Connected Region vs Simply connected region: Theorem 15.7 in Larson Calculus

Question

Theorem 15.7 in Larson's Calculus 9th Edition describes conditions for conservative vector fields over a region R. In the Theorem it is stated that R must be "open connected" but does not state that the region must be "simply" connected. In attempting to understand why the classic example $F=\left<\frac{y}{x^2+y^2},\frac{-x}{x^2+y^2}\right>$ for a loop around the origin doesn't satisfy the theorem, it seems to me that the "hole" at the origin is the problem, but I don't see why a donut shape in $R^2$ isn't connected. I only see that it isn't "simply connected", but this isn't stated in the theorem. Do I misunderstand the meaning of "open connected", or should the theorem state "simply connected? Theorem as stated in Larson, 9th, p1088

Answer 1

The responses by saulspatz clarified my misunderstanding. The theorem I quoted provides sufficient conditions for the three statements to be equivalent but not (as I mistakenly read) sufficient to claim any one of these to be true. A correct reading would be that, if the conditions are met, either all three are false, or all three are true.

The key step is to determine if one of them is true, or false. In most cases, I would imagine, the one to consider is #1 (conservative). The definition of a conservative field F is one for which there exists a differentiable function whose gradient is F. The $f(x,y)$ I proposed is not differentiable at the origin, and indeed does not exist along the x-axis. If one chooses a region in the plane where it is differentiable, then F has the properties of a conservative field.

For the particular exercise, it is shown that despite curl(F) being zero wherever defined, the line integral for a closed loop containing the origin and passing through the x-axis does not yield 0 as might be expected.