# A question about path orientation in Green's Theorem

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1. Jun 8, 2015

### kostoglotov

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

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?

Last edited: Jun 8, 2015
2. Jun 8, 2015

### wabbit

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.