# 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

use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F and the curve C. $F=\left({y}^{2}-{x}^{2}\right)i+\left({x}^{2}+{y}^{2}\right)j$
C: The triangle bounded by y = 0, x = 3, and y = x
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Step 1
Given: $F=\left({y}^{2}-{x}^{2}\right)i+\left({x}^{2}+{y}^{2}\right)j$
C: The triangle bounded by y=0, x=3, and y=x.
Step 2
Counterclockwise circulation:
To find counterclockwise circulation use, ${\oint }_{C}F\cdot Tds=\int {\int }_{R}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dxdy$.
Here, $F=\left({y}^{2}-{x}^{2}\right)i+\left({x}^{2}+{y}^{2}\right)j$
$⇒M={y}^{2}-{x}^{2},N={x}^{2}+{y}^{2}$
$⇒\frac{\partial M}{\partial y}=2y,\frac{\partial N}{\partial x}=2x$
We have $\left\{R\mid 0\le x\le 3,0\le y\le x\right\}$.
Therefore, the counterclockwise circulation is given by
${\oint }_{C}F\cdot Tds={\int }_{0}^{3}{\int }_{0}^{x}\left(2x-2y\right)dydx$
$={\int }_{0}^{3}\left[{\int }_{0}^{x}\left(2x-2y\right)dy\right]dx$
$={\int }_{0}^{3}{\left[2xy-\frac{2{y}^{2}}{2}\right]}_{0}^{x}dx$
$={\int }_{0}^{3}{\left[2xy-{y}^{2}\right]}_{0}^{x}dx$
$={\int }_{0}^{3}\left[\left(2{x}^{2}-{x}^{2}\right)-0\right]dx$
$={\int }_{0}^{3}{x}^{2}dx$
$={\left[\frac{{x}^{3}}{3}\right]}_{0}^{3}$
$=\frac{{\left(3\right)}^{3}}{3}-0$
=9
Hence, the counterclockwise circulation = 9.
Step 3
Outward flux:
To find outward flux use, ${\oint }_{C}F\cdot nds=\int {\int }_{R}\left(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}\right)dxdy$.
Here, $F=\left({y}^{2}-{x}^{2}\right)i+\left({x}^{2}+{y}^{2}\right)j$