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

\(\displaystyle=-{9}{x}\pi{r}^{{2}}\)

\(\displaystyle=-{9}{x}\pi{x}{\left(\sqrt{{{3}}}\right)}^{{2}}\)

\(\displaystyle{I}=-{27}\pi\)

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

\(\displaystyle=-{9}{x}\pi{r}^{{2}}\)

\(\displaystyle=-{9}{x}\pi{x}{\left(\sqrt{{{3}}}\right)}^{{2}}\)

\(\displaystyle{I}=-{27}\pi\)