Step 1

Green theorem (Outward flux):

\(\displaystyle\oint_{{C}}{F}.{n}{d}{s}=\int\int_{{R}}{\left(\frac{{\partial{M}}}{{\partial{x}}}+\frac{{\partial{N}}}{{\partial{y}}}\right)}{d}{x}{\left.{d}{y}\right.}\)

Green's theorem (circulation form):

\(\displaystyle\oint_{{C}}{F}.{d}{r}=\int\int_{{R}}{\left(\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}\right)}{d}{x}{\left.{d}{y}\right.}\)

Step 2

Given that,

\(F=(x−y)i+(y−x)j\)

Step 3

Evaluate

Outward flux:

\(\displaystyle\oint_{{C}}{F}.{n}{d}{s}=\int\int_{{R}}{\left(\frac{{\partial{M}}}{{\partial{x}}}+\frac{{\partial{N}}}{{\partial{y}}}\right)}{d}{x}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{0}}^{{1}}}{\left(\frac{\partial}{{\partial{x}}}{\left({x}-{y}\right)}+\frac{\partial}{{\partial{y}}}{\left({y}-{x}\right)}\right)}{d}{x}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{0}}^{{1}}}{\left({1}+{1}\right)}{d}{x}{\left.{d}{y}\right.}\)

\(\displaystyle={2}{\int_{{0}}^{{1}}}{\left.{d}{x}\right.}{\int_{{0}}^{{1}}}{\left.{d}{y}\right.}\)

\(\displaystyle={2}{{\left({x}\right)}_{{0}}^{{1}}}{{\left({y}\right)}_{{0}}^{{1}}}\)

\(=2\)

Hence outward flux is 2.

Step 4

Evaluate

Counterclockwise Circulation:

\(\displaystyle\oint_{{C}}{F}.{d}{r}=\int\int_{{R}}{\left(\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}\right)}{d}{x}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{0}}^{{1}}}{\left(\frac{\partial}{{\partial{x}}}{\left({y}-{x}\right)}-\frac{\partial}{{\partial{y}}}{\left({x}-{y}\right)}\right)}{d}{x}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{0}}^{{1}}}{\left(-{1}+{1}\right)}{d}{x}{\left.{d}{y}\right.}\)

\(=0\)

Hence Counterclockwise Circulation: is 0.