$\mathbf{F}(x,y,z)=5\mathbf{i}+13\mathbf{j}+2\mathbf{k}$, where S is the unit sphere, with n pointing outwards.

Consider the given question,

$\mathbf{F}(x,y,z)=5\mathbf{i}+13\mathbf{j}+2\mathbf{k}$

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

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

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

Where, role="math" localid="1650344210652" $D=\left\{\right(u,v)\mid 0\le u\le \pi ,0\le v\le 2\pi \}$.

Now, find role="math" localid="1650344418737" $r_u$,

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

Find the value of $r_v$,

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

Now, find $||r_u\times r_v||$,

role="math" localid="1650344705531"

On solving the above equation,

$\begin{array}{r}∥{\mathbf{r}}_u\times {\mathbf{r}}_v∥=∥\left({\sin }^{2}u\cos v,{\sin }^{2}us\mathrm{in}v,\sin u\cos u\right.>∥\\ =\sqrt{{\left({\sin }^{2}u\cos v\right)}^2+{\left({\sin }^{2}u\sin v\right)}^2+(\sin u\cos u{)}^2}\\ \\ =\sqrt{{\sin }^{4}u\left(1\right)+{\sin }^{2}u{\cos }^{2}u}\\ \begin{array}{r}=\sqrt{{\sin }^{2}u\left({\sin }^{2}u+{\cos }^{2}u\right)}\\ =\sqrt{{\sin }^{2}u\left(1\right)}\end{array}\\ =\sin u\end{array}$

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

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

Substitute the values in equation (i), with the following region of integration,

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

The flux of the vector field through the surface S is given below,

$$\begin{gathered} {\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\int }_D \left(\mathbf{F}\right(x,y,z)\cdot \mathbf{n})∥{\mathbf{r}}_u\times {\mathbf{r}}_v∥dA \\ ={\int }_0^{\pi } {\int }_0^{2\pi } \left(\frac{\left(5{\sin }^{2}u\cos v+13{\sin }^{2}u\sin v+2\sin u\cos u\right)}{\sin u}\right)\sin udvdu \\ ={\int }_0^{\pi } {\left[5{\sin }^{2}u\sin v-13{\sin }^{2}u\cos v+2v\sin u\cos u\right]}_0^{2\pi }du \\ ={\int }_0^{\pi } \left[\left(5{\sin }^{2}u\sin 2\pi -13{\sin }^{2}u\cos 2\pi +2\cdot 2\pi \cdot \sin u\cos u\right)\left.-\left(5{\sin }^{2}u\sin 0-13\sin 2z\cos 0+2\cdot 0\cdot \sin u\cos u\right)\right]du\right. \end{gathered}$$

On solving the above equation,

$$\begin{gathered} {\int }_S \mathbf{F}(x,y,z)\cdot \mathbf{n}dS={\int }_0^{\pi } \left[\left(5{\sin }^{2}u\cdot 0-13{\sin }^{2}u\cdot 1+4\pi \sin u\cos u\right)-\left(5{\sin }^{2}u\cdot 0-13{\sin }^{2}u\cdot 1+0\right)\right]du \\ =4\pi {\int }_0^{\pi } \sin u\cos udu \\ =4\pi {\left[\frac{1}{2}{\sin }^{2}u\right]}_0^{\pi } \\ =4\pi \left[\frac{1}{2}{\sin }^2\pi -\frac{1}{2}{\sin }^{2}0\right] \\ =4\pi \left[\frac{1}{2}\cdot 0^2-\frac{1}{2}\cdot 0^2\right] \\ =0 \end{gathered}$$