91影视

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

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

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

${鈭�}_s\mathbf{F}(x,y,z)路\mathbf{n}dS={鈭�}_0\left(\mathbf{F}\right(x,y,z)路\mathbf{n})\left(\sqrt{{\left(\frac{鈭�z}{鈭�x}\right)}^2+{\left(\frac{鈭�z}{鈭�y}\right)}^2+1}dA\right.$

The surface is portion of cone as below:

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

Differentiating partially

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

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

Also,

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

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

Hence, we get

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

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

$=\sqrt{2}$

As the $n$is pointing outwards.

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

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

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

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

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 $\mathbf{F}(x,y,z)路\mathbf{n}$

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

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

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

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

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

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

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

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

$=0$

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

Therefore,

${鈭�}_s\mathbf{F}(x,y,z)路\mathbf{n}dS={鈭�}_D\left(\mathbf{F}\right(x,y,z)路\mathbf{n})\left(\sqrt{{\left(\frac{鈭�z}{鈭�x}\right)}^2+{\left(\frac{鈭�z}{鈭�y}\right)}^2+1}dA\right.$

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

$=0$

Hence, required outward flux is zero.