Singularities and conservative vector fields

[NewGuy]

I have a question regarding conservative vectorfields and singularities.
Suppose we have a vectorfield who is defined everywhere in R^2 except at the origin where it has a singularity, and suppose it's curl is zero. We then have that it is conservative in every open, simply connected subset in R^2 not containing the origin.
Now if we have a closed curve around the origin we can't generally say that the curve integral is zero right? That depends on the given vectorfield? If we have calculated the curve integral around a circle with centre in the origin and shown that it is zero, can we then say that every closed curve around the origin has curve integral zero?
Suppose for example we have a square around the origin. Can we calculate the curveintegral for the whole square by adding the curveintegrals along the 4 sides?

 

[HallsofIvy]

Yes, if you have two distinct paths, both inclosing a singularity, then you can construct a line from one (call it C_a) to the other (C_B) and just a slight distance away, a second line back to the first singularity Consider the integration from any point on C_A, counterclockwise around C_A to the first "cutoff", along it to C_B, around C_B, clockwise to the second "cutoff", along it back to C_A, and finally around C_A to the starting point. We know have integrals: I_A' is almost I_A,the integral around A- it does not include that small section of A between the two "cutoffs". I₁ is the integral along the first "cutoff" from C_A to C_B. I_B' is almost -I_B, the negative of the integral around C_B- it does not include the small section between the "cutoffs" (negative because we have to go clockwise around B). Finally, I₂ is the integral along the second "cutoff" from C_B back to C_A. Because there in no singularity inside that path the integral is 0: I_A'+ I₁- I_B'+ I₂= 0.

Now, take the limit as we move the two "cutoffs" toward each other. Since, in the limit, they become the same path but integrated in opposite directions, in the limit, I<sub>1[/sub[+ I2= 0. Also, the left out sections between the two "cutoffs" disappear: in the limit IA'= IA, the integral all the way around A, and IB'= IB, the integral all the way around B. Since there is no singularity between the two cutoffs, the left side of the equation remains 0. In the limit, IA- IB= 0 so IA= IA.

As long as, moving one closed path to another, we do not cross any singularities, the integrals are the same.

The last part, "Suppose for example we have a square around the origin. Can we calculate the curveintegral for the whole square by adding the curveintegrals along the 4 sides?" is a completely different question but, of course, it is true. That's just saying that the sum of the two partial integrals is equal to the entire integral. I would think you did that back in basic calculus. Did you never do a path integral of a broken line?</sub>

 

[NewGuy]

As long as, moving one closed path to another, we do not cross any singularities, the integrals are the same.

This is a nice thing indeed! If we have that the curve integral around the singularity is 2 pi that is true for all closed paths around the singularity? I suspect that it is a requirement that curl(F) is zero everywhere along the path? Does the same thing apply to R^3?

Regarding the square: I meant a square around the singularity. I know that in a curve integral you can add the different paths together, I just didn't know if you could around a singularity.

So if we somehow can show that a vector field is the gradient of some function (by guessing at an appropriate function for example), it is true that the curve integral is zero for all closed paths, regardless if they surround the singularity?