91影视

The given vector field is$\mathbf{F}(x,y,z)=x^2\mathbf{i}+y^2\mathbf{j}+z^2\mathbf{k}$

The surface S is the top half of the unit sphere centered at the origin.

Let S be graph of $z=z(x,y)$, flux is

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

As it is given that

$z=\sqrt{1-x^2-y^2}$

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

role="math" localid="1653290817110" $\frac{鈭�z}{鈭�x}=\frac{-x}{\sqrt{1-x^2-y^2}}$

Also

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

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

Therefore,

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

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

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

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

$n$should be pointing outwards.

If $z=z(x,y)$, then

$v=\left(-\frac{鈭�z}{鈭�x},-\frac{鈭�z}{鈭�y},1\right)$should be normal to surface.

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

Hence, normal vector

$n=\frac{1}{\left|v\right|}v$

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

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

The value is

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

$=\sqrt{1-x^2-y^2}\left(\frac{x^3}{\sqrt{1-x^2-y^2}}+\frac{y^3}{\sqrt{1-x^2-y^2}}+z^2\right)$

Using $\mathbf{F}(x,y,z)路\mathbf{n}=\sqrt{1-x^2-y^2}\left(\frac{x^3}{\sqrt{1-x^2-y^2}}+\frac{y^3}{\sqrt{1-x^2-y^2}}+z^2\right)$and

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

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

$={鈭�}_D\sqrt{1-x^2-y^2}\left(\frac{x^3}{\sqrt{1-x^2-y^2}}+\frac{y^3}{\sqrt{1-x^2-y^2}}+z^2\right)\left(\frac{1}{\sqrt{1-x^2-y^2}}\right)dA$

As $z=\sqrt{1-x^2-y^2}$

$鈬�z^2=1-x^2-y^2$

Integral becomes

role="math" localid="1653291868686" ${鈭�}_{S}F(x,y,z)路\mathbf{n}dS={鈭�}_D\left(\frac{x^3}{\sqrt{1-x^2-y^2}}+\frac{y^3}{\sqrt{1-x^2-y^2}}+z^2\right)dA$

role="math" localid="1653291916934" $={鈭�}_D\left(\frac{x^3}{\sqrt{1-x^2-y^2}}+\frac{y^3}{\sqrt{1-x^2-y^2}}+\left(1-x^2-y^2\right)\right)dA$

$={鈭�}_D\left(\frac{x^3}{\sqrt{1-\left(x^2+y^2\right)}}+\frac{y^3}{\sqrt{1-\left(x^2+y^2\right)}}+1-\left(x^2+y^2\right)\right)dA$

The region D is unit disk in cartesian plane centered at origin

Hence, region of integration in polar coordinates is

$D=\left\{\right(r,胃)鈭�0鈮�r鈮�1,0鈮�胃鈮�2蟺\}$

Here, $x=r\cos 胃$and $y=r\sin 胃$

$鈬�x^2+y^2=r^2$

The flux of vector through surface is ${鈭�}_s\mathbf{F}(x,y,z)路\mathbf{n}dS$

$={鈭�}_D\left(\frac{x^3}{\sqrt{1-\left(x^2+y^2\right)}}+\frac{y^3}{\sqrt{1-\left(x^2+y^2\right)}}+1-\left(x^2+y^2\right)\right)dA$

role="math" localid="1653292327284" $={鈭�}_0^{2\mathrm{蟺}}{鈭�}_0^1\left(\frac{r^3{\cos }^3胃}{\sqrt{1-r^2}}+\frac{r^3{\sin }^3胃}{\sqrt{1-r^2}}+1-r^2\right)rdrd胃$

role="math" localid="1653292317909" $={鈭�}_0^{2\mathrm{蟺}}{鈭�}_0^1\left(\frac{r^4}{\sqrt{1-r^2}}{\cos }^3胃+\frac{r^4}{\sqrt{1-r^2}}{\sin }^3胃+r-r^3\right)drd胃$

role="math" localid="1653292363249" $={鈭�}_0^{2蟺}{鈭�}_0^1\left(\left({\cos }^3胃+{\sin }^3胃\right)\frac{r^4}{\sqrt{1-r^2}}+r-r^3\right)drd胃$

