91影视

The goal is to calculate the line integral.${鈭�}_C\mathbf{F}(x,y,z)脳d\mathbf{r}$, where the curve C$is defined as follows:

The surface is intersected by curve C.$z=e^{-\left(x^2+y^2\right)x^2+y^2=9}$

To calculate this integral, use Stokes' Theorem. According to Stokes' Theorem,

"Assume S is an oriented, smooth or piecewise-smooth surface enclosed by a curve C. Assume n is an oriented unit normal vector of S and C has a parametrization that traverses C counterclockwise with respect to $\mathbf{n}$.

If 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 on S, then,

localid="1650793402578" ${鈭�}_C鈥�\mathbf{F}(x,y,z)脳d\mathbf{r}={鈭�}_S鈥�\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}dS$

First, determine the vector field's curl.

First, determine the vector field's curl.$\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined as follows:

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

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

First, determine the vector field's curl. will be,

$\mathrm{curl}鈦�\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{鈭�}}{\mathrm{鈭�}x}& \frac{\mathrm{鈭�}}{\mathrm{鈭�}y}& \frac{\mathrm{鈭�}}{\mathrm{鈭�}z}\\ 3x+3& 4x+\ln 鈦�\left(y^2+1\right)鈭�z& 2x+y\end{array}\right|$

$=\left[\frac{\mathrm{鈭�}}{\mathrm{鈭�}y}(2x+y)鈭�\frac{\mathrm{鈭�}}{\mathrm{鈭�}z}\left(4x+\ln 鈦�\left(y^2+1\right)鈭�z\right)\right]\mathbf{i}鈭�\left[\frac{\mathrm{鈭�}}{\mathrm{鈭�}x}(2x+y)鈭�\frac{\mathrm{鈭�}}{\mathrm{鈭�}z}(3x+3)\right]\mathbf{j}$

$+\left[\frac{\mathrm{鈭�}}{\mathrm{鈭�}x}\left(4x+\ln 鈦�\left(y^2+1\right)鈭�z\right)鈭�\frac{\mathrm{鈭�}}{\mathrm{鈭�}y}(3x+3)\right]\mathbf{k}$

$=[1鈭�(鈭�1\left)\right]\mathbf{i}鈭�[2鈭�0]\mathbf{j}+[4鈭�0]\mathbf{k}$

$=2i-2j+4k$

=$=鉄�2,鈭�2,4鉄�.$

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

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

is normal to the surface

Now, for $z=e^{-\left(x^2+y^2\right)}$,It produces the vector perpendicular to this surface shown below. ;

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

role="math" localid="1650791535292"

$=\left(2\right)脳\left(2x{e}^{鈭�\left(x^2+y^2\right)}\right)+(鈭�2)脳\left(2y{e}^{鈭�\left(x^2+y^2\right)}\right)+\left(4\right)脳\left(1\right)$

$=4x{e}^{鈭�\left(x^2+y^2\right)}鈭�4y{e}^{鈭�\left(x^2+y^2\right)}+4$

$=4(x鈭�y)e^{鈭�\left(x^2+y^2\right)}+4$

The surface is intersected by curve C. $z=e^{-\left(x^2+y^2\right)}$and the cylinder $x^2+y^2=9$, as a result, integration zone D is the circle-described disk. $x^2+y^2=9$in the $xy$-plane.

The region of integration is described in polar coordinates as follows.,

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

$x=r\cos 鈦�胃,y=r\sin 鈦�胃,x^2+y^2=r^2$

and

$dA=rdrd胃$

Now, apply Stokes' Theorem (1) to determine the integral.$鈭�\mathbf{F}(x,y,z)脳d\mathbf{r}$as follows:

$鈭�\mathbf{F}(x,y,z)脳d\mathbf{r}$

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

$={鈭�}_D鈥�\left(4(x鈭�y)e^{鈭�\left(x^2+y^2\right)}+4\right)dA$

$={鈭�}_0^{2蟺}鈥�{鈭�}_0^3鈥�\left(4(r\cos 鈦�胃鈭�r\sin 鈦�胃)e^{鈭�r^2}+4\right)rdrd胃$

$={鈭�}_0^{2蟺}鈥�{鈭�}_0^3鈥�\left(4r{e}^{鈭�r^2}(\cos 鈦�胃鈭�\sin 鈦�胃)+4\right)rdrd胃$

$={鈭�}_0^{2蟺}鈥�{鈭�}_0^3鈥�\left(4r^2{e}^{鈭�r^2}(\cos 鈦�胃鈭�\sin 鈦�胃)+4r\right)drd胃$

$={鈭�}_0^3鈥�\left[{鈭�}_0^{2蟺}鈥�\left(4r^2{e}^{鈭�r^2}(\cos 鈦�胃鈭�\sin 鈦�胃)+4r\right)d胃\right]dr$

role="math" localid="1650793299338" $={鈭�}_0^3鈥�{\left[4r^2{e}^{鈭�r^2}(\sin 鈦�胃+\cos 鈦�胃)+4r胃\right]}_0^{2蟺}dr$

$={鈭�}_0^{2蟺}鈥�{鈭�}_0^3鈥�\left(4r^2{e}^{鈭�r^2}(\cos 鈦�胃鈭�\sin 鈦�胃)+4r\right)drd胃$

role="math" localid="1650791972329" $={鈭�}_0^3鈥�\left[\left(4r^2{e}^{鈭�r^2}(\sin 鈦�2蟺+\cos 鈦�2蟺)+4r脳2蟺\right)鈭�\left(4r^2{e}^{鈭�r^2}(\sin 鈦�0+\cos 鈦�0)+4r脳0\right)\right]dr$

$={鈭�}_0^3鈥�\left[\left(4r^2{e}^{鈭�r^2}(0+1)+8蟺r\right)鈭�\left(4r^2{e}^{鈭�r^2}(0+1)+0\right)\right]dr$

$={鈭�}_0^3鈥�\left[\left(4r^2{e}^{鈭�r^2}(0+1)+8蟺r\right)鈭�\left(4r^2{e}^{鈭�r^2}(0+1)+0\right)\right]dr$

$={鈭�}_0^3鈥�\left[\left(4r^2{e}^{鈭�r^2}+8蟺r\right)鈭�\left(4r^2{e}^{鈭�r^2}\right)\right]dr$

$={鈭�}_0^3鈥�8蟺rdr$

$={\left[4蟺r^2\right]}_0^3$

$=4蟺(3{)}^2鈭�4蟺(0{)}^2$

$=36蟺$