Step 1

Given that, \(\displaystyle{F}={\left[{x}^{{2}}{y}^{{2}},-\frac{{x}}{{y}^{{2}}}\right]}\) and \(\displaystyle{R}:{1}\le{x}^{{2}}+{y}^{{2}}\le{4},{x}\ge{0},{y}\ge{x}\).

Here,

\(\displaystyle{F}_{{1}}={x}^{\gamma}{y}^{\gamma}{1}\le{r}\le{2}\)

\(\displaystyle{F}_{{2}}=-\frac{{x}}{{y}^{{2}}}\frac{\pi}{{4}}\le{0}\le\frac{\pi}{{2}}\)

\(\displaystyle\frac{{{d}{F}_{{2}}}}{{{\left.{d}{x}\right.}}}=-\frac{{1}}{{y}^{{2}}}\)

\(\displaystyle\frac{{{d}{F}_{{1}}}}{{{\left.{d}{y}\right.}}}={2}{y}{x}^{{2}}\)

Step 2

Then, by Green's theorem,

\(\displaystyle\int_{{C}}{F}{d}{r}=\int\int_{{R}}{\left(\frac{{{d}{F}_{{2}}}}{{{\left.{d}{x}\right.}}}-\frac{{{d}{F}_{{1}}}}{{{\left.{d}{y}\right.}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

Consider the polar coordinates.

\(\displaystyle{x}={r}{\cos{{0}}}\)

\(\displaystyle{y}={r}{\sin{{0}}}\)

j = r

That is, dx dy = r dr d0

Step 3

Evaluate the integral as follows.

\(\displaystyle\int_{{C}}{F}{d}{r}={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\int_{{1}}^{{2}}}{\left[-\frac{{1}}{{{r}^{{2}}{{\sin}^{{2}}{0}}}}-{2}{r}^{{3}}{{\cos}^{{2}}{0}}{\sin{{0}}}\right]}{r}{d}{r}{d}{0}\)

\(\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{{\left[\frac{{-{\ln{{r}}}}}{{{{\sin}^{{2}}{0}}}}-\frac{{{2}{r}^{{5}}{{\cos}^{{2}}{0}}}}{{5}}{\sin{{0}}}\right]}_{{1}}^{{2}}}{d}{0}\)

\(\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\left(\frac{{-{\ln{{2}}}}}{{{{\sin}^{{2}}{0}}}}-\frac{{{2}{\left({31}\right)}{{\cos}^{{2}}{0}}}}{{5}}\right)}{\sin{{0}}}{)}{d}{0}\)

\(\displaystyle={\ln{{2}}}{\left(\frac{{\cot{\pi}}}{{4}}-\frac{{\cot{\pi}}}{{4}}\right)}+\frac{{62}}{{5.3}}{\left({{\cos}^{{3}}{\left(\frac{\pi}{{2}}\right)}}-{{\cos}^{{4}}{\left(\frac{\pi}{{4}}\right)}}\right)}\)

\(\displaystyle=-{\ln{{2}}}=\frac{{31}}{{{15}\sqrt{{{2}}}}}\)

Therefore, \(\displaystyle\int_{{C}}{F}{d}{r}=-{\ln{{2}}}-\frac{{31}}{{{15}\sqrt{{{2}}}}}\)

Given that, \(\displaystyle{F}={\left[{x}^{{2}}{y}^{{2}},-\frac{{x}}{{y}^{{2}}}\right]}\) and \(\displaystyle{R}:{1}\le{x}^{{2}}+{y}^{{2}}\le{4},{x}\ge{0},{y}\ge{x}\).

Here,

\(\displaystyle{F}_{{1}}={x}^{\gamma}{y}^{\gamma}{1}\le{r}\le{2}\)

\(\displaystyle{F}_{{2}}=-\frac{{x}}{{y}^{{2}}}\frac{\pi}{{4}}\le{0}\le\frac{\pi}{{2}}\)

\(\displaystyle\frac{{{d}{F}_{{2}}}}{{{\left.{d}{x}\right.}}}=-\frac{{1}}{{y}^{{2}}}\)

\(\displaystyle\frac{{{d}{F}_{{1}}}}{{{\left.{d}{y}\right.}}}={2}{y}{x}^{{2}}\)

Step 2

Then, by Green's theorem,

\(\displaystyle\int_{{C}}{F}{d}{r}=\int\int_{{R}}{\left(\frac{{{d}{F}_{{2}}}}{{{\left.{d}{x}\right.}}}-\frac{{{d}{F}_{{1}}}}{{{\left.{d}{y}\right.}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

Consider the polar coordinates.

\(\displaystyle{x}={r}{\cos{{0}}}\)

\(\displaystyle{y}={r}{\sin{{0}}}\)

j = r

That is, dx dy = r dr d0

Step 3

Evaluate the integral as follows.

\(\displaystyle\int_{{C}}{F}{d}{r}={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\int_{{1}}^{{2}}}{\left[-\frac{{1}}{{{r}^{{2}}{{\sin}^{{2}}{0}}}}-{2}{r}^{{3}}{{\cos}^{{2}}{0}}{\sin{{0}}}\right]}{r}{d}{r}{d}{0}\)

\(\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{{\left[\frac{{-{\ln{{r}}}}}{{{{\sin}^{{2}}{0}}}}-\frac{{{2}{r}^{{5}}{{\cos}^{{2}}{0}}}}{{5}}{\sin{{0}}}\right]}_{{1}}^{{2}}}{d}{0}\)

\(\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\left(\frac{{-{\ln{{2}}}}}{{{{\sin}^{{2}}{0}}}}-\frac{{{2}{\left({31}\right)}{{\cos}^{{2}}{0}}}}{{5}}\right)}{\sin{{0}}}{)}{d}{0}\)

\(\displaystyle={\ln{{2}}}{\left(\frac{{\cot{\pi}}}{{4}}-\frac{{\cot{\pi}}}{{4}}\right)}+\frac{{62}}{{5.3}}{\left({{\cos}^{{3}}{\left(\frac{\pi}{{2}}\right)}}-{{\cos}^{{4}}{\left(\frac{\pi}{{4}}\right)}}\right)}\)

\(\displaystyle=-{\ln{{2}}}=\frac{{31}}{{{15}\sqrt{{{2}}}}}\)

Therefore, \(\displaystyle\int_{{C}}{F}{d}{r}=-{\ln{{2}}}-\frac{{31}}{{{15}\sqrt{{{2}}}}}\)