Use Green’s Theorem to evaluate the integrals:

Find ${\int }_{C}F.dr$, where $F\left(x,y\right)=\left(x+2y\right)i+\left(x-2y\right)j$and $C$is the boundary of the region bounded by the curves $x=y^2,x=4$, traversed counterclockwise.

The objective is to evaluate the integral ${\int }_{C}F.dr$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 $a\le t\le b$.

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,

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

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 $\frac{\partial F_2}{\partial x}$and $\frac{\partial F_1}{\partial y}$.

Then,

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

and,

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

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

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

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:

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

Then, evaluate the integral (2) as follows:

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