Step 1

Let's compare the given integrand with the standard form.

\(\displaystyle{\left({2}{x}+{\ln{{2}}}\right)}{\left.{d}{y}\right.}-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}{\left.{d}{x}\right.}={Q}.{\left.{d}{y}\right.}+{P}.{\left.{d}{x}\right.}\)

Hence, \(\displaystyle{Q}={2}{x}+{\ln{{2}}}\)

and \(\displaystyle{P}=-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}\)

Step 2

Please see the white board.

\(\displaystyle{P}=-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}\)

\(\displaystyle\Rightarrow\frac{{\partial{P}}}{{\partial{y}}}=-{16}{y}\)

\(\displaystyle{Q}={\left({2}{x}+{\ln{{2}}}\right)}\)

\(\displaystyle\Rightarrow\frac{{\partial{Q}}}{{\partial{x}}}={2}\)

Step 3

Limits of integration for x will be from x = 3 to x = 6 (the x corrdinates of the square)

and limits of integration for y will be from y = 1 to y = 4 (the y coordinates of the square)

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

\(\displaystyle=\int\int_{{R}}{\left({2}+{16}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{1}}^{{4}}}{\left({2}+{16}{y}\right)}{\left.{d}{y}\right.}{\int_{{3}}^{{6}}}{\left.{d}{x}\right.}\)

\(\displaystyle={{\left[{2}{y}+{8}{y}^{{2}}\right]}_{{1}}^{{4}}}\times{{\left[{x}\right]}_{{3}}^{{6}}}\)

\(\displaystyle={126}\times{3}={378}\)

Step 4

Hence, the final answer is:

Given integral = 378

Let's compare the given integrand with the standard form.

\(\displaystyle{\left({2}{x}+{\ln{{2}}}\right)}{\left.{d}{y}\right.}-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}{\left.{d}{x}\right.}={Q}.{\left.{d}{y}\right.}+{P}.{\left.{d}{x}\right.}\)

Hence, \(\displaystyle{Q}={2}{x}+{\ln{{2}}}\)

and \(\displaystyle{P}=-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}\)

Step 2

Please see the white board.

\(\displaystyle{P}=-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}\)

\(\displaystyle\Rightarrow\frac{{\partial{P}}}{{\partial{y}}}=-{16}{y}\)

\(\displaystyle{Q}={\left({2}{x}+{\ln{{2}}}\right)}\)

\(\displaystyle\Rightarrow\frac{{\partial{Q}}}{{\partial{x}}}={2}\)

Step 3

Limits of integration for x will be from x = 3 to x = 6 (the x corrdinates of the square)

and limits of integration for y will be from y = 1 to y = 4 (the y coordinates of the square)

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

\(\displaystyle=\int\int_{{R}}{\left({2}+{16}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{1}}^{{4}}}{\left({2}+{16}{y}\right)}{\left.{d}{y}\right.}{\int_{{3}}^{{6}}}{\left.{d}{x}\right.}\)

\(\displaystyle={{\left[{2}{y}+{8}{y}^{{2}}\right]}_{{1}}^{{4}}}\times{{\left[{x}\right]}_{{3}}^{{6}}}\)

\(\displaystyle={126}\times{3}={378}\)

Step 4

Hence, the final answer is:

Given integral = 378