Here we have to evalute the integral \(\displaystyle\int{\left({6}{y}+{x}\right)}{\left.{d}{x}\right.}+{\left({y}+{2}{x}\right)}{\left.{d}{y}\right.}\) over the regin bounced by \(\displaystyle{y}={x}^{2}{\quad\text{and}\quad}{x}={y}^{2}\) say the regin D.By Greens theorem

\(\displaystyle\int{\left({6}{y}+{x}\right)}{\left.{d}{x}\right.}+{\left({y}+{2}{x}\right)}{\left.{d}{y}\right.}=\int\int_{{D}}{\left(\frac{\partial}{{\partial{x}}}{\left({y}+{2}{x}\right)}-\frac{\partial}{{\partial{y}}}\right)}{\left({6}{y}+{x}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle{\int_{{0}}^{{1}}}{\int_{{{y}={x}^{2}}}^{{{y}=\sqrt{{x}}}}}{\left(\frac{\partial}{{\partial{x}}}{\left({y}+{2}{x}\right)}-\frac{\partial}{{\partial{y}}}{\left({6}{y}+{x}\right)}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{{y}={x}^{2}}}^{{{y}=\sqrt{{x}}}}}{\left({2}-{6}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle-{4}{\int_{{0}}^{{1}}}{\int_{{{y}={x}^{2}}}^{{{y}=\sqrt{{x}}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle-{4}{\int_{{0}}^{{1}}}{\left(\sqrt{{x}}-{x}^{2}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle=-{4}\frac{{{x}^{{\frac{3}{{2}}}}}}{{\frac{3}{{2}}}}-\frac{{{x}^{3}}}{{3}}{{|}_{{0}}^{{1}}}\) \(\displaystyle=-{4}{\left(\frac{2}{{3}}-\frac{1}{{3}}\right)}\)

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

\(\displaystyle\int{\left({6}{y}+{x}\right)}{\left.{d}{x}\right.}+{\left({y}+{2}{x}\right)}{\left.{d}{y}\right.}=\int\int_{{D}}{\left(\frac{\partial}{{\partial{x}}}{\left({y}+{2}{x}\right)}-\frac{\partial}{{\partial{y}}}\right)}{\left({6}{y}+{x}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle{\int_{{0}}^{{1}}}{\int_{{{y}={x}^{2}}}^{{{y}=\sqrt{{x}}}}}{\left(\frac{\partial}{{\partial{x}}}{\left({y}+{2}{x}\right)}-\frac{\partial}{{\partial{y}}}{\left({6}{y}+{x}\right)}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{{y}={x}^{2}}}^{{{y}=\sqrt{{x}}}}}{\left({2}-{6}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle-{4}{\int_{{0}}^{{1}}}{\int_{{{y}={x}^{2}}}^{{{y}=\sqrt{{x}}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle-{4}{\int_{{0}}^{{1}}}{\left(\sqrt{{x}}-{x}^{2}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle=-{4}\frac{{{x}^{{\frac{3}{{2}}}}}}{{\frac{3}{{2}}}}-\frac{{{x}^{3}}}{{3}}{{|}_{{0}}^{{1}}}\) \(\displaystyle=-{4}{\left(\frac{2}{{3}}-\frac{1}{{3}}\right)}\)

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