Use Green’s Theorem to evaluate the integrals:

Find ${\int }_{C}F.dr$, where role="math" localid="1650475436711" $F\left(x,y\right)=-4x^{2}y+4x{y}^{2}j$and role="math" $C$is the unit circle traversed counterclockwise.

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

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,

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

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

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

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

Then,

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

and,

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

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

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

Changing integral (2) to polar coordinates.

$x=r\cos \left(\theta \right),y=r\sin \left(\theta \right),x^2+y^2=r^2,$and

$dA=rdrd\theta$.

Evaluate the integral,

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