91影视

Consider the following vector field

$\mathbf{F}(x,y,z)=-y{z}^2\mathbf{i}+x{z}^2\mathbf{j}+3^{-x=z}\mathbf{k}$

The goal is to evaluate the integral ${鈭�}_{s}curl\mathbf{F}(x,y,z)路\mathbf{n}dS$,where the surface $S$is s defined as follows:

The surface $S$of the unit sphere that lies below the$xy$-plane and inside the cylinder $x^2+y^2=\frac{1}{9}$with its normal pointing outwards is the cap.

Find the curl of the vector field $\mathbf{F}(x,y,z)=-y{z}^2\mathbf{i}+x{z}^2\mathbf{j}+3^{-rx}\mathbf{k}$.

The curl of a vector field$\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined as follows:

$curl\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{鈭�}{鈭�x}& \frac{鈭�}{鈭�y}& \frac{鈭�}{鈭�z}\\ F_1(x,y,z)& F_2(x,y,z)& F_3(x,y,z)\end{array}\right|$

$=\left[\frac{鈭�F_3}{鈭�y}-\frac{鈭�F_2}{鈭�z}\right]\mathbf{i}-\left[\frac{鈭�F_3}{鈭�x}-\frac{鈭�F_1}{鈭�z}\right]\mathbf{j}+\left[\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right]\mathbf{k}\text{.}$

The curl of a vector field is $\mathbf{F}(x,y,z)=-y{z}^2\mathbf{i}+x{z}^2\mathbf{j}+3^{-xz}\mathbf{k}$will be,

$curl\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{鈭�}{鈭�x}& \frac{鈭�}{鈭�y}& \frac{鈭�}{鈭�z}\\ -y{z}^2& x{z}^2& 3^{-xz}\end{array}\right|$

$=\left[\frac{鈭�\left(3^{-pz}\right)}{鈭�y}-\frac{鈭�\left(x{z}^2\right)}{鈭�z}\right]\mathbf{i}-\left[\frac{鈭�\left(3^{-nz}\right)}{鈭�x}-\frac{鈭�\left(-y{z}^2\right)}{鈭�z}\right]\mathbf{j}+\left[\frac{鈭�\left(x{z}^2\right)}{鈭�x}-\frac{鈭�\left(-y{z}^2\right)}{鈭�y}\right]\mathbf{k}$

$=\left[\left(-3^{-9x}xz\ln 3\right)-2xz\right]\mathbf{i}-\left[\left(-3^{-ng}yz\ln 3\right)-(-2yz)\right]\mathbf{j}+\left[z^2-\left(-z^2\right)\right]\mathbf{k}$

$=\left[-3^{-9z}\ln 3-2\right]xz\mathbf{i}+\left[3^{-nz}\ln 3-2\right]yz\mathbf{j}+2z^2\mathbf{k}$

The choice$\mathbf{n}$should point outwards.

If $z=z(x,y),$then the vector is,

$\mathbf{n}=\left(-\frac{鈭�z}{鈭�x},-\frac{鈭�z}{鈭�y},1\right)$

is normal to surface.

Here, the surface $S$is in the unit sphere of the cap, in the following $xy$-plane, its equation銆�-

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

Now, $z=-\sqrt{1-x^2-y^2}$, this vector perpendicular to this plane.

$\mathbf{n}=\left(-\frac{\widehat{鈭�z}}{鈭�x},-\frac{鈭�z}{鈭�y},1\right)$

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

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

the value of $curl\mathbf{F}(x,y,z)路\mathbf{n}$will be,

$=\left(-3^{-5z}\ln 3-2\right)xz路\frac{x}{z}+\left(3^{-9z}\ln 3-2\right)yz+\frac{y}{z}+2z^2路1$

$=\left(-3^{-9x}\ln 3-2\right)x^2+\left(3^{-9x}\ln 3-2\right)y^2+2z^2$

$=-3^{-nx}{x}^2\ln 3-2x^2+3^{-mx}{y}^2\ln 3-2y^2+2z^2$

$=3^{-5x}\left(-x^2+y^2\right)\ln 3-2\left(x^2+y^2-z^2\right)$

Because $z=-\sqrt{1-x^2-y^2}$or $z^2=1-x^2-y^2$, so the value of $curl\mathbf{F}(x,y,z)路\mathbf{n}$is,

