Can Green's Theorem be used to evaluate line integrals over circles?

[SapphireLFC]

Homework Statement

Solve: ∫(-ydx+xdy)/(x²+y²) counterclockwise around x²+y²=4

Homework Equations

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

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?

 

[LCKurtz]

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

 

[SapphireLFC]

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

 

[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.

 

[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?

 

[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.

 

[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.

 

[LCKurtz]

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.

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.