# Use Green's Theorem to find int_C vec(F) * d vec(r) where vec(F) = <<y^3, - x^3>> and C is the circle x^2+y^2=3.

Use Green's Theorem to find $$\displaystyle\int_{{C}}\vec{{{F}}}\cdot{d}\vec{{{r}}}$$ where $$\displaystyle\vec{{{F}}}={\left\langle{y}^{{3}},-{x}^{{3}}\right\rangle}$$ and C is the circle $$\displaystyle{x}^{{2}}+{y}^{{2}}={3}$$.

2020-11-09

Step 1
The given vector is
$$\displaystyle{F}={y}^{{3}}{i}-{x}^{{3}}{j}$$
Step 2
$$\displaystyle{I}=\int_{{C}}\vec{{{F}}}\cdot{d}\vec{{{r}}}$$
but d $$\displaystyle\vec{{{r}}}={\left.{d}{x}\right.}\hat{{{i}}}+{\left.{d}{y}\right.}\hat{{{j}}}$$
$$\displaystyle{I}=\int_{{C}}{\left({y}^{{3}}\hat{{{i}}}-{x}^{{3}}\hat{{{j}}}\right)}.{\left({\left.{d}{x}\right.}\hat{{{i}}}+{\left.{d}{y}\right.}\hat{{{j}}}\right)}$$
$$\displaystyle=\int_{{C}}{\left({y}^{{3}}{\left.{d}{x}\right.}-{x}^{{3}}{\left.{d}{y}\right.}\right)}$$
The given circle
$$\displaystyle{x}^{{2}}+{y}^{{2}}={3}$$
$$\displaystyle{x}^{{2}}+{y}^{{2}}={\left(\sqrt{{{3}}}\right)}^{{2}}$$
$$\displaystyle{r}{a}{d}{i}{u}{s}=\sqrt{{{3}}}$$
Apply Green Theorem
$$\displaystyle\int_{{C}}{\left({M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}\right)}=\int\int{\left(\frac{{\partial{N}}}{{\partial{x}}}-\frac{{{D}{e}{l}{M}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle\Rightarrow{I}=\int\int{\left(-{3}{x}^{{2}}-{3}{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle=\int\int-{3}{\left({x}^{{2}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle=-{3}\int\int{\left({x}^{{2}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle=-{9}\int\int{\left.{d}{x}\right.}{\left.{d}{y}\right.}\because{x}^{{2}}+{y}^{{2}}={3}$$
$$=-9 x$$ Area of circle
$$\displaystyle=-{9}{x}\pi{r}^{{2}}$$
$$\displaystyle=-{9}{x}\pi{x}{\left(\sqrt{{{3}}}\right)}^{{2}}$$
$$\displaystyle{I}=-{27}\pi$$