use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. F = (y^2 - x^2)i + (x^2 + y^2)j C: The triangle bounded by y = 0, x = 3, and y = x

Question
use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. \(\displaystyle{F}={\left({y}^{{2}}-{x}^{{2}}\right)}{i}+{\left({x}^{{2}}+{y}^{{2}}\right)}{j}\)
C: The triangle bounded by y = 0, x = 3, and y = x

Answers (1)

2020-11-02
Step 1
Given: \(\displaystyle{F}={\left({y}^{{2}}−{x}^{{2}}\right)}{i}+{\left({x}^{{2}}+{y}^{{2}}\right)}{j}\)
C: The triangle bounded by y=0, x=3, and y=x.
Step 2
Counterclockwise circulation:
To find counterclockwise circulation use, \(\displaystyle\oint_{{C}}{F}\cdot{T}{d}{s}=\int\int_{{R}}{\left(\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\).
Here, \(\displaystyle{F}={\left({y}^{{2}}-{x}^{{2}}\right)}{i}+{\left({x}^{{2}}+{y}^{{2}}\right)}{j}\)
\(\displaystyle\Rightarrow{M}={y}^{{2}}-{x}^{{2}},{N}={x}^{{2}}+{y}^{{2}}\)
\(\displaystyle\Rightarrow\frac{{\partial{M}}}{{\partial{y}}}={2}{y},\frac{{\partial{N}}}{{\partial{x}}}={2}{x}\)
We have \(\displaystyle{\left\lbrace{R}{\mid}{0}\le{x}\le{3},{0}\le{y}\le{x}\right\rbrace}\).
Therefore, the counterclockwise circulation is given by
\(\displaystyle\oint_{{C}}{F}\cdot{T}{d}{s}={\int_{{0}}^{{3}}}{\int_{{0}}^{{x}}}{\left({2}{x}-{2}{y}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{\left[{\int_{{0}}^{{x}}}{\left({2}{x}-{2}{y}\right)}{\left.{d}{y}\right.}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{{\left[{2}{x}{y}-\frac{{{2}{y}^{{2}}}}{{2}}\right]}_{{0}}^{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{{\left[{2}{x}{y}-{y}^{{2}}\right]}_{{0}}^{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{\left[{\left({2}{x}^{{2}}-{x}^{{2}}\right)}-{0}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{x}^{{2}}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left[\frac{{x}^{{3}}}{{3}}\right]}_{{0}}^{{3}}}\)
\(\displaystyle=\frac{{{\left({3}\right)}^{{3}}}}{{3}}-{0}\)
=9
Hence, the counterclockwise circulation = 9.
Step 3
Outward flux:
To find outward flux use, \(\displaystyle\oint_{{C}}{F}\cdot{n}{d}{s}=\int\int_{{R}}{\left(\frac{{\partial{M}}}{{\partial{x}}}+\frac{{\partial{N}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\).
Here, \(\displaystyle{F}={\left({y}^{{2}}-{x}^{{2}}\right)}{i}+{\left({x}^{{2}}+{y}^{{2}}\right)}{j}\)
\(\displaystyle\Rightarrow{M}={y}^{{2}}-{x}^{{2}},{N}={x}^{{2}}+{y}^{{2}}\)
\(\displaystyle\Rightarrow\frac{{\partial{M}}}{{\partial{x}}}=-{2}{x},\frac{{\partial{N}}}{{\partial{y}}}={2}{y}\)
We have \(\displaystyle{\left\lbrace{R}{\mid}{0}\le{x}\le{3},{0}\le{y}\le{x}\right\rbrace}\).
Therefore, the outward flux is given by
\(\displaystyle\oint_{{C}}{F}\cdot{n}{d}{s}={\int_{{0}}^{{3}}}{\int_{{0}}^{{x}}}{\left(-{2}{x}+{2}{y}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{\left[{\int_{{0}}^{{x}}}{\left(-{2}{x}+{2}{y}\right)}{\left.{d}{y}\right.}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{{\left[-{2}{x}{y}+\frac{{{2}{y}^{{2}}}}{{2}}\right]}_{{0}}^{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{{\left[-{2}{x}{y}+{y}^{{2}}\right]}_{{0}}^{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}{\left[{\left(-{2}{x}^{{2}}+{x}^{{2}}\right)}-{0}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{3}}}-{x}^{{2}}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left[-\frac{{x}^{{3}}}{{3}}\right]}_{{0}}^{{3}}}\)
\(\displaystyle=-\frac{{\left({3}\right)}^{{3}}}{{3}}-{0}\)
=-9
Hence, the outward flux =- 9.
0

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