Step 1

Consider the equation

Step 2

\(\displaystyle{F}{\left({x},{y}\right)}={\left({x}{y}^{{2}}+{8}{x},{x}^{{2}}{y}-{8}{y}\right)}\)

\(\displaystyle\oint{F}.{n}{d}{s}==\int\int\nabla\times{D}{d}{A}\)

\(\displaystyle\oint{F}.{n}{d}{s}=\int\int_{{S}}{\left({8}{x}{\left.{d}{y}\right.}+{y}\right)}={3}\pi\)

\(\displaystyle\int{\left({8}{x},{y}\right)}.{\left({x},{y}\right)}{d}{s}=\int{8}{x}{\left.{d}{y}\right.}+{y}{\left(-{\left.{d}{x}\right.}\right)}\)

\(\displaystyle{x}={\cos{{0}}}{y}={\sin{{0}}}{d}{s}={r}{d}{0}\Rightarrow{d}{s}={d}{0}\) because the radius is 1

\(\displaystyle{\left.{d}{x}\right.}=-{y}{d}{0}{\left.{d}{y}\right.}={\cos{{0}}}{d}{0}\)

\(\displaystyle\oint{\left(-{y},{8}{x}\right)}.{\left.{d}{x}\right.}\)

\(\displaystyle\int\int_{{S}}\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}={3}\)

(the area) \(\displaystyle={3}{\left(\pi{\left({1}\right)}^{{2}}\right)}={3}\pi\)

Consider the equation

Step 2

\(\displaystyle{F}{\left({x},{y}\right)}={\left({x}{y}^{{2}}+{8}{x},{x}^{{2}}{y}-{8}{y}\right)}\)

\(\displaystyle\oint{F}.{n}{d}{s}==\int\int\nabla\times{D}{d}{A}\)

\(\displaystyle\oint{F}.{n}{d}{s}=\int\int_{{S}}{\left({8}{x}{\left.{d}{y}\right.}+{y}\right)}={3}\pi\)

\(\displaystyle\int{\left({8}{x},{y}\right)}.{\left({x},{y}\right)}{d}{s}=\int{8}{x}{\left.{d}{y}\right.}+{y}{\left(-{\left.{d}{x}\right.}\right)}\)

\(\displaystyle{x}={\cos{{0}}}{y}={\sin{{0}}}{d}{s}={r}{d}{0}\Rightarrow{d}{s}={d}{0}\) because the radius is 1

\(\displaystyle{\left.{d}{x}\right.}=-{y}{d}{0}{\left.{d}{y}\right.}={\cos{{0}}}{d}{0}\)

\(\displaystyle\oint{\left(-{y},{8}{x}\right)}.{\left.{d}{x}\right.}\)

\(\displaystyle\int\int_{{S}}\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}={3}\)

(the area) \(\displaystyle={3}{\left(\pi{\left({1}\right)}^{{2}}\right)}={3}\pi\)