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 square with vertices at $\left(0,0\right),\left(0,2\right),\left(2,0\right)$, and $\left(2,2\right)$, traversed counterclockwise.

Green's Theorem states that,

RLet $R$be a region in the plane with a smooth boundary curve localid="1650885413080" $C$oriented counterclockwise by localid="1650885459096" $r\left(t\right)=\left(x\left(t\right),y\left(t\right)\right)$for localid="1650885476447" $a鈮�t鈮�b$.

If a vector field localid="1650885503704" $F\left(x,y\right)=\left(F_1\left(x,y\right),F_2\left(x,y\right)\right)$is defined on $R$, then,

localid="1650885548801" ${鈭�}_{C}F.dr=鈭�{鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA..........\left(1\right)$

For the vector field,$F\left(x,y\right)=y\ln {\left(x+1\right)}^2+j$

$$\begin{gathered} F_1\left(x,y\right)=y\ln {\left(x+1\right)}^2 \\ {F}_2\left(x,y\right)=1 \end{gathered}$$

Now, first, find localid="1650885695009" $\frac{鈭�F_2}{鈭�x}$and $\frac{鈭�F_1}{鈭�y}$.

Then,

localid="1650907384165" $$\begin{gathered} \frac{鈭�F_2}{鈭�x}=\frac{鈭�}{鈭�x}\left(1\right) \\ =0 \end{gathered}$$

and,

$$\begin{gathered} \frac{鈭�F_1}{鈭�y}=\frac{鈭�}{鈭�y}\left(y\ln {\left(x+1\right)}^2\right) \\ =\ln {\left(x+1\right)}^2 \end{gathered}$$

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

$$\begin{gathered} {鈭�}_{C}F.dr=鈭�{鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA \\ =鈭�{鈭�}_R\left(0-\ln {\left(x+1\right)}^2\right)dA \\ =鈭�{鈭�}_R\left(-\ln {\left(x+1\right)}^2\right)dA \\ =-鈭�{鈭�}_R\ln {\left(x+1\right)}^{2}dA...............\left(2\right) \end{gathered}$$

Here, the boundary curve $C$is the boundary of the square with vertices at $\left(0,0\right),\left(0,2\right),\left(2,0\right),\left(2,2\right)$, traversed counterclockwise.

So, the region $R$bounded by the square with vertices at$\left(0,0\right),\left(0,2\right),\left(2,0\right),\left(2,2\right)$, as shown in the following figure:

Viewed it as a x-or y-simple region, then the region of the integration will be,

$R=\left\{\left(x,y\right)I0鈮�x鈮�2,0鈮�y鈮�2\right\}$.

Then, evaluate the integral (2) as follows:

$$\begin{gathered} {鈭�}_{C}F.dr=鈭�{鈭�}_R\ln {\left(x+1\right)}^{2}dA \\ =-{鈭�}_0^2{鈭�}_0^2\ln {\left(x+1\right)}^{2}dydx \\ =-{鈭�}_0^2\ln {\left(x+1\right)}^2\left[{鈭�}_0^{2}dy\right]dx \\ =-{鈭�}_0^2\ln {\left(x+1\right)}^2{\left[y\right]}_0^{2}dx \\ =-{鈭�}_0^{2}2\ln {\left(x+1\right)}^{2}dx \\ =-{\left[2\left(x+1\right)\left(-2+\ln {\left(x+1\right)}^2\right)\right]}_0^2 \\ =-\left[2\left(2+1\right)\left(-2+\ln {\left(2+1\right)}^2\right)-2\left(0+1\right)\left(-2+\ln {\left(0+1\right)}^2\right)\right] \\ =-\left[6\left(-2+\ln \left(9\right)\right)-2\left(-2+\ln \left(1\right)\right)\right] \\ =-\left[6\left(-2+\ln \left(9\right)\right)-2\left(-2+0\right)\right] \\ =-\left[6\left(-2+\ln \left(9\right)\right)+4\right] \\ =-\left[-12+6\ln \left(9\right)+4\right] \\ =-\left[-8+6\ln \left(9\right)\right] \\ =8-6\ln \left(9\right) \end{gathered}$$