A question about path orientation in Green's Theorem

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SUMMARY

In the discussion regarding path orientation in Green's Theorem, it is established that for non-simply-connected regions, the curves must be oriented such that the region D remains on the left side during traversal. This orientation is crucial as it affects the sign of the line integral, particularly when considering the traversal of curves C1 and C2. The discussion highlights that reversing the orientation leads to subtracting instead of adding, which is essential for maintaining the integrity of the integral over the closed contour formed by connecting the curves.

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kostoglotov
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So if we have a non-simply-connected region, like this one

fVoIlJb.png


to apply Green's Theorem we must orient the C curves so that the region D is always on the left of the curve as the curve is traversed.

Why is this? I have seen some proofs of Green's Theorem for simply connected regions, and I understand why going the opposite direction along a curve gives you the negative of a line integral.

I guess what I'm asking for is a little insight into what's happening here.

Can I imagine that I'm walking along some path, circling some geometric center of a region, measuring the region to my left and summing all the areas or the function values as I go along? So then I were to traverse C2 from the above diagram, I am subtracting instead of adding? So we then need to reverse the orientation?

I can accept the rules, the conditions and can read some proofs, but I don't feel like I understand it very well.

Thanks.

edit: it has something to do with what happens when we divide the region in two doesn't it?
 
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it has something to do with what happens when we divide the region in two doesn't it?
If you connect the two circles by a line L, you get a closed connected contour going once around the whole area (C1+L+C2-L), provided you orient C1 and C2 as you did i.e. relative to the bounded region. The integral over the connecting line cancels out since you travel it once in each direction, so the result follows from Green's formula for a simply connected region.
 
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