# Green theorem of curve

Use Green's Theorem to evaluate the integral $$\int\limits_C \left(y-x\right) \mathrm dx+\left(2x-y\right) \mathrm dy$$ for the path C defined as $x=2\cos\theta \;\text{and}\; y=2\sin\theta.$

Here is my attempt at setting up the integral: 4∫ ((2cos(θ))^2-(sin(θ))^2)

Could anyone tell me if i am headed in the right direction? Thanks.

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...please...? What have you done so far? – DonAntonio Jul 4 '13 at 21:46
The answer is so easy if you read the related book once. – Mahdi Khosravi Jul 4 '13 at 21:50
To be honest, I am a completely lost on setting this up. My assumption is to substitute x in the integral as 2cos(θ) and y as 2sin(θ. Similarly changing dx to -2sin(θ)dθ and dy to 2cos(θ)dθ. – TonyP Jul 4 '13 at 22:28
@DonAntonio what is the right answer? – jain smit Jul 6 '13 at 18:44
@jainsmit, check the answer I just posted now. – DonAntonio Jul 6 '13 at 21:47

You need to use the following vector form of the Green's function

$$\oint P(x,y) dx + \oint Q(x,y) dy = \oint F.dr,$$

where $F = Pi+Qj$. Now, your contour is the parametrized circle with radius $2$

$$r = xi+yj = 2\cos(\theta)i+2\sin(\theta)j\implies dr = (-4\sin(\theta)i+4\cos(\theta)j)d\theta,$$

and

$$F = Pi + Qj = = (y-x)i+( 2x-y )j$$

$$F = (2\sin(\theta)-2\cos(\theta))i+ (4\cos(\theta)-\sin(\theta) )j .$$

So, the result follows from evaluating the integral

$$\int_{0}^{2\pi} F.dr = \dots.$$

I leave the rest for you.

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Thanks for helping. Greatly appreciate it – TonyP Jul 4 '13 at 23:42
@TonyP: You are very welcome. – Mhenni Benghorbal Jul 5 '13 at 0:05
what did you get for this answer? just wanted to see if i got the same – jain smit Jul 6 '13 at 18:02

This is what I get, with $\,C_2:=(2\cos\theta\,,\,2\sin\theta)=\{(x,y)\in\Bbb R^2\;;\;x^2+y^2=4\}\;$:

$$\int\limits_C(y-x)dx+(2x-y)dy=\int\int\limits_{C_2}(2-1)dA=4\pi$$

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I find it is easier to use the other form of greens theorem for a vector field given by $$\vec{F} = P \hat{\imath} + Q \hat{\jmath}$$ Where Greens theorem is: $$\oint_C P dx + Q dy = \int \int_A \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} dA$$ Where A is the area enclosed by the curve, this effectively turns a one dimensional line integral over a boundary into a double integral over the area enclosed. Using that $$P= y-x \rightarrow \frac{\partial P}{\partial y}=1$$ $$Q= 2x-y \rightarrow \frac{\partial Q}{\partial x}=2$$ Then the appropriate integral using greens theorem should be $$\int \int dA$$ Where dA will be determined by the curve enclosing the area (a circle of radius 2)

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Thank you for sharing this alternative method! It cleans up well. – TonyP Jul 5 '13 at 21:21
No problem TonyP – Dan Jul 6 '13 at 12:28
@Dan what is the final answer? – jain smit Jul 6 '13 at 18:13
When I did it I got $12 \pi$ since $\int \int dA$ for a circle of radius 2 is $4 \pi$. I was going to let the person it it out a bit before letting on to the final answer. – Dan Jul 6 '13 at 20:39
I think there's a mistake here: $$P=y-x\implies\frac{\partial P}{\partial y}=1$$so you're off by a factor of $\,3\,$: it should be, imho, simply $\,1\,$ – DonAntonio Jul 6 '13 at 21:46