Find the outward flux of the given vector field through the specified surface.

$\mathbf{F}(x,y,z)=⟨x,y,z⟩$and S is the sphere with equation$x^2+y^2+z^2=9$

The given vector field is $\mathbf{F}(x,y,z)=(x,y,z)$. The surface S is the sphere with equation$x^2+y^2+z^2=9$with$n$pointing outwards.

The surface S parametrized by $\mathrm{r}(u,v)$for $(u,v)\in D$, then flux through a surface is

localid="1653381021945" ${\int }_s\mathbf{F}(x,y,z)·\mathbf{n}dS={\iint }_0\left(\mathbf{F}\right(x,y,z)·\mathbf{n})||{\mathbf{r}}_u\times {\mathbf{r}}_v||dA$

Hence, region D is given by

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

Solving for ${\mathbf{r}}_u$, we get,

${\mathbf{r}}_u=\frac{\partial }{\partial u}\mathbf{r}(u,v)$

$=\frac{\partial }{\partial u}⟨3\sin u\cos v,3\sin u\sin v,3\cos u⟩$

$=⟨3\cos u\cos v,3\cos u\sin v,-3\sin u⟩$

Also,

${\mathrm{r}}_{\mathrm{v}}=\frac{\partial }{\partial v}\mathbf{r}(u,v)$

$=\frac{\partial }{\partial v}⟨3\sin u\cos v,3\sin u\sin v,3\cos u⟩$

$=⟨-3\sin u\sin v,3\sin u\cos v,0⟩$

Taking cross product, we get

${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨3\cos u\cos v,3\cos u\sin v,-3\sin u⟩\times ⟨-3\sin u\sin v,3\sin u\cos v,0⟩$

$=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 3\cos u\cos v& 3\cos u\sin v& -3\sin u\\ -3\sin u\sin v& 3\sin u\cos v& 0\end{array}\right|$

localid="1653286894662" $=\left[\right(3cosusinv\left)\right(0)-(-3sinu\left)\right(3sinucosv\left)\right]\mathrm{i}-\left[\right(3\mathrm{cosucosv}\left)\right(0)-(-3\mathrm{sinu}\left)\right(-3\mathrm{sinusinv}\left)\right]\mathrm{j}+\left[\right(3\mathrm{cosucosv}\left)\right(3\mathrm{sinucosv})-(3\mathrm{cosusinv}\left)\right(-3\mathrm{sinusinv}\left)\right]\mathrm{k}$

$=9{\sin }^{2}u\cos v\mathbf{i}+9{\sin }^{2}u\sin v\mathbf{j}+9\sin u\cos u\left({\cos }^{2}v+{\sin }^2\right)v\mathbf{k}$

$=9{\sin }^{2}u\cos v\mathbf{i}+9{\sin }^{2}u\sin v\mathbf{j}+9\sin u\cos uk$

Therefore,

localid="1653287040742"

$=\sqrt{81{\sin }^{4}u\left({\cos }^{2}v+{\sin }^{2}v\right)+81{\sin }^{2}u{\cos }^{2}u}$

Solving the above equation gives,

$=\sqrt{81{\sin }^{4}u\left(1\right)+81{\sin }^{2}u{\cos }^{2}u}$

$=\sqrt{81{\sin }^{2}u\left({\sin }^{2}u+{\cos }^{2}u\right)}$

$=\sqrt{81{\sin }^{2}u}$

$=9\sin u$

The desired vector is,

$n=\frac{r_{\mathrm{s}}\times {\mathbf{r}}_{\mathrm{v}}}{||{\mathrm{r}}_{\mathrm{w}}\times {\mathrm{T}}_{\mathrm{w}}||}$

$=\frac{\left(9{\sin }^{2}u\cos v,9{\sin }^{2}u\sin v,9\sin u\cos u\right)}{9\sin u}$

For parametrization, $\mathrm{r}(u,v)=⟨3\sin u\cos v,3\sin u\sin v,3\cos u⟩$

vector field corresponds to,

$\mathbf{F}\left(x\right(u,v),y(u,v),z(u,v\left)\right)=⟨3\sin u\cos v,3\sin u\sin v,3\cos u⟩$

Hence,

$F·n=(3\sin u\cos v,3\sin u\sin v,3\cos u⟩·⟨\sin u\cos v,\sin u\sin v,\cos u⟩$

$=3{\sin }^{2}u{\cos }^{2}v+3{\sin }^{2}u{\sin }^{2}v+3{\cos }^{2}u$

$=3{\sin }^{2}u\left({\cos }^{2}v+{\sin }^{2}v\right)+3{\cos }^{2}u$

$=3\left({\sin }^{2}u+{\cos }^{2}u\right)$

$=3$

Using value of $\mathbf{F}(x,y,z)·\mathbf{n}=3\text{and}||{\mathbf{r}}_w\times {\mathbf{r}}_{\mathrm{v}}||=9\sin u$with region of integration,

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

We get

${\int }_S\mathbf{F}(x,y,z)·\mathbf{n}dS=={\iint }_D\left(\mathbf{F}\right(x,y,z)·\mathbf{n})||{\mathbf{r}}_z\times {\mathbf{r}}_v||dA$

$={\int }_0^{\pi /2}{\int }_0^{2\pi }\left(3\right)\left(9\sin u\right)dvdu$

$={\int }_0^{\pi }\left(27\sin u\right)\left[{\int }_0^{2\pi }dv\right]du$

$={\int }_0^{\pi }\left(27\sin u\right)[2\pi -0]du$

$=54\pi {\int }_0^{\mathrm{\pi }}\sin udu$

$=54\pi [-\cos u{]}_0^{\mathrm{\pi }}$

$=54\pi \left[\right(-(-1))-(-1\left)\right]$

$=54\pi \left(2\right)$

$=108\mathrm{\pi }$

Hence, the outward flux is $108\mathrm{\pi }$.