use Green’s Theorem to find the counterclockwise circulation and outward flux for the field F...

Step 1
Given: ${\displaystyle F=(y^2-x^2)i+(x^2+y^2)j}$
C: The triangle bounded by y=0, x=3, and y=x.
Step 2
Counterclockwise circulation:
To find counterclockwise circulation use, ${\displaystyle {\oint }_{C}F\cdot Tds=\int {\int }_R(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dxdy}$.
Here, ${\displaystyle F=(y^2-x^2)i+(x^2+y^2)j}$
${\displaystyle \Rightarrow M=y^2-x^2,N=x^2+y^2}$
${\displaystyle \Rightarrow \frac{\partial M}{\partial y}=2y,\frac{\partial N}{\partial x}=2x}$
We have ${\displaystyle \{R\mid 0\le x\le 3,0\le y\le x\}}$.
Therefore, the counterclockwise circulation is given by
${\displaystyle {\oint }_{C}F\cdot Tds={\int }_0^3{\int }_0^x(2x-2y)dydx}$
${\displaystyle ={\int }_0^3\left[{\int }_0^x(2x-2y)dy\right]dx}$
${\displaystyle ={\int }_0^3{[2xy-\frac{2y^2}{2}]}_0^{x}dx}$
${\displaystyle ={\int }_0^3{[2xy-y^2]}_0^{x}dx}$
${\displaystyle ={\int }_0^3[(2x^2-x^2)-0]dx}$
${\displaystyle ={\int }_0^3{x}^{2}dx}$
${\displaystyle ={\left[\frac{x^3}{3}\right]}_0^3}$
${\displaystyle =\frac{{\left(3\right)}^3}{3}-0}$
=9
Hence, the counterclockwise circulation = 9.
Step 3
Outward flux:
To find outward flux use, ${\displaystyle {\oint }_{C}F\cdot nds=\int {\int }_R(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y})dxdy}$.
Here, ${\displaystyle F=(y^2-x^2)i+(x^2+y^2)j}$

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