91Ó°ÊÓ

The objective is to evaluate the integral ${∫}_{C}F.dr$by using Green's Theorem, where $C$is the boundary of the region in the third quadrant bounded by the x-axis, for$x∈\left[-1,0\right]$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 $a≤t≤b$.

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,

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

For the vector field, $F\left(x,y\right)=x{e}^{x^2+y^2}i+\left(10\sin \left(y\right)+x\right)j$

$$\begin{gathered} F_1\left(x,y\right)=x{e}^{x^2+y^2} \\ {F}_2\left(x,y\right)=10\sin \left(y\right)+x \end{gathered}$$

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

Then,

$$\begin{gathered} \frac{∂F_2}{∂x}=\frac{∂}{∂x}\left(10\sin \left(y\right)+x\right) \\ =1 \end{gathered}$$

and,

$$\begin{gathered} \frac{∂F_1}{∂y}=\frac{∂}{∂y}\left(x{e}^{x^2+y^2}\right) \\ =2xy{e}^{x^2+y^2} \end{gathered}$$

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

$$\begin{gathered} {∫}_{C}F.dr=∫{∫}_R\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA \\ =∫{∫}_R\left(1-2xy{e}^{x^2+y^2}\right)dA...............\left(2\right) \end{gathered}$$

Here, the boundary curve $C$is the boundary of the region in the third quadrant bounded by $y=x$and the x-axis, for $x∈\left[-1,0\right]$, traversed counterclockwise.

So the region $R$bounded by $y=x$and the x-axis for $x∈\left[-1,0\right]$in the third quadrant as shown in the following figure,

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

$R=\left\{\left(x,y\right)I-1≤x≤0,x≤y≤0\right\}$.

Then, evaluate the integral (2) as follows,

$$\begin{gathered} {∫}_{C}F.dr=∫{∫}_R\left(1-2yx{e}^{x^2+y^2}\right)dA \\ ={∫}_{-1}^0{∫}_x^0\left(1-2yx{e}^{x^2+y^2}\right)dydx \\ ={∫}_{-1}^0\left[{∫}_x^0\left(1-2yx{e}^{x^2+y^2}\right)dy\right]dx \\ ={∫}_{-1}^0{\left(y-x{e}^{x^2+y^2}\right)}_x^{0}dx \\ ={∫}_{-1}^0\left(-x{e}^{x^2}-x+x{e}^{2x^2}\right)dx \\ ={\left[-\frac{1}{2}{e}^{x^2}-\frac{x^2}{2}+\frac{1}{4}{e}^{2x^2}\right]}_{-1}^0 \\ =\frac{1}{4}\left(1+2e-e^2\right) \end{gathered}$$