91Ó°ÊÓ

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 $(u,v)∈D$, then the flux of $\mathit{F}\left(x,y,z\right)$ through S is given below,

data-custom-editor="chemistry" ${∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={âˆ\neg }_D \left(\mathbf{F}\right(x,y,z)â‹\dots \mathbf{n})∥{\mathbf{r}}_u×{\mathbf{r}}_v∥dA\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{}\mathit{\left(}i\mathit{\right)}$

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

Where, role="math" localid="1650344210652" $D=\left\{\right(u,v)âˆ\pounds 0≤u≤\mathrm{Ï€},0≤v≤2\mathrm{Ï€}\}$.

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

$\begin{array}{r}{\mathbf{r}}_u=\frac{\mathrm{∂}}{\mathrm{∂}u}\mathbf{r}(u,v)\\ =\frac{\mathrm{∂}}{\mathrm{∂}u}⟨\sin u\cos v,\sin u\sin v,\cos uâŸ\copyright \\ =⟨\cos u\cos v,\cos u\sin v,−\sin uâŸ\copyright \end{array}$

Find the value of $r_v$,

$\begin{array}{r}{\mathbf{r}}_v=\frac{\mathrm{∂}}{\mathrm{∂}v}\mathbf{r}(u,v)\\ =\frac{\mathrm{∂}}{\mathrm{∂}v}⟨\sin u\cos v,\sin u\sin v,\cos uâŸ\copyright \\ =⟨−\sin u\sin v,\sin u\cos v,0âŸ\copyright \end{array}$

Now, find $||r_u×r_v||$,

role="math" localid="1650344705531"

On solving the above equation,

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

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,

$D=\left\{\right(u,v)âˆ\pounds 0≤u≤\mathrm{Ï€},0≤v≤2\mathrm{Ï€}\}$

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

$$\begin{gathered} {∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={∫}_D \left(\mathbf{F}\right(x,y,z)â‹\dots \mathbf{n})∥{\mathbf{r}}_u×{\mathbf{r}}_v∥dA \\ ={∫}_0^{\mathrm{Ï€}} {∫}_0^{2\mathrm{Ï€}} \left(\frac{\left(5{\sin }^{2}u\cos v+13{\sin }^{2}u\sin v+2\sin u\cos u\right)}{\sin u}\right)\sin udvdu \\ ={∫}_0^{\mathrm{Ï€}} {\left[5{\sin }^{2}u\sin v−13{\sin }^{2}u\cos v+2v\sin u\cos u\right]}_0^{2\mathrm{Ï€}}du \\ ={∫}_0^{\mathrm{Ï€}} \left[\left(5{\sin }^{2}u\sin 2\mathrm{Ï€}−13{\sin }^{2}u\cos 2\mathrm{Ï€}+2â‹\dots 2\mathrm{Ï€}â‹\dots \sin u\cos u\right)\left.−\left(5{\sin }^{2}u\sin 0−13\sin 2z\cos 0+2â‹\dots 0â‹\dots \sin u\cos u\right)\right]du\right. \end{gathered}$$

On solving the above equation,

$$\begin{gathered} {∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={∫}_0^{\mathrm{Ï€}} \left[\left(5{\sin }^{2}uâ‹\dots 0−13{\sin }^{2}uâ‹\dots 1+4\mathrm{Ï€}\sin u\cos u\right)−\left(5{\sin }^{2}uâ‹\dots 0−13{\sin }^{2}uâ‹\dots 1+0\right)\right]du \\ =4\mathrm{Ï€}{∫}_0^{\mathrm{Ï€}} \sin u\cos udu \\ =4\mathrm{Ï€}{\left[\frac{1}{2}{\sin }^{2}u\right]}_0^{\mathrm{Ï€}} \\ =4\mathrm{Ï€}\left[\frac{1}{2}{\sin }^2\mathrm{Ï€}−\frac{1}{2}{\sin }^{2}0\right] \\ =4\mathrm{Ï€}\left[\frac{1}{2}â‹\dots 0^2−\frac{1}{2}â‹\dots 0^2\right] \\ =0 \end{gathered}$$