin u,3\cos u,0鉄�\\ \begin{array}{r}{\mathbf{r}}_v=\frac{\mathrm{鈭�}}{\mathrm{鈭�}v}\mathbf{r}(u,v)\\ =\frac{\mathrm{鈭�}}{\mathrm{鈭�}v}鉄�3\cos u,3\sin u,v鉄�\\ =鉄�0,0,1鉄�.\end{array}\end{array}$

Find $||r_u脳r_v||$,

$$\begin{gathered} {\mathbf{r}}_{\mathrm{u}}脳{\mathbf{r}}_{\mathrm{v}}=鉄�鈭�3\sin \mathrm{u},3\cos \mathrm{u},0鉄�脳鉄�0,0,1鉄� \\ =\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 鈭�3\sin \mathrm{u}& 3\cos \mathrm{u}& 0\\ 0& 0& 1\end{array}\right| \\ =\left[\right(3\cos \mathrm{u}\left)\right(0)鈭�0鈰�0]\mathbf{i}鈭�\left[\right(鈭�3\sin \mathrm{u}\left)\right(1)鈭�0鈰�0]\mathbf{j}+\left[\right(鈭�3\sin \mathrm{u}\left)\right(0)鈭�(3\cos \mathrm{u}\left)\right(0\left)\right]\mathbf{k} \\ =0\mathbf{i}+3\sin \mathrm{u}\mathbf{j}+0\mathbf{k} \\ =鉄�0,3\sin \mathrm{u},0鉄� \end{gathered}$$

Then, $\begin{array}{r}鈭�{\mathbf{r}}_u脳{\mathbf{r}}_v鈭�=鈭�鉄�0,3\sin u,0鉄�鈭�\\ =\sqrt{0^2+(3\sin u{)}^2+0^2}\\ =3\sin u\end{array}$

The choice of n should have pointing upwards. The desired normal vector will be,

$\begin{array}{r}\mathbf{n}=\frac{{\mathbf{r}}_u脳{\mathbf{r}}_v}{鈭�{\mathbf{r}}_w脳{\mathbf{r}}_v鈭�}\\ =\frac{鉄�0,3\sin u,0鉄�}{3\sin u}\end{array}$

For the parametrization $\mathbf{r}(u,v)=鉄�3\cos u,3\sin u,v鉄�$, the following vector field will be,

Then, the value of $\mathit{F}\left(x,y,z\right)\mathbf{.}\mathit{n}$will be,

Substitute the values in equation (i), with the following region of integration, $D=\left\{\right(u,v)鈭�0鈮�u鈮�2蟺,2鈮�v鈮�4\}$.

Then, the flux of the vector field through the surface Sis given below,

role="math" localid="1650346265761" $\begin{array}{r}{鈭�}_S鈥�\mathbf{F}(x,y,z)鈰�\mathbf{n}dS={鈭�}_D鈥�\left(\mathbf{F}\right(x,y,z)鈰�\mathbf{n})鈭�{\mathbf{r}}_u脳{\mathbf{r}}_v鈭�dA\\ ={鈭�}_0^{2蟺}鈥�{鈭�}_2^4鈥�(3\cos u)(3\sin u)dvdu\end{array}$

On solving the above equation,

$\begin{array}{r}{鈭�}_S鈥�\mathbf{F}(x,y,z)鈰�\mathbf{n}dS={鈭�}_0^{2蟺}鈥�(9\sin u\cos u)[v{]}_2^{4}du\\ \begin{array}{r}={鈭�}_0^{2蟺}鈥�18\sin u\cos udu\\ ={\left[9{\sin }^{2}u\right]}_0^{2蟺}\\ =9{\sin }^{2}2蟺鈭�9{\sin }^{2}0\\ =9(0{)}^2鈭�9(0{)}^2\\ =0\end{array}\end{array}$