# Green's Theorem over a circle

1. Feb 10, 2013

### SapphireLFC

1. The problem statement, all variables and given/known data
Solve: ∫(-ydx+xdy)/(x2+y2) counterclockwise around x2+y2=4

2. Relevant equations
Greens Theorem:
∫Pdx + Qdy = ∫∫(dQ/dx - dP/dy)dxdy

3. The attempt at a solution
Using Greens Theorem variables, I get that:
P = -y/(x2+y2) and
Q=x/(x2+y2)

and thus dQ/dx = (y2-x2)/(y2+x2)2

and dP/dx = (y2-x2)/(y2+x2)2

So, ∫∫(dQ/dx - dP/dy)dxdy = ∫∫( (y2-x2)/(y2+x2)2 - (y2-x2)/(y2+x2)2)dxdy

... which means I'm integrating 0 (which can't be right as that would equal 0 over a definite integral). Not sure where I've gone wrong! Can anyone spot an error?

2. Feb 10, 2013

### LCKurtz

What's wrong with zero? You have something against it??

3. Feb 10, 2013

### SapphireLFC

Not at all! Just doesn't seem like the correct solution in this instance

4. Feb 10, 2013

### LCKurtz

Well, looking more closely at your problem, I see that you have a singularity at (0,0). Perhaps you should try evaluating the integral directly.

5. Feb 10, 2013

### SapphireLFC

I don't understand what you mean by evaluating it directly. My method was to use greens theorem and then (providing it didn't give me 0!) convert to polar coordinates and integrate over r and theta, thus the singularity wouldn't affect anything. Is this wrong?

6. Feb 10, 2013

### LCKurtz

Since you are integrating over a region that encloses a singularity, the hypotheses of Green's theorem are not fulfilled. Your function isn't even defined at (0,0). So you can't use Green's theorem. So just work the line integral out. I would suggest using the polar angle $\theta$ as your parameter.

7. Feb 10, 2013

### SapphireLFC

Right, ok. Thank you very much for your help!

Sorry to be such a pain but I'm still a little baffled about the zero I'm getting. Just as a side note, suppose I was integrating that function over a region with no singularities (say circle centre (3,0), rad 1) would I still not be able to use Green's Theorem? If I did, I'd still be integrating zero, which still seems wrong in context.

8. Feb 10, 2013

### LCKurtz

Yes, you could use Green's theorem in that case. And you would get zero. And there is nothing wrong with getting zero in a line integral. It happens all the time, for example when integrating around a closed loop in a conservative force field (under appropriate hypotheses). I don't know what "context" is bothering you.