91影视

The objective is to evaluate the integral${鈭�}_{C}F.dr$by using the Green's theorem whererole="math" localid="1650478060605" $F\left(x,y\right)=\left(y^3+3x^{2}y\right)i+2xj$.

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 $a鈮�t鈮�b$.

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,

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

For the vector field $F\left(x,y\right)=\left(y^3+3x^{2}y\right)i+2xj$.

$$\begin{gathered} F_1\left(x,y\right)=y^3+3x^{2}y \\ {F}_2\left(x,y\right)=2x \end{gathered}$$

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

Then,

$$\begin{gathered} \frac{鈭�F_2}{鈭�x}=\frac{鈭�}{鈭�x}\left(2x\right) \\ =2 \end{gathered}$$

and,

$$\begin{gathered} \frac{鈭�F_1}{鈭�y}=\frac{鈭�}{鈭�y}\left(y^3+3x^{2}y\right) \\ =3y^2+3x^2 \end{gathered}$$

Use Green's theorem to evaluate the integral ${鈭�}_{C}F.dr$as follows,

$$\begin{gathered} {鈭�}_{C}F.dr=鈭�{鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA \\ =鈭�{鈭�}_R\left(0-\left(3y^2+3x^2\right)\right)dA \\ =鈭�{鈭�}_R-\left(3y^2+3x^2\right)dA \\ =-3鈭�{鈭�}_R\left(x^2+y^2\right)dA..........\left(2\right) \end{gathered}$$

Changing integral (2) to polar coordinates,

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

$dA=rdrd胃$.

Evaluate the integral,

$$\begin{gathered} {鈭�}_{C}F.dr=-3鈭�{鈭�}_R\left(x^2+y^2\right)dA \\ =-3{鈭�}_0^{\mathrm{蟺}}{鈭�}_0^1\left(r^2\right)rdrd胃 \\ =-3{鈭�}_0^{\mathrm{蟺}}{鈭�}_0^1{r}^{3}drd胃 \\ =-3{鈭�}_0^{\mathrm{蟺}}\left[{鈭�}_0^1{r}^{3}dr\right]d胃 \\ =-3{鈭�}_0^{\mathrm{蟺}}{\left[\frac{r^4}{4}\right]}_0^{1}d胃 \\ =-3{鈭�}_0^{\mathrm{蟺}}\left[\frac{1}{4}-\frac{0}{4}\right]d胃 \\ =-3{鈭�}_0^{\mathrm{蟺}}\frac{1}{4}d胃 \\ =-\frac{3}{4}{鈭�}_0^{\mathrm{蟺}}d胃 \\ =-\frac{3}{4}{\left[胃\right]}_0^{\mathrm{蟺}} \\ =-\frac{3}{4}\left[\mathrm{蟺}-0\right] \\ =-\frac{3\mathrm{蟺}}{4} \end{gathered}$$