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May 9th, 2017, 09:41 AM   #1
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Green's Theorem and Line Integrals

Hey Everybody,
So I'm doing some line integrals and checking my work with Green's Theorem, but I'm not getting the same answer. Here's the problem:
Evaluate the line integral F · dr over C, where C is the counterclockwise oriented triangle with vertices (0, 0),(2, 0), and (2, 2) and
$\displaystyle
F(x, y) = <5 − 2xy − y^2, 2xy − x^2>
$
The answer is 16/3, which I get when I use Green's Theorem, but when I try to directly calculate the line integral I get 14/3. Here is my process:

First I parametrize the region C.
x = x
y = x
where 0<= x <= 2

Then I solve:
r(x) = <x,x>
r'(x) = <1, 1>
F(r(x)) = <5 - 2x^2 - x^2, 2x^2 - x^2>
F(r(x)) = <5 - 3x^2, x^2>
F(r(x)) · r'(x) = 5 - x^2

The integral of that with respect to x is 14/3

Please let me know what I am doing wrong. Thanks for your help!
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May 9th, 2017, 11:33 AM   #2
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I'm not really sure what you're doing there.

There are three pieces to the curve being integrated over.

$r1=(2t,0),~t \in [0,1]$
$r2=(2,0)+(0,2t),~t \in [0,1]$
$r3=(2,2)+(-2t,-2t),~t \in [0,1]$

$\dot{r1} = (2,0)$
$\dot{r2}=(0,2)$
$\dot{r3}=(-2,-2)$

$f=(5-2xy-y^2,~2xy - x^2)$

on $r1,~f = (5, -4t^2)$

$\dot{r1}\cdot f = 10$

$\ell_1 = \displaystyle \int_0^1~10~dt = 10$

I leave it to you to finish the other two pieces.

$\dfrac {16}{3}$ is the correct answer.

Last edited by romsek; May 9th, 2017 at 11:54 AM.
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May 9th, 2017, 01:34 PM   #3
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Just an FYI, that dot above r1, r2, and r3 means the first derivative. I didn't know that for the longest time, until I started a physics class.

I'd also recommend in the future you don't try to parameterize with a variable that already exists in the original function, it'll only lead to tears in the future. It seems to be common convention when doing line integrals of one parameter to use variable t.

Romsek has it dead on with his solution. Remember when doing line integrals you have to follow the line itself and in this example, it tells you have follow it in a counter-clockwise direction. When doing line integral of force field, orientation is very very important (it is the direction you are traveling with or against those forces).

Also, remember to check for conservative fields, closed loop line integral in a conservative field is always 0.

if $\displaystyle \frac{\partial f}{\partial y}=\frac{\partial g}{\partial x}$ when $\displaystyle F =(f(x,y),g(x,y))$, then the field is conservative.

$\displaystyle \oint_{C} F\cdot dr=0$, if F is conservative.

Hope this helps you.

Last edited by phrack999; May 9th, 2017 at 01:38 PM.
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