Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
...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 $2$

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


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

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

So, the result follows from evaluating the integral

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

I leave the rest for you.

share|cite|improve this answer
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 $\,C_2:=(2\cos\theta\,,\,2\sin\theta)=\{(x,y)\in\Bbb R^2\;;\;x^2+y^2=4\}\;$:


share|cite|improve this answer

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)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.