91影视

$\mathbf{F}(x,y,z)=(2x鈭�3y+4z)\mathbf{i}+(5x+y鈭�z)\mathbf{j}+(x+4y+2z)\mathbf{k}$

The goal is to calculate the line integral.localid="1650767287662" ${鈭�}_C\mathbf{F}(x,y,z)脳d\mathbf{r}$, where the curve C is defined as follows:

The curve C is the plane curve.localid="1650706544139" $x-y+z=20$and that lies above the curves $y=4$and $y=x^2$in the$xy$-plane, traversed counterclockwise direction with respect to $\mathbf{n}=鉄�1,-1,1鉄�$.

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 is an oriented unit normal vector of S and C has a parametrization that traverses C counterclockwise with respect to.

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="1650767307394" ${鈭�}_C鈥�\mathbf{F}(x,y,z)脳d\mathbf{r}={鈭�}_S鈥�\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}dS=$

$\mathbf{F}=(2x鈭�3y+4z)\mathbf{i}+(5x+y鈭�z)\mathbf{j}+(x+4y+2z)\mathbf{k}$

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

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

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

The vector field's curl is then applied.$\mathbf{F}=(2x-3y+4z)\mathbf{i}+(5x+y-z)\mathbf{j}+(x+4y+2z)\mathbf{k}$will be,

$\mathrm{curl}鈦�\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{鈭�}}{\mathrm{鈭�}x}& \frac{\mathrm{鈭�}}{\mathrm{鈭�}y}& \frac{\mathrm{鈭�}}{\mathrm{鈭�}z}\\ (2x鈭�3y+4z)& (5x+y鈭�z)& (x+4y+2z)\end{array}\right|$

$=\left[\frac{\mathrm{鈭�}}{\mathrm{鈭�}y}(x+4y+2z)鈭�\frac{\mathrm{鈭�}}{\mathrm{鈭�}z}(5x+y鈭�z)\right]\mathbf{i}$

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

$+\left[\frac{\mathrm{鈭�}}{\mathrm{鈭�}x}(5x+y鈭�z)鈭�\frac{\mathrm{鈭�}}{\mathrm{鈭�}y}(2x鈭�3y+4z)\right]\mathbf{k}$

$=[4鈭�(鈭�1\left)\right]\mathbf{i}鈭�[1鈭�4]\mathbf{j}+[5鈭�(鈭�3\left)\right]\mathbf{k}$

role="math" localid="1650708411537" $=5\mathbf{i}+3\mathbf{j}+8\mathbf{k}$

$=鉄�5,3,8鉄�$

The normal vector is $\mathbf{n}=鉄�1,-1,1鉄�$

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

localid="1650765726819" $\mathrm{curl}鈦�\mathbf{F}(x,y,z)鈰�\mathbf{n}=鉄�5,3,8鉄�脳鉄�1,鈭�1,1鉄�$

localid="1650765645431" $$\begin{gathered} =5脳1+3脳(鈭�1)+8脳1 \\ =5-3+8 \\ =10 \end{gathered}$$

The curve C is located above the curves.$y=4$and $y=x^2$in the $xy$-plane, as shown in the figure below, traversed counterclockwise:

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

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

$={鈭�}_D鈥�10dA$

$={鈭�}_{鈭�2}^2鈥�{鈭�}_{x^2}^4鈥�10dydx$

$={鈭�}_{鈭�2}^2鈥�\left[{鈭�}_{x^2}^4鈥�10dy\right]dx$

$={鈭�}_{鈭�2}^2鈥�10[y{]}_{x^2}^{4}dx$

$={鈭�}_{鈭�2}^2鈥�10\left(4鈭�x^2\right)dx$

$=10{\left[4x鈭�\frac{x^3}{3}\right]}_{鈭�2}^2$

localid="1650765592856" $=10\left[\left(4脳2鈭�\frac{2^3}{3}\right)鈭�\left(4(鈭�2)鈭�\frac{(鈭�2{)}^3}{3}\right)\right]$

$=10\left(\frac{32}{3}\right)$

$=\frac{320}{3}.$