91Ó°ÊÓ

The objective is to evaluate the integral ${∫}_{C}F.dr$using the Green's theorem where $F\left(x,y\right)-4x^{2}yi+4x{y}^{2}j$.

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,

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

For the vector field $F\left(x,y\right)=-4x^{2}yi+4x{y}^{2}j.$

$F_1\left(x,y\right)=-4x^{2}y$and $F_2\left(x,y\right)=4x{y}^2$

Now finding $\frac{∂F_2}{∂x}$and $\frac{∂F_1}{∂y}$.

Then,

$$\begin{gathered} \frac{∂F_2}{∂x}=\frac{∂}{∂x}\left(4x{y}^2\right) \\ =4y^2 \end{gathered}$$

and,

$$\begin{gathered} \frac{∂F_1}{∂y}=\frac{∂}{∂y}\left(-4x^{2}y\right) \\ =-4x^2 \end{gathered}$$

Use Green's Theorem to evaluate the integral ${∫}_{C}F.dr$as follows,

$$\begin{gathered} {∫}_{C}F.dr=∫{∫}_C\left(\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y}\right)dA \\ =∫{∫}_C\left(4y^2-\left(-4x^2\right)\right)dA \\ =∫{∫}_C\left(4y^2+4x^2\right)dA \\ =∫{∫}_{C}4\left(x^2+y^2\right)dA..........\left(2\right) \end{gathered}$$

Changing integral (2) to polar coordinates.

$x=r\cos \left(\mathrm{θ}\right),y=r\sin \left(\mathrm{θ}\right),x^2+y^2=r^2,$and

$dA=rdrd\mathrm{θ}$.

Evaluate the integral,

$$\begin{gathered} {∫}_{C}F.dr=∫{∫}_{R}4\left(x^2+y^2\right)dA \\ ={∫}_0^{2\mathrm{π}}{∫}_0^{1}4\left(r^2\right)rrdrd\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}{∫}_0^{1}4r^{3}drrd\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}\left[{∫}_0^{1}4r^{3}dr\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}{\left[r^4\right]}_0^{1}d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}\left[1^4-0^4\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}d\mathrm{θ} \\ ={\left[\mathrm{θ}\right]}_0^{2\mathrm{π}} \\ =2\mathrm{π}-0 \\ =2\mathrm{π} \end{gathered}$$