Question

[calculus 3- line integrals] Vector field F(x,y) = (-y/(x^2+y^2)i + x/(x^2+y^2)j. The curl of this vector field is 0. Let c1 be the path from (0,1) to (0,-1) along the right half of the unit circle, and C2 be the path from (0,1) to (0,-1) along the left half of the unit circle, calculate the path integral of F along c1 and calculate the path integral of F along c2.

Answer 1

So I got c1 = -pi and c2 = pi.

My friend got that they were -2 and 2.

Answer 2

Anyone mind going through the problem, and seeing if either of us were right? If neither of us were, help plz cause we're both lost.

Answer 3

Wait, the paths don't make sense.

Answer 4

You start at \((0,1)\) and then go to \((0,-1)\), but then you jump to \((0,1)\) all of the sudden.

Answer 5

Wait, they are two different paths. I see now.

Answer 6

Yeah. You're going down two different sides of the unit circle. Basically 2 calculations.

Answer 7

Since the vector field is conservative, it should be path independent.

Answer 8

From what I understand it's going to be the exact same problem for both but with different bounds?

Answer 9

Isn't that for the complete circle though?

Answer 10

We're doing two half circles.

Answer 11

No, for a closed path, the integral evaluates to 0.

Answer 12

For a non-closed path, it can be non-zero result, but it only depends on the start and finish point. The path doesn't matter.

Answer 13

you can go straight from (0,1) to (0,-1) along y axis instead

Answer 14

Closed paths evaluate to 0 because it's no different than having not moved at all, due to path independence.

Answer 15

Right, but since we're not doing a full rotation for either I don't think it's going to evaluate to 0.

I figure the two are going to be negatives of eachother because they're going in opposite directions? Might be completely off here.

Answer 16

Path independence means that direction doesn't matter either.