The line integral of ∮{y/(x²+y²)dx-x/(x²+y²)dy} around the unit circle around the origin traversed once in the counter clockwise =? 0 or -2π
if I put x=cosθ and y=sinθ ; dx=-sinθdθ, dy=cosθdθ where 0≤θ≤2π and evaluated the integral I will get -2π as the answer.
However if I evaluated the integral using Green's theorem I will get 0 as the answer.
I also checked the curl of the vector f= y/(x²+y²)i-x/(x²+y²)j which is equal to zero. That implies the line integral around the unit circle should be zero as the vector filed is irrotational.
My question is why I'm getting -2π as the answer when i put x=cosθ and y=sinθ ; dx=-sinθdθ, dy=cosθdθ where 0≤θ≤2π and evaluated the integral?
Can anybody answer the question? Thanks in advance.
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Verified answer
Note that P(x, y) = y/(x²+y²) and Q(x, y) = -x/(x²+y²) are discontinuous at (0, 0), which is contained inside the region of integration. So, P and Q are not differentiable at (0, 0) and Green's Theorem does not apply.
I hope this helps!