Question

use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. F = (x - y)i + ( y - x)jC: The square bounded by x = 0, x = 1, y = 0, and y = 1

use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. \(F = (x - y)i + ( y - x)j\)
C: The square bounded by \(x = 0, x = 1, y = 0,\ and\ y = 1\)

Answers (1)

2021-02-01

Step 1
Green theorem (Outward flux):
\(\displaystyle\oint_{{C}}{F}.{n}{d}{s}=\int\int_{{R}}{\left(\frac{{\partial{M}}}{{\partial{x}}}+\frac{{\partial{N}}}{{\partial{y}}}\right)}{d}{x}{\left.{d}{y}\right.}\)
Green's theorem (circulation form):
\(\displaystyle\oint_{{C}}{F}.{d}{r}=\int\int_{{R}}{\left(\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}\right)}{d}{x}{\left.{d}{y}\right.}\)
Step 2
Given that,
\(F=(x−y)i+(y−x)j\)
Step 3
Evaluate
Outward flux:
\(\displaystyle\oint_{{C}}{F}.{n}{d}{s}=\int\int_{{R}}{\left(\frac{{\partial{M}}}{{\partial{x}}}+\frac{{\partial{N}}}{{\partial{y}}}\right)}{d}{x}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{0}}^{{1}}}{\left(\frac{\partial}{{\partial{x}}}{\left({x}-{y}\right)}+\frac{\partial}{{\partial{y}}}{\left({y}-{x}\right)}\right)}{d}{x}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{0}}^{{1}}}{\left({1}+{1}\right)}{d}{x}{\left.{d}{y}\right.}\)
\(\displaystyle={2}{\int_{{0}}^{{1}}}{\left.{d}{x}\right.}{\int_{{0}}^{{1}}}{\left.{d}{y}\right.}\)
\(\displaystyle={2}{{\left({x}\right)}_{{0}}^{{1}}}{{\left({y}\right)}_{{0}}^{{1}}}\)
\(=2\)
Hence outward flux is 2.
Step 4
Evaluate
Counterclockwise Circulation:
\(\displaystyle\oint_{{C}}{F}.{d}{r}=\int\int_{{R}}{\left(\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}\right)}{d}{x}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{0}}^{{1}}}{\left(\frac{\partial}{{\partial{x}}}{\left({y}-{x}\right)}-\frac{\partial}{{\partial{y}}}{\left({x}-{y}\right)}\right)}{d}{x}{\left.{d}{y}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{0}}^{{1}}}{\left(-{1}+{1}\right)}{d}{x}{\left.{d}{y}\right.}\)
\(=0\)
Hence Counterclockwise Circulation: is 0.

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