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 portion of the cone with equation$z=\sqrt{x^2+y^2}\mathrm{for}z\le 3$.

It is given that $\mathbf{F}(x,y,z)=⟨x,y,z⟩$

Equation of surface is $z=\sqrt{x^2+y^2}\text{for}z\le 3$with$n$pointing outwards.

Flux $\mathbf{F}(x,y,z)$for surface of graph $z=z(x,y)$is

${\int }_s\mathbf{F}(x,y,z)·\mathbf{n}dS={\iint }_0\left(\mathbf{F}\right(x,y,z)·\mathbf{n})\left(\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1}dA\right.$

The surface is portion of cone as below:

$z=\sqrt{x^2+y^2}$

Differentiating partially

role="math" localid="1653284481822" $\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left(\sqrt{x^2+y^2}\right)$

$=\frac{x}{\sqrt{x^2+y^2}}$

Also,

$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left(\sqrt{x^2+y^2}\right)$

$=\frac{y}{\sqrt{x^2+y^2}}$

Hence, we get

$\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1}=\sqrt{{\left(\frac{x}{\sqrt{x^2+y^2}}\right)}^2+{\left(\frac{y}{\sqrt{x^2+y^2}}\right)}^2+1}$

$=\sqrt{\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}+1}$

$=\sqrt{2}$

As the $n$is pointing outwards.

If role="math" localid="1653284771657" $z=z(x,y)$then

$\left(-\frac{\partial }{\partial x},-\frac{\partial z}{\partial y},1\right)$. It is normal to the surface.

For $z=\sqrt{x^2+y^2}$, the vector perpendicular to the surface is $v=\left(-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right)$

$=\left(-\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}},1\right)$

Now, normal vector is

role="math" localid="1653285460245" $n=\frac{1}{\left|v\right|}v$

$\left.=\frac{1}{\left(-\frac{y}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}},1\right.}\right)\left(-\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}},1\right)$

$=\frac{1}{{\left.\sqrt{\left(\sqrt{x^2+y^2}\right.}\right)}^2+{\left(\sqrt{x^2+y^2}\right)}^2+1}\left(-\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}},1\right)$

$=\frac{1}{\sqrt{2}}\left(-\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}},1\right)$

Calculating $\mathbf{F}(x,y,z)·\mathbf{n}$

$\mathbf{F}(x,y,z)·\mathbf{n}=⟨x,y,z⟩·\frac{1}{\sqrt{2}}\left(-\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}}·1\right)$

$=\frac{1}{\sqrt{2}}(x,y,z)·\left\{-\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}},1\right)$

$=\frac{1}{\sqrt{2}}\left(-\frac{x^2}{\sqrt{x^2+y^2}}-\frac{y^2}{\sqrt{x^2+y^2}}+z\right)$

$=\frac{1}{\sqrt{2}}\left(-\frac{\left(x^2+y^2\right)}{\sqrt{x^2+y^2}}+z\right)$

$=\frac{1}{\sqrt{2}}\left(-\sqrt{x^2+y^2}+z\right)$

Since $z=\sqrt{x^2+y^2}$

$\mathbf{F}(x,y,z)·\mathbf{n}=\frac{1}{\sqrt{2}}\left(-\sqrt{x^2+y^2}+z\right)$

$=\frac{1}{\sqrt{2}}(-z+z)$

$=0$

Using $\mathbf{F}(x,y,z)·\mathbf{n}=0$and $\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1}=\sqrt{2}$

Therefore,

${\int }_s\mathbf{F}(x,y,z)·\mathbf{n}dS={\iint }_D\left(\mathbf{F}\right(x,y,z)·\mathbf{n})\left(\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1}dA\right.$

$={\iint }_D\left(0\right)\left(\sqrt{2}\right)dA$

$=0$

Hence, required outward flux is zero.