Step 1

The give integral is \(\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}\) ….(1)

where C is the boundary of the region in first quadrant bounded by \(\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}\).

From greens theorem we know that

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

Comparing we get

\(\displaystyle{P}={e}^{{x}}+{y}^{{2}}{\quad\text{and}\quad}{Q}={e}^{{y}}+{x}^{{2}}\)

So,

\(\displaystyle\frac{{\partial{P}}}{{\partial{y}}}={2}{y}{\quad\text{and}\quad}\frac{{\partial{Q}}}{{\partial{x}}}={2}{x}\) ...(3)

Step 2

From (1), (2) and (3) we get

\(\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}=\int\int_{{D}}{2}{\left({x}-{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\) ...(4)

Now, we will find the region of integration.

\(\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}\)

this gives

\(\displaystyle{x}^{{2}}={4}\)

\(x=+-2\)

Since the region lies on first quadrant so we leave negative sign.

Therefore, x varies from 0 to 2 and y varies from 4 to x2

Now, from (4)

Step 4

\(\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}={\int_{{{x}={0}}}^{{2}}}{\int_{{{y}={4}}}^{{{x}^{{2}}}}}{2}{\left({x}-{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

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

\(\displaystyle={2}{\int_{{{x}={0}}}^{{2}}}{\left({x}\times{x}^{{2}}-\frac{{{\left({x}^{{2}}\right)}^{{2}}}}{{2}}\right)}-{4}{x}+{8}{)}{\left.{d}{x}\right.}\)

\(\displaystyle={2}{\int_{{{x}={0}}}^{{2}}}{\left({x}^{{3}}-\frac{{x}^{{4}}}{{2}}-{4}{x}+{8}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle={2}{{\left(\frac{{x}^{{4}}}{{4}}-\frac{{x}^{{5}}}{{10}}-{2}{x}^{{2}}+{8}{x}\right)}_{{0}}^{{2}}}\)

\(\displaystyle={2}{\left(\frac{{{4}-{16}}}{{{5}-{8}+{16}}}\right)}\)

\(\displaystyle={2}{\left({12}-\frac{{16}}{{5}}\right)}\)

\(\displaystyle={2}{\left(\frac{{{60}-{16}}}{{5}}\right)}\)

\(\displaystyle={2}{\left(\frac{{44}}{{5}}\right)}\)

\(\displaystyle=\frac{{88}}{{5}}\)