# When can you use Green's Theorