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...
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?
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 =...