Step 1

Given,

\(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}\), Where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3),(0,3)

Step 2

Green's theorem:

Let C be the simply connected region with piecewise smooth boundary C oriented counterclockwise,

If M and N are the first partial derivative in an open region containing C, then

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

Consider, \(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}\)

M=xy

\(\displaystyle{N}={x}^{{2}}\)

\(\displaystyle\frac{{\partial{M}}}{{\partial{y}}}={x}\)

\(\displaystyle\frac{{\partial{N}}}{{\partial{x}}}={2}{x}\)

Limits of x are \(\displaystyle{0}\le{x}\le{2}\)

Limits of y are \(\displaystyle{0}\le{y}\le{3}\)

Now the integral becomes,

\(\displaystyle{\int_{{0}}^{{3}}}{\int_{{0}}^{{2}}}{\left({2}{x}-{x}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{0}}^{{3}}}{\int_{{0}}^{{2}}}{\left({x}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{3}}}{{\left(\frac{{x}^{{2}}}{{2}}\right)}_{{0}}^{{2}}}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{3}}}{\left(\frac{{2}^{{2}}}{{2}}-\frac{{0}}{{2}}\right)}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{3}}}\frac{{1}}{{4}}{\left.{d}{y}\right.}\)

\(\displaystyle=\frac{{1}}{{4}}{\int_{{0}}^{{3}}}{\left.{d}{y}\right.}\)

\(\displaystyle=\frac{{1}}{{4}}{{\left({y}\right)}_{{0}}^{{3}}}\)

\(\displaystyle=\frac{{1}}{{4}}{\left({3}-{0}\right)}\)

\(\displaystyle=\frac{{3}}{{4}}\)

Therefore, \(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}=\frac{{3}}{{4}}\)

Given,

\(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}\), Where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3),(0,3)

Step 2

Green's theorem:

Let C be the simply connected region with piecewise smooth boundary C oriented counterclockwise,

If M and N are the first partial derivative in an open region containing C, then

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

Consider, \(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}\)

M=xy

\(\displaystyle{N}={x}^{{2}}\)

\(\displaystyle\frac{{\partial{M}}}{{\partial{y}}}={x}\)

\(\displaystyle\frac{{\partial{N}}}{{\partial{x}}}={2}{x}\)

Limits of x are \(\displaystyle{0}\le{x}\le{2}\)

Limits of y are \(\displaystyle{0}\le{y}\le{3}\)

Now the integral becomes,

\(\displaystyle{\int_{{0}}^{{3}}}{\int_{{0}}^{{2}}}{\left({2}{x}-{x}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{0}}^{{3}}}{\int_{{0}}^{{2}}}{\left({x}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{3}}}{{\left(\frac{{x}^{{2}}}{{2}}\right)}_{{0}}^{{2}}}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{3}}}{\left(\frac{{2}^{{2}}}{{2}}-\frac{{0}}{{2}}\right)}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{0}}^{{3}}}\frac{{1}}{{4}}{\left.{d}{y}\right.}\)

\(\displaystyle=\frac{{1}}{{4}}{\int_{{0}}^{{3}}}{\left.{d}{y}\right.}\)

\(\displaystyle=\frac{{1}}{{4}}{{\left({y}\right)}_{{0}}^{{3}}}\)

\(\displaystyle=\frac{{1}}{{4}}{\left({3}-{0}\right)}\)

\(\displaystyle=\frac{{3}}{{4}}\)

Therefore, \(\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}=\frac{{3}}{{4}}\)