91影视

$\mathbf{F}(x,y,z)=(3x+y鈭�z)\mathbf{i}+(4y鈭�2z)\mathbf{j}+(x鈭�3z)\mathbf{k}$

The goal is to calculate the line integral.localid="1650767632481" ${鈭�}_C\mathbf{F}(x,y,z)脳d\mathbf{r}$, where the curve C is defined as follows:

The curve C is the paraboloid curve.$z=x^2+y^2$and that lies above the unit circle, traversed counterclockwise with respect to the normal vector pointing outward.

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

"Assume that S is an oriented, smooth or piecewise-smooth surface bounded by a curve C.. Assume is an oriented unit normal vector of S and C with a parametrization that traverses C counterclockwise with respect to 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="1650767639116" ${鈭�}_C鈥�\mathbf{F}(x,y,z)脳d\mathbf{r}={鈭�}_S鈥�\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}dS.$

First find the curl of the vector field $\mathbf{F}(x,y,z)=(3x+y-z)\mathbf{i}+(4y-2z)\mathbf{j}+(x-3z)\mathbf{k}$

$=\left[\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}y}鈭�\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}z}\right]\mathbf{i}鈭�\left[\frac{\mathrm{鈭�}{F}_3}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}z}\right]\mathbf{j}+\left[\frac{\mathrm{鈭�}{F}_2}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}{F}_1}{\mathrm{鈭�}y}\right]\mathbf{k}$first, determine the vector field's curl.

$\mathbf{F}(x,y,z)=(3x+y-z)\mathbf{i}+(4y-2z)\mathbf{j}+(x-3z)\mathbf{k}$

A 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:

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

Then, A vector field's curl$\mathbf{F}(x,y,z)=(3x+y-z)\mathbf{i}+(4y-2z)\mathbf{j}+(x-3z)\mathbf{k}$will be,

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

$\begin{array}{r}=\left[\frac{\mathrm{鈭�}(x鈭�3z)}{\mathrm{鈭�}y}鈭�\frac{\mathrm{鈭�}(4y鈭�2z)}{\mathrm{鈭�}z}\right]\mathbf{i}鈭�\left[\frac{\mathrm{鈭�}(x鈭�3z)}{\mathrm{鈭�}x}鈭�\frac{\mathrm{鈭�}(3x+y鈭�z)}{\mathrm{鈭�}z}\right]\mathbf{j}\\ \end{array}$

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

localid="1650767650444" $=0鈭�(鈭�2)]\mathbf{i}鈭�1鈭�(鈭�1\left)\right]\mathbf{j}+[0鈭�1]k$

localid="1650766213891" $=2i+2j+2k$

localid="1650766235827" $<2,-2,1>$

The letter n should be pointing outwards.

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

is the normal to the surface

Now, for $z=x^2+y^2$, It produces the perpendicular vector to this surface;

$=鉄�鈭�2x,鈭�2y,1鉄�$

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

localid="1650766271635" $\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}=鉄�2,鈭�2,鈭�1鉄�脳鉄�鈭�2x,鈭�2y,1鉄�$

localid="1650767664870" $$\begin{gathered} =\left(2\right)脳(鈭�2x)+(鈭�2)脳(鈭�2y)+(鈭�1)脳\left(1\right) \\ -4x+4y-1 \end{gathered}$$

The region of integration D is the unit disk in the x y-plane because the curve C lies above the unit circle when traversed counterclockwise.

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

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

In this case

$x=r\cos 鈦�胃,y=r\sin 鈦�胃,x^2+y^2=r^2$

And

$dA=rdrd胃$

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

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

$={鈭�}_D鈥�(鈭�4x+4y鈭�1)dA$

$={鈭�}_0^{2蟺}鈥�{鈭�}_0^1鈥�(鈭�4r\cos 鈦�胃+4r\sin 鈦�胃鈭�1)rdrd胃$

$={鈭�}_0^{2蟺}鈥�{鈭�}_0^1鈥�\left(\right(\sin 鈦�胃鈭�\cos 鈦�胃)4r鈭�1)rdrd胃$

$={鈭�}_0^{2蟺}鈥�{鈭�}_0^1鈥�\left((\sin 鈦�胃鈭�\cos 鈦�胃)4r^2鈭�r\right)drd胃$

$={鈭�}_0^{2蟺}鈥�\left[{鈭�}_0^1鈥�\left((\sin 鈦�胃鈭�\cos 鈦�胃)4r^2鈭�r\right)dr\right]d胃$

$={鈭�}_0^{2蟺}鈥�{\left[(\sin 鈦�胃鈭�\cos 鈦�胃)\frac{4r^3}{3}鈭�\frac{r^2}{2}\right]}_0^{1}d胃$

localid="1650766301574" $={鈭�}_0^4鈥�\left[\left((\sin 鈦�胃鈭�\cos 鈦�胃)\frac{4脳1^3}{3}鈭�\frac{1^2}{2}\right)鈭�\left((\sin 鈦�胃鈭�\cos 鈦�胃)\frac{4脳0^3}{3}鈭�\frac{0^2}{2}\right)\right]d胃$

$={鈭�}_0^{2蟺}鈥�\left((\sin 鈦�胃鈭�\cos 鈦�胃)\frac{4}{3}鈭�\frac{1}{2}\right)d胃$

$={鈭�}_0^{2蟺}鈥�\left(\frac{4}{3}\sin 鈦�胃鈭�\frac{4}{3}\cos 鈦�胃鈭�\frac{1}{2}\right)d胃$

$={\left[鈭�\frac{4}{3}\cos 鈦�胃+\frac{4}{3}\sin 鈦�胃鈭�\frac{1}{2}胃\right]}_0^{2蟺}$localid="1650766362431" $=\left(鈭�\frac{4}{3}\cos 鈦�2蟺+\frac{4}{3}\sin 鈦�2蟺鈭�\frac{1}{2}脳2蟺\right)鈭�\left(鈭�\frac{4}{3}\cos 鈦�0+\frac{4}{3}\sin 鈦�0鈭�\frac{1}{2}脳0\right)$

localid="1650766338659" $$\begin{gathered} =\left(鈭�\frac{4}{3}脳1+\frac{4}{3}脳0鈭�\frac{1}{2}脳2蟺\right)鈭�\left(鈭�\frac{4}{3}脳1+\frac{4}{3}脳0鈭�\frac{1}{2}脳0\right) \\ =-\mathrm{蟺} \end{gathered}$$