91影视

Consider the vector field below:

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

The goal is to calculate the line integral. ${鈭�}_C鈥�\mathbf{F}(x,y,z)鈰�d\mathbf{r}$,where the curve $C$has the following definition:

The curve $C$is the region's boundary that lies above the square in the using $xy$-plane and with vertices $(3,5,0),(3,7,0),(4,5,0)$, and $(4,7,0)$, it's in the plane, localid="1650771295676" $z=12鈭�x鈭�y$with an upward-pointing normal vector

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

Let $S$be a curved, oriented, smooth, or piecewise-smooth surface with a $C$curve bounded by it.

Assume that $n$is an oriented unit normal vector of $S$and $C$that its parametrization travels $C$in a clockwise path 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,

Let the following equation be ($1$)

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

Find the vector field's curl first. $\mathbf{F}(x,y,z)=(y鈭�z)\mathbf{i}+2x\mathbf{j}+5xz\mathbf{k}$

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

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

Then, The curl of a vector field $\mathbf{F}(x,y,z)=(y鈭�z)\mathbf{i}+2x\mathbf{j}+5xz\mathbf{k}$will be,

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

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

$=0\mathbf{i}鈭�[5z鈭�(鈭�1\left)\right]\mathbf{j}+[2鈭�1]\mathbf{k}$

$=0\mathbf{i}鈭�(5z+1)\mathbf{j}+1\mathbf{k}$

A plane's normal vector $ax+by+cz=d$is as follows:

$\mathbf{n}=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$

Hence, the planes's normal vector $z=12鈭�x鈭�y$or $x+y+z=12$is,

$\mathbf{n}=1\mathbf{i}+1\mathbf{j}+1\mathbf{k}$.

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

Curl $\mathbf{F}(x,y,z)鈰�\mathbf{n}=(0\mathbf{i}鈭�(5z+1)\mathbf{j}+1\mathbf{k})鈰�(\mathbf{i}+1\mathbf{j}+1\mathbf{k})$

$=鈭�(5z+1)+1$

$=鈭�5z$.

Because $z=12鈭�x鈭�y$, so the value of$\mathrm{curl}鈦�\mathbf{F}(x,y,z)脳\mathbf{n}$will be,

$\mathrm{curl}\mathbf{F}(x,y,z)鈰�\mathbf{n}=鈭�5z$

$=鈭�5(12鈭�x鈭�y)$

$=鈭�(60鈭�5x鈭�5y)$

$=5x+5y鈭�60$

The surface $S$is defined and bounded by $C$, where $C$is the border of the region above the square in the $xy$-plane and has vertices $(3,5,0),(3,7,0),(4,5,0)$and $(4,7,0)$.

As a result, the $D$integration region is the square with vertices. $(3,5,0),(3,7,0),(4,5,0)$and $(4,7,0)$in the$xy$-plane, as seen in the diagram:

Here, the region of integration will be.

$D=\left\{\right(x,y)鈭�3鈮�x鈮�4,5鈮�y鈮�7\}$

Now evaluate the integral using Stokes' Theorem ($1$) ${鈭�}_C鈥�\mathbf{F}(x,y,z)鈰�d\mathbf{r}$as follows:

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

$={鈭�}_D鈥�(5x+5y鈭�60)dA$

$={鈭�}_5^7鈥�{鈭�}_3^4鈥�(5x+5y鈭�60)dxdy$

$={鈭�}_5^7鈥�\left[{鈭�}_3^4鈥�(5x+5y鈭�60)dx\right]dy$

$={鈭�}_5^7鈥�{\left[\frac{5x^2}{2}+5xy鈭�60x\right]}_3^{4}dy$

$={鈭�}_5^7鈥�\left[\left(\frac{5鈰�4^2}{2}+5鈰�4鈰�y鈭�60鈰�4\right)鈭�\left(\frac{5鈰�3^2}{2}+5鈰�3鈰�y鈭�60鈰�3\right)\right]dy$

$={鈭�}_5^7鈥�\left[(40+20y鈭�240)鈭�\left(\frac{45}{2}+15y鈭�180\right)\right]dy$

$={鈭�}_5^7鈥�\left(5y鈭�\frac{85}{2}\right)dy$

$={\left[\frac{5y^2}{2}鈭�\frac{85y}{2}\right]}_5^7$

$=\left(\frac{5鈰�7^2}{2}鈭�\frac{85鈰�7}{2}\right)鈭�\left(\frac{5鈰�5^2}{2}鈭�\frac{85鈰�5}{2}\right)$

$=鈭�25$.

Thus, the required integral is localid="1650712355476" ${鈭�}_C鈥�\mathbf{F}(x,y,z)鈰�dr=鈭�25.$