$={鈭�}_0^{2\mathrm{蟺}}{\left[\left({\cos }^3胃+{\sin }^3胃\right)\left(\sqrt{1-r^2}\left(-\frac{3r}{8}-\frac{r^3}{4}\right)+\frac{3}{8}{\sin }^{-1}r\right)+\frac{r^2}{2}-\frac{r^4}{4}\right]}_0^{1}d胃$

role="math" localid="1653292448580" $={鈭�}_0^{2\mathrm{蟺}}\left[\left({\cos }^3胃+{\sin }^3胃\right)\left(\sqrt{1-1^2}\left(-\frac{3\left(1\right)}{8}-\frac{1^3}{4}\right)+\frac{3}{8}{\sin }^{-1}1\right)+\frac{1^2}{2}-\frac{1^4}{4}\right)$

$\left.-\left(\left({\cos }^3胃+{\sin }^3胃\right)\left(\sqrt{1-0^2}\left(-\frac{3\left(0\right)}{8}-\frac{0^3}{4}\right)+\frac{3}{8}{\sin }^{-1}0\right)+\frac{0^2}{2}-\frac{0^4}{4}\right)\right]d胃$

$={鈭�}_0^{2\mathrm{蟺}}\left(\frac{3蟺}{16}\left({\cos }^3胃+{\sin }^3胃\right)+\frac{1}{4}\right)d胃$

Using ${\cos }^3胃=\frac{3}{4}\cos 胃+\frac{1}{4}\cos 3胃$and ${\sin }^3胃=\frac{3}{4}\sin 胃-\frac{1}{4}\sin 3胃$, we get

${鈭�}_S\mathbf{F}(x,y,z)路\mathbf{n}dS=={鈭�}_0^{2蟺}\left(\frac{3蟺}{16}\left({\cos }^3胃+{\sin }^3胃\right)+\frac{1}{4}\right)d胃$

$={鈭�}_0^{2蟺}\left(\frac{3蟺}{16}\left(\frac{3}{4}\cos 胃+\frac{1}{4}\cos 3胃+\frac{3}{4}\sin 胃-\frac{1}{4}\sin 3胃\right)+\frac{1}{4}\right)d胃$

$=\frac{3蟺}{16}{鈭�}_0^{22}\left(\frac{3}{4}\cos 胃+\frac{1}{4}\cos 3胃+\frac{3}{4}\sin 胃-\frac{1}{4}\sin 3胃\right)d胃+\frac{1}{4}{鈭�}_0^{2蟺}d胃$

$=\frac{3蟺}{16}\left[\left(\frac{3}{4}\sin 2蟺+\frac{1}{12}\sin 6蟺-\frac{3}{4}\cos 2蟺+\frac{1}{12}\cos 6蟺\right)\right.$

$\left.-\left(\frac{3}{4}\sin 0+\frac{1}{12}\sin 0-\frac{3}{4}\cos 0+\frac{1}{12}\cos 0\right)\right]+\frac{1}{4}[2蟺-0]$

$=\frac{3蟺}{16}\left[\left(\frac{3}{4}\left(0\right)+\frac{1}{12}\left(0\right)-\frac{3}{4}\left(1\right)+\frac{1}{12}\left(1\right)\right)-\left(\frac{3}{4}\left(0\right)+\frac{1}{12}\left(0\right)-\frac{3}{4}\left(1\right)+\frac{1}{12}\left(1\right)\right)\right]+\frac{1}{4}\left(2蟺\right)$

$=\frac{3蟺}{16}\left(0\right)+\frac{1}{4}\left(2蟺\right)$

$=\frac{\mathrm{蟺}}{2}$

Hence, the outward flux of vector field is$\frac{\mathrm{蟺}}{2}$