Kye

Answered

2020-11-01

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

C: The triangle bounded by y = 0, x = 3, and y = x

Answer & Explanation

sweererlirumeX

Expert

2020-11-02Added 91 answers

Step 1

Given:$F=({y}^{2}-{x}^{2})i+({x}^{2}+{y}^{2})j$

C: The triangle bounded by y=0, x=3, and y=x.

Step 2

Counterclockwise circulation:

To find counterclockwise circulation use,${\oint}_{C}F\cdot Tds=\int {\int}_{R}(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dxdy$ .

Here,$F=({y}^{2}-{x}^{2})i+({x}^{2}+{y}^{2})j$

$\Rightarrow M={y}^{2}-{x}^{2},N={x}^{2}+{y}^{2}$

$\Rightarrow \frac{\partial M}{\partial y}=2y,\frac{\partial N}{\partial x}=2x$

We have$\{R\mid 0\le x\le 3,0\le y\le x\}$ .

Therefore, the counterclockwise circulation is given by

${\oint}_{C}F\cdot Tds={\int}_{0}^{3}{\int}_{0}^{x}(2x-2y)dydx$

$={\int}_{0}^{3}\left[{\int}_{0}^{x}(2x-2y)dy\right]dx$

$={\int}_{0}^{3}{[2xy-\frac{2{y}^{2}}{2}]}_{0}^{x}dx$

$={\int}_{0}^{3}{[2xy-{y}^{2}]}_{0}^{x}dx$

$={\int}_{0}^{3}[(2{x}^{2}-{x}^{2})-0]dx$

$={\int}_{0}^{3}{x}^{2}dx$

$={\left[\frac{{x}^{3}}{3}\right]}_{0}^{3}$

$=\frac{{\left(3\right)}^{3}}{3}-0$

=9

Hence, the counterclockwise circulation = 9.

Step 3

Outward flux:

To find outward flux use,${\oint}_{C}F\cdot nds=\int {\int}_{R}(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y})dxdy$ .

Here,$F=({y}^{2}-{x}^{2})i+({x}^{2}+{y}^{2})j$

Given:

C: The triangle bounded by y=0, x=3, and y=x.

Step 2

Counterclockwise circulation:

To find counterclockwise circulation use,

Here,

We have

Therefore, the counterclockwise circulation is given by

=9

Hence, the counterclockwise circulation = 9.

Step 3

Outward flux:

To find outward flux use,

Here,

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