$curl\mathbf{F}(x,y,z)路\mathbf{n}=3^{-ygz}\left(-x^2+y^2\right)\ln 3-2\left(x^2+y^2-z^2\right)$

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

$=3^{x\sqrt{1-\left(x^2+y^2\right)}}\left(-x^2+y^2\right)\ln 3-4\left(x^2+y^2\right)+2$

The surface $S$is the cap of the unit sphere that lies below the$xy$-plane and inside the cylinder is. $x^2+y^2=\frac{1}{9}$The region of integration$D$is the disk in the $xy$-plane determined by the circle. $x^2+y^2=\frac{1}{9}$

In polar coordinates, the region of integration is described as follows:

$D=\left\{(r,胃)鈭�0鈮�r鈮�\frac{1}{3},0鈮�胃鈮�2蟺\right\}$

In this case, $x=r\cos 胃,y=r\sin 胃,x^2+y^2=r^2$, and

$dA=rdrd胃$

The required integral is,

${鈭�}_{S}curl\mathbf{F}(x,y,z)路\mathbf{n}dS$

role="math" localid="1650742091772" $={鈭�}_D\left(3^{xy\sqrt{1-\left(x^2+y^2\right)}}\left(-x^2+y^2\right)\ln 3-4\left(x^2+y^2\right)+2\right)dA$

$={鈭�}_0^{2r}{鈭�}_0^{1/3}\left(3^{\cos 胃\sin 胃\sqrt{1-r^2}}\left(-r^2{\cos }^2胃+r^2{\sin }^2胃\right)\ln 3-4r^2+2\right)rdrd胃$

$\left.={鈭�}_0^{2r1/3}{鈭�}_0^{1/2r^2\sqrt{1-r^3}\sin 胃\cos 胃}{r}^2\left({\sin }^2胃-{\cos }^2胃\right)\ln 3-4r^2+2\right)rdrd胃$

$\left.={鈭�}_0^{2r1/3}{鈭�}_0^{\frac{1}{2}{r}^2\sqrt{1-r^2}\sin 2胃}{r}^3(-\cos 2胃)\ln 3-4r^3+2r\right)drd胃$$\left.={鈭�}_0^{1/32r}{鈭�}_0^{1/r^2\sqrt{1-r^2}\sin 2胃}{r}^3(-\cos 2胃)\ln 3-4r^3+2r\right)d胃dr$

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

$={鈭�}_0^{13}{\left[-\frac{r}{\sqrt{1-r^2}}{3}^{\frac{1}{2}{r}^2\sqrt{1-r^2}\sin 2胃}-4r^3胃+2r胃\right]}_0^{2r}dr$

$={鈭�}_0^{13}\left[\left(-\frac{r}{\sqrt{1-r^2}}{3}^{\frac{1}{2}{r}^2\sqrt{1-r^2}\sin 4蟺}-4r^3路2蟺+2r路2蟺\right)\right.$

role="math" localid="1650742530918" $\left.-\left(-\frac{r}{\sqrt{1-r^2}}{3}^{\frac{1}{2}{r}^2\sqrt{1-r^2}\sin 0}-4r^3路0+2r路0\right)\right]dr$

$={鈭�}_0^{1/3}\left(\left(-\frac{r}{\sqrt{1-r^2}}{3}^{\frac{1}{2}路\sqrt{1-r^2-0}}-8蟺r^3+4蟺r\right)-\left(-\frac{r}{\sqrt{1-r^2}}{3}^{\frac{1}{r^2}\sqrt{1-r^2-0}}-0\right)dr\right.$

$={鈭�}_0^{13}\left(-8蟺r^3+4蟺r\right)dr$

$={\left[-2蟺r^4+2蟺r^2\right]}_0^{1/3}$

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

$=\left(-\frac{2蟺}{81}+\frac{2蟺}{9}\right)-0$

$=\frac{16蟺}{81}$

Therefore, the required integral found out is${鈭�}_{s}curl\mathbf{F}(x,y,z)路\mathbf{n}dS=\frac{16蟺}{81}$.