# Use Green's theorem to calculate line integral

I've the areas

$C1: x^{2}+y^2 = 1$

$C2: r = \frac{2}{\sqrt{2-cos(\theta)}}$

$R$ is the area between C1 and C2, both curves are oriented with positive direction/flow (don't really know the translation of this).

I've found that:

$\oint\limits_{C1} \frac{-y}{(x^{2}+y^2)^2} dx +\frac{x}{(x^{2}+y^2)^2} dy = 2\pi$

and

$\iint\limits_R \frac{1}{(x^{2}+y^2)^2}dxdy = \frac{\pi}{2}$

How can I use these two results to calculate:

$\oint\limits_{C2} \frac{-y}{(x^{2}+y^2)^2} dx +\frac{x}{(x^{2}+y^2)^2} dy$

I'm sure there is a smart way where you subtract 1 result from the other I just can't see it.

Edit: So what I've tried to do is taking $\oint\limits_{C2} = \oint\limits_{C1} + Area(R)$ what I ended up with was something like R is negative because I've switched the orientation so it goes counterclockwise $2\pi-2\frac{\pi}{2}=\pi$ This however doesn't seem quite correct

• Another NTNU student who is late at turning in their homework. Hopefully this comment will make people avoiding answering the question before the deadline (tommorow at midnight) – N3buchadnezzar Mar 15 '15 at 16:03
• Not really late, just not able to do this correctly apprently. I know the answer should be $\pi$ but I'm unable to see how. Any tips are appriciated – user123577 Mar 15 '15 at 16:33
• So deadline is finished so I wonder how could you actually do this. I guess you could take the line integral of R and subtract C1 to get the whole area of 2? – user123577 Mar 17 '15 at 11:35

You are asked to calculate the integral around the closed curve $C_2$. Now one can proceed as you did in the first part and try to calculate the integral directly. This gives you a horrible mess. Since $$x = r \cos \theta = \frac{2 \cos \theta}{\sqrt{2-\cos \theta}} \quad \text{and} \quad y = r \sin \theta = \frac{2 \sin \theta}{\sqrt{2-\cos \theta}}$$ Inserting this into the line integral makes it just worse since $$\frac{1}{(x^2+y^2)^2} = \frac{1}{16}\left(\frac{1}{\cos^2 \theta}-\frac{1}{2 \cos \theta}\right)^2$$ Expanding and figuring out $\mathrm{d}x$ does not make it any nicer and leads to more complicated expressions. So the idea is to use another method to integrate the curve integral. The natural method is then to use Greens theorem, since we are integrating over a simple closed curve.

But wait a minute mister, there is another restriction to greens theorem. The function must have continous partial derivatives everywhere in the interior of the curve.

To see that the function does not have continous partial derivatives we can either calculate them directly. For example for $L$ we have

$$\frac{\partial L}{\partial x} \left( \frac{x}{(x^2+y^2)^2}\right) = \frac{y^2 - 3x^2}{(x^2+y^2)^2}$$

Following the line $y=0$ or $x=0$ towards the origo, one sees that the integral tends to infinity and hence does not have continuous partial derivatives. It is enough to see that one partial derivative is not continuous for the theorem to fail. A perhaps simpler idea is to say that since the function is not continuous, it can not possibly have continuous derivatives. Changing to polar coordinates we have $$\lim_{\substack{x\to 0 \\ y\to 0}} \frac{x}{(x^2+y^2)^2} = \lim_{\substack{r \to 0 \\ \theta \in [0,2\pi]}} \frac{r \cos \theta}{(r^2)^2}$$ Which we sees blows up for nearly all theta (Remember that it is enough to find one $\theta \in [0,2\pi]$ such that the limit blows up). Hence you need to add a curve to avoid the origin. A clever choice here is to use the curve $C_1$ (since you have already calculated this line-integral). Then use the ideas below for creating such a curve.

See for example Extensions and applications of Green's theorem. by Jeremy Orloff.

Or this example on stack.exchange Taking the line integral of a region with holes with Green's Theorem

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• Good thing I wrote the wrong domain of C2 and thanks for the tips :) I wrote 2/2cos(theta) which became quite difficult – user123577 Mar 15 '15 at 17:16
• Sorry to bother you again but, why is the domain of C2 not connected? $r=\frac{2}{\sqrt{2-cos(\theta)}$ seems to be connected. – user123577 Mar 15 '15 at 17:53
• @user123577, I will edit my question to answer your concerns. What I wrote above is the correct way to tackle the problem, but the reasoning with simply connected domain is false. – N3buchadnezzar Mar 15 '15 at 18:41
• Could you answer with the solution now since the deadline is over. I just want to make sure if I did it correctly? – user123577 Mar 17 '15 at 11:29