91Ó°ÊÓ

Consider the vector field below:

The goal is to calculate the line integral ${∫}_C\mathbf{F}(x,y,z)·d\mathbf{r}$, wherethe$C$curve is defined as follows:

The curve$C$is the triangle with vertices$\left(2,0,0\right),\left(0,6,0\right)and\left(0,0,3\right)$,and $n$points upwards.

This problem is presented in the book. WRONG.

Correct Issue:

Consider the vector field below:

$\mathbf{F}(x,y,z)=\left(y^2-x^2\right)\mathbf{i}+\left(z^2-y^2\right)\mathbf{j}+\left(x^2-z^2\right)\mathbf{k}\text{.}$

The goal is to calculate the line integral, which has the following definition:

${∫}_C\mathbf{F}(x,y,z)·d\mathbf{r}$where the Curve $C$

The triangle with vertices $(2,0,0),(0,6,0)$and$(0,0,3)$, $n$points upwards is the curve $C$

"Let$S$be an oriented, smooth or piecewise-smooth surface bounded by a curve$C$," Stokes' Theorem says. Assume that$n$is an oriented unit normal vector of$S$and$C$that its parametrization travels in a clockwise path with respect to$n$.

If a vector field exists,

$\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined on$S$, then

${∫}_C\mathbf{F}(x,y,z)·d\mathbf{r}={âˆ\neg }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS$

$\mathbf{F}(x,y,z)=\left(y^2-x^2\right)\mathbf{i}+\left(z^2-y^2\right)\mathbf{j}+\left(x^2-z^2\right)\mathbf{k}$

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:

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

$=\left[\frac{∂F_3}{∂y}-\frac{∂F_2}{∂z}\right]\mathbf{i}-\left[\frac{∂F_3}{∂x}-\frac{∂F_1}{∂z}\right]\mathbf{j}+\left[\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right]\mathbf{k}.$

After that, the vector field's curl will be,

$\mathbf{F}(x,y,z)=\left(y^2-x^2\right)\mathbf{i}+\left(z^2-y^2\right)\mathbf{j}+\left(x^2-z^2\right)\mathbf{k}$

$curl\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ y^2-x^2& z^2-y^2& x^2-z^2\end{array}\right|$

$$\begin{gathered} =\left[\frac{∂}{∂y}\left(x^2-z^2\right)-\frac{∂}{∂z}\left(z^2-y^2\right)\right]\mathbf{i}-\left[\frac{∂}{∂x}\left(x^2-z^2\right)-\frac{∂}{∂z}\left(y^2-x^2\right)\right]\mathbf{j} \\ \mathbf{+}\left[\frac{∂}{∂x}\left(z^2-y^2\right)-\frac{∂}{∂y}\left(y^2-x^2\right)\right]\mathbf{k} \\ \mathbf{=}\mathbf{[}\mathbf{0}\mathbf{-}\mathbf{2}\mathbf{z}\mathbf{]}\mathbf{i}\mathbf{-}\mathbf{[}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{0}\mathbf{]}\mathbf{j}\mathbf{+}\mathbf{[}\mathbf{0}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{]}\mathbf{k} \\ \mathbf{=}\mathbf{-}\mathbf{2}\mathbf{zi}\mathbf{-}\mathbf{2}\mathbf{xj}\mathbf{-}\mathbf{2}\mathbf{yk} \\ \mathbf{=}\mathbf{⟨}\mathbf{-}\mathbf{2}\mathbf{z}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{âŸ\copyright } \end{gathered}$$

As illustrated in the diagram, the curve$C$is a triangle with vertices $(2,0,0)(0,6,0)$and$(0,0,3)$

This plane's equation is as follows:

$$\begin{gathered} \frac{x}{2}+\frac{y}{6}+\frac{z}{3}=1or \\ 3x+y+2z=6 \end{gathered}$$

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

$\mathbf{n}=⟨a,b,câŸ\copyright$

As a result, the plane's normal vector$3x+y+2z=6$is$\mathbf{n}=⟨3,1,2âŸ\copyright$, and the value of $curl\mathbf{F}(x,y,z)·\mathbf{n}$will be.

role="math" localid="1652185126029" $$\begin{gathered} curl\mathbf{F}(x,y,z)·\mathbf{n}=⟨-2z,-2x,-2yâŸ\copyright ·⟨3,1,2âŸ\copyright \\ curl\mathit{F}(x,y,z)·\mathit{n}=(-2z,-2x,-2y)·1+(-2y)·2 \\ =-6z-2x-4y \end{gathered}$$

Because$$\begin{gathered} 3x+y+2z=6orz=\frac{1}{2}(6-3x-y),so \\ curl\mathit{F}(x,y,z)·n=-6z-2x-4y \\ =-6\left(\frac{1}{2}\right(6-3x-y\left)\right)-2x-4y \\ =-18+9x+3y-2x-4y \\ =7x-y-18 \\ curl\mathit{F}(x,y,z)·n=-6z-2x-4y \end{gathered}$$

The surface S is bounded by$C$, where$C$is the triangle with vertices$(2,0,0),(0,6,0)$and$(0,0,3)$

The plane equation through$(2,0,0),(0,6,0)$and$(0,0,3)$is,

$3x+y+2z=6\text{.}$

As a result, the integration$D$region is a triangle with vertices$(0,0,0),(2,0,0)$$(0,6,0)$in thelocalid="1652188238264" $xy$-plane. In other words, the integration zone is defined by the -x axis, y -axis, as $3x+y=6$illustrated in the diagram:

The irrigation will be:

$D=\left\{\right(x,y)âˆ\pounds 0≤x≤2,0≤y≤6-3x\}$

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

role="math" localid="1652187288398" $$\begin{gathered} {∫}_c\mathit{F}(x,y,z)·d\mathit{r}=∫{∫}_{S}curl\mathit{F}(x,y,z)·\mathit{n}dS \\ =∫{∫}_D(7x-y-18)dA \\ ={∫}_0^2\left[{∫}_0^{6-3x}(7x-y-18)dy\right]dx \\ ={∫}_0^2{{\left[7xy-\frac{y^2}{2}-18y\right]}_0}^{6-3x}dx \\ ={∫}_0^2\left[\left(7x\right(6-3x)-\frac{{(6-3x)}^2}{2}-18(6-3x\left)\right)-(7x·0-\frac{0^2}{2}-18·0)\right]dx \\ ={∫}_0^2(42x-21x^2-18+18x-\frac{9}{2}{x}^2-108+54x)dx \\ ={{\left[-\frac{17}{2}{x}^3+57x^2-126x\right]}_0}^2 \\ =(-\frac{17}{2}·2^3+57·2^2-126·2)-(-\frac{17}{2}·0^3+57·0^2-126·0) \\ =-92 \end{gathered}$$

As a result, the necessary integral is.

${∫}_C\mathbf{F}(x,y,z)·d\mathbf{r}=-92$