# 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

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.

### Relevant Questions

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
Apply Green’s theorem to find the outward flux for the field
$$\displaystyle{F}{\left({x},{y}\right)}={{\tan}^{{−{1}}}{\left(\frac{{y}}{{x}}\right)}}{i}+{\ln{{\left({x}^{{2}}+{y}^{{2}}\right)}}}{j}$$
Use Green’s Theorem to evaluate around the boundary curve C of the region R, where R is the triangle formed by the point (0, 0), (1, 1) and (1, 3).
Find the work done by the force field F(x,y)=4yi+2xj in moving a particle along a circle $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ from(0,1)to(1,0).
Use Stokes' Theorem to evaluate $$\displaystyle\int_{{C}}{F}\cdot{d}{r}$$ where C is oriented counterclockwise as viewed from above.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}+{y}^{{2}}\right)}{i}+{\left({y}+{z}^{{2}}\right)}{j}+{\left({z}+{x}^{{2}}\right)}{k}$$,
C is the triangle with vertices (3,0,0),(0,3,0), and (0,0,3).
Set up the integral for the divergence theorem both ways. Then find the flux.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}\hat{{{i}}}+{x}{y}\hat{{{j}}}+{2}{x}{z}\hat{{{k}}}$$
E is the cube bounded by the planes x = 0, x = 3, y = 0, y = 3, and z = 0, z = 3.
Use Green's Theorem to evaluate $$\displaystyle\int_{{C}}\vec{{{F}}}\cdot{d}\vec{{{r}}}$$ where $$\displaystyle\vec{{{F}}}{\left({x},{y}\right)}={x}{y}^{{2}}{i}+{\left({1}-{x}{y}^{{3}}\right)}{j}$$ and C is the parallelogram with vertices (-1,2), (-1,-1),(1,1)and(1,4).
The orientation of C is counterclockwise.
Let C be the ellipse contained in the xy plane whose equation is $$\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}$$, oriented clockwise. The force field F described by $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}$$, moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: $$\displaystyle{z}={4}-{4}{x}^{{2}}-{y}^{{2}},{z}\ge{0}$$ (with ascending orientation, that is, the component in the z direction equal to 1) and assuming $$\displaystyle\pi={3.14}$$, we can state what:
a) It is not necessary to apply Stokes' Theorem, as C is a closed curve and therefore W = 0.
b) Inverting the orientation of the surface S, we can apply Stokes' Theorem and conclude that W = 12.56.
c) We can apply Stokes' Theorem and conclude that W = 6.28
d) We can apply Stokes' Theorem and conclude that W = 12.56.
Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.
$$\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{\sin{{y}}}\right\rangle}$$ , S is the boundary of the region bounded by the planes x = 1, y = 0, $$\displaystyle{y}=\frac{\pi}{{2}},{z}={0}$$, and z = x.
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}$$
S is the surface of the solid bounded by the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ and the plane z = 9.
$$\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}$$