Is every closed path integral zero?

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SUMMARY

A conservative field is defined by the property that the line integral of every closed path is zero. However, non-conservative fields can also exhibit closed paths where the integral evaluates to zero, as demonstrated by the field (\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}) in \mathbb{R}^2. Specific paths, such as those not enclosing the origin (0,0), can yield a zero integral, but this does not imply that the field is conservative. Therefore, all closed path integrals must vanish for a field to be classified as conservative.

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Red_CCF
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Hi

I was wondering, one way that a conservative field can be found is if the line integral of any closed path is 0. However, what if I have a non-conservative field, I travel in a circle in a clockwise manner back to my starting point, then travel along the same path in a counter clockwise manner back to the starting point, isn't this path closed and the line integral of it zero but yet the field is not conservative?

Any help is appreciated
 
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Hi Red CCF! :smile:

A field is convervative if the integral of every closed path vanishes. It's not because the integral of one special closed path vanishes, that all the integrals vanish.

For example, the field

(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})

is a famous example of a nonconservative field in \mathbb{R}^2. Nevertheless, there are paths over which the integral vanished. Your path is one example, another example is the path formed by the circle with radius (10,10) and radius 1. In general, any path in which (0,0) is not in the interior of the path, will have an integral that vanishes.

So while a lot of integrals may vanish over a certain field, we need them all to vanish in order for the field to be conservative!
 

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