91影视

The vector field,

the aim is to show the line integral,

Using stoke theorem,

if the 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 on $S$then ,

localid="1650786965580" ${鈭�}_C鈥�\mathbf{F}(x,y,z)脳d\mathbf{r}={鈭�}_S鈥�\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}d{S}^n$.

The curl of the vector field is given below:

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

$=\left[\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}y}鈭�\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}z}\right]\mathbf{i}鈭�\left[\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}z}\right]\mathbf{j}+\left[\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}y}\right]\mathbf{k}$

The normal vector of a plane $ax+by+cz=d$is shown below

$n=<a,b,c>$

The normal vector of a plane $x=r$is

$n=<r,0,0>$

The value of $\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}$will be,

role="math" localid="1650787297112" $=\left(\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}y}鈭�\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}z}\right)脳r+\left(\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}z}鈭�\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}x}\right)脳0+\left(\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}y}\right)0$

$=\left(\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}y}鈭�\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}z}\right)r_{.}$

Evaluating through stokes theorem,

${鈭�}_C鈥�\mathbf{F}(x,y,z)脳d\mathbf{r}={鈭�}_S鈥�\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}dS$

$={鈭�}_D鈥�\left(\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}y}鈭�\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}z}\right)rdA$

Though the normal vector of a plane $y=r$is,

$\mathbf{n}=鉄�0,r,0鉄�$

Value of $\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}$is,

$=\left(\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}y}鈭�\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}z}\right)脳0+\left(\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}z}鈭�\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}x}\right)脳r+\left(\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}y}\right)0$

$=\left(\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}z}鈭�\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}x}\right)r^{鈭�}$

Evaluating the integral${鈭�}_C鈥�\mathbf{F}(x,y,z)脳d\mathbf{r}$as below

${鈭�}_C鈥�\mathbf{F}(x,y,z)脳d\mathbf{r}={鈭�}_S鈥�\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}dS$

$={鈭�}_D鈥�\left(\frac{\mathrm{鈭�}{F}_1}{\widehat{C}z}鈭�\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}x}\right)rdA$

Hence it is proved.