$\mathbf{F}(x,y,z)=-y\mathbf{i}+x\mathbf{j}-e^{yz}\mathbf{k}$, where S is the cylinder with equation $x^2+y^2=9$from $z=2,4$, with n pointing outwards.

Consider the given question,

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

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

${\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})∥{\mathbf{r}}_u\times {\mathbf{r}}_v∥dA......\left(i\right)$

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

Where, $D=\left\{\right(u,v)\mid 0\le u\le 2\pi ,2\le v\le 4\}$.

Now, find $r_u,r_v$,

$\begin{array}{r}{\mathbf{r}}_u=\frac{\mathrm{\partial }}{\mathrm{\partial }u}\mathbf{r}(u,v)\\ =\frac{\mathrm{\partial }}{\mathrm{\partial }u}⟨3\cos u,3\sin u,v⟩\\ =⟨-3\sin u,3\cos u,0⟩\\ \begin{array}{r}{\mathbf{r}}_v=\frac{\mathrm{\partial }}{\mathrm{\partial }v}\mathbf{r}(u,v)\\ =\frac{\mathrm{\partial }}{\mathrm{\partial }v}⟨3\cos u,3\sin u,v⟩\\ =⟨0,0,1⟩.\end{array}\end{array}$

Find $||r_u\times r_v||$,

$$\begin{gathered} {\mathbf{r}}_{\mathrm{u}}\times {\mathbf{r}}_{\mathrm{v}}=⟨-3\sin \mathrm{u},3\cos \mathrm{u},0⟩\times ⟨0,0,1⟩ \\ =\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -3\sin \mathrm{u}& 3\cos \mathrm{u}& 0\\ 0& 0& 1\end{array}\right| \\ =\left[\right(3\cos \mathrm{u}\left)\right(0)-0\cdot 0]\mathbf{i}-\left[\right(-3\sin \mathrm{u}\left)\right(1)-0\cdot 0]\mathbf{j}+\left[\right(-3\sin \mathrm{u}\left)\right(0)-(3\cos \mathrm{u}\left)\right(0\left)\right]\mathbf{k} \\ =0\mathbf{i}+3\sin \mathrm{u}\mathbf{j}+0\mathbf{k} \\ =⟨0,3\sin \mathrm{u},0⟩ \end{gathered}$$

Then, $\begin{array}{r}∥{\mathbf{r}}_u\times {\mathbf{r}}_v∥=\parallel ⟨0,3\sin u,0⟩\parallel \\ =\sqrt{0^2+(3\sin u{)}^2+0^2}\\ =3\sin u\end{array}$

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

$\begin{array}{r}\mathbf{n}=\frac{{\mathbf{r}}_u\times {\mathbf{r}}_v}{∥{\mathbf{r}}_w\times {\mathbf{r}}_v∥}\\ =\frac{⟨0,3\sin u,0⟩}{3\sin u}\end{array}$

For the parametrization $\mathbf{r}(u,v)=⟨3\cos u,3\sin u,v⟩$, 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, $D=\left\{\right(u,v)\mid 0\le u\le 2\pi ,2\le v\le 4\}$.

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

role="math" localid="1650346265761" $\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\iint }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})∥{\mathbf{r}}_u\times {\mathbf{r}}_v∥dA\\ ={\int }_0^{2\pi } {\int }_2^4 (3\cos u)(3\sin u)dvdu\end{array}$

On solving the above equation,

$\begin{array}{r}{\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\int }_0^{2\pi } (9\sin u\cos u)[v{]}_2^{4}du\\ \begin{array}{r}={\int }_0^{2\pi } 18\sin u\cos udu\\ ={\left[9{\sin }^{2}u\right]}_0^{2\pi }\\ =9{\sin }^{2}2\pi -9{\sin }^{2}0\\ =9(0{)}^2-9(0{)}^2\\ =0\end{array}\end{array}$