91影视

The objective is to evaluate the integral ${鈭�}_{C}F.dr$by using Green's Theorem, where $C$is the boundary of the region bounded by the curves $x=y^2,x=4$,traversed counterclockwise.

Green's Theorem states that,

Let $R$be a region in the plane with a smooth boundary curve $C$oriented counterclockwise by $r\left(t\right)=\left(x\left(t\right),y\left(t\right)\right)$for $a鈮�t鈮�b$.

If a vector field $F\left(x,y\right)=\left(F_1\left(x,y\right),F_2\left(x,y\right)\right)$is defined on $R$, then,

${鈭�}_{C}F.dr=鈭�{鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA.......\left(1\right)$

For the vector field $F\left(x,y\right)=\left(x+2y\right)i+\left(x-2y\right)j$

$$\begin{gathered} F_1\left(x,y\right)=x+2y \\ {F}_2\left(x,y\right)=x-2y \end{gathered}$$

Now, first, find $\frac{鈭�F_2}{鈭�x}$and $\frac{鈭�F_1}{鈭�y}$.

Then,

$$\begin{gathered} \frac{鈭�F_2}{鈭�x}=\frac{鈭�}{鈭�x}\left(x-2y\right) \\ =1 \end{gathered}$$

and,

$$\begin{gathered} \frac{鈭�F_1}{鈭�y}=\frac{鈭�}{鈭�y}\left(x+2y\right) \\ =2 \end{gathered}$$

Now, use Green's Theorem (1) to evaluate the integral ${鈭�}_{C}F.dr$as follows:

$$\begin{gathered} {鈭�}_{C}F.dr=鈭�{鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA \\ =鈭�{鈭�}_R\left(1-2\right)dA \\ =鈭�{鈭�}_R-1dA \\ =-鈭�{鈭�}_{R}dA............\left(2\right) \end{gathered}$$

Here, the boundary curve $C$is the boundary of the region bounded by the curves $x=y^2,x=4$, traversed counterclockwise.

So the region $R$bounded by the curves $x=y^2,x=4$as shown in the following figure:

Viewed it as an x-simple region, then the region of integration will be:

$R=\left\{\left(x,y\right)I{y}^2鈮�x鈮�4,-2鈮�y鈮�2\right\}$.

Then, evaluate the integral (2) as follows:

$$\begin{gathered} {鈭�}_{C}F.dr=-鈭�{鈭�}_{R}dA \\ =-{鈭�}_{-2}^2{鈭�}_{y^2}^{4}dxdy \\ =-{鈭�}_{-2}^2\left[{鈭�}_{y^2}^{4}dx\right]dy \\ =-{鈭�}_{-2}^2{\left[x\right]}_{y^2}^{4}dy \\ =-{鈭�}_{-2}^2\left[4-y^2\right]dy \\ =-{\left[4y-\frac{y^3}{3}\right]}_{-2}^2 \\ =-\left[\left(4脳2-\frac{2^3}{3}\right)-\left(4脳\left(-2\right)-\frac{{\left(-2\right)}^3}{3}\right)\right] \\ =-\left[\frac{16}{3}-\left(-\frac{16}{3}\right)\right] \\ =-\frac{32}{3} \end{gathered}$$