Step 1
We are given that \(\displaystyle{F}{\left({x},{y}\right)}={{\tan}^{{-{1}}}{\left(\frac{{y}}{{x}}\right)}}{i}+{\ln{{\left({x}^{{2}}+{y}^{{2}}\right)}}}{j}\) and the curve C defined as the boundary of the given region.
We will be using Green's theorem to find the outward Flux.
Let \(\displaystyle{P}{\left({x},{y}\right)}={{\tan}^{{-{1}}}{\left(\frac{{y}}{{x}}\right)}}{\quad\text{and}\quad}{Q}{\left({x},{y}\right)}={\ln{{\left({x}^{{2}}+{y}^{{2}}\right)}}}\).
Using the first version of Green's theorem, we get
Flux = \(\displaystyle\int\int_{{R}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\).
Step 2
Now,
\(\displaystyle\frac{{\partial{Q}}}{{\partial{x}}}=\frac{{1}}{{{x}^{{2}}+{y}^{{2}}}}{2}{x}=\frac{{{2}{x}}}{{{x}^{{2}}+{y}^{{2}}}}\)
And \(\displaystyle\frac{{\partial{P}}}{{\partial{y}}}=\frac{{1}}{{x}}\frac{{1}}{{{1}+{\left(\frac{{y}}{{x}}\right)}^{{2}}}}=\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}\).
So, Flux = \(\displaystyle\int\int_{{R}}{\left(\frac{{{2}{x}}}{{{x}^{{2}}+{y}^{{2}}}}-\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle=\int\int_{{R}}\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
In polar form, \(\displaystyle{x}={r}{\cos{{0}}}{\quad\text{and}\quad}{y}={r}{\sin{{0}}}\).
Jacobean = \(\displaystyle{\left[\begin{array}{cc} {\cos}&{\sin{{0}}}\\-{r}{\sin{{0}}}&{r}{\cos{{0}}}\end{array}\right]}={r}\).
Thus, Flux = \(\displaystyle\int\int_{{R}}\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{0}}^{\pi}}{\int_{{1}}^{{2}}}\frac{{{r}{\cos{{0}}}}}{{r}^{{2}}}{r}{d}{r}{d}{0}\)
\(\displaystyle={\int_{{0}}^{\pi}}{\int_{{1}}^{{2}}}{\cos{{0}}}{d}{r}{d}{0}\)
\(\displaystyle={\int_{{0}}^{\pi}}{\cos{{0}}}{d}{0}\)
\(\displaystyle={{\left[{\sin{{0}}}\right]}_{{0}}^{\pi}}\)
=0
We are given that \(\displaystyle{F}{\left({x},{y}\right)}={{\tan}^{{-{1}}}{\left(\frac{{y}}{{x}}\right)}}{i}+{\ln{{\left({x}^{{2}}+{y}^{{2}}\right)}}}{j}\) and the curve C defined as the boundary of the given region.
We will be using Green's theorem to find the outward Flux.
Let \(\displaystyle{P}{\left({x},{y}\right)}={{\tan}^{{-{1}}}{\left(\frac{{y}}{{x}}\right)}}{\quad\text{and}\quad}{Q}{\left({x},{y}\right)}={\ln{{\left({x}^{{2}}+{y}^{{2}}\right)}}}\).
Using the first version of Green's theorem, we get
Flux = \(\displaystyle\int\int_{{R}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\).
Step 2
Now,
\(\displaystyle\frac{{\partial{Q}}}{{\partial{x}}}=\frac{{1}}{{{x}^{{2}}+{y}^{{2}}}}{2}{x}=\frac{{{2}{x}}}{{{x}^{{2}}+{y}^{{2}}}}\)
And \(\displaystyle\frac{{\partial{P}}}{{\partial{y}}}=\frac{{1}}{{x}}\frac{{1}}{{{1}+{\left(\frac{{y}}{{x}}\right)}^{{2}}}}=\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}\).
So, Flux = \(\displaystyle\int\int_{{R}}{\left(\frac{{{2}{x}}}{{{x}^{{2}}+{y}^{{2}}}}-\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle=\int\int_{{R}}\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
In polar form, \(\displaystyle{x}={r}{\cos{{0}}}{\quad\text{and}\quad}{y}={r}{\sin{{0}}}\).
Jacobean = \(\displaystyle{\left[\begin{array}{cc} {\cos}&{\sin{{0}}}\\-{r}{\sin{{0}}}&{r}{\cos{{0}}}\end{array}\right]}={r}\).
Thus, Flux = \(\displaystyle\int\int_{{R}}\frac{{x}}{{{x}^{{2}}+{y}^{{2}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{0}}^{\pi}}{\int_{{1}}^{{2}}}\frac{{{r}{\cos{{0}}}}}{{r}^{{2}}}{r}{d}{r}{d}{0}\)
\(\displaystyle={\int_{{0}}^{\pi}}{\int_{{1}}^{{2}}}{\cos{{0}}}{d}{r}{d}{0}\)
\(\displaystyle={\int_{{0}}^{\pi}}{\cos{{0}}}{d}{0}\)
\(\displaystyle={{\left[{\sin{{0}}}\right]}_{{0}}^{\pi}}\)
=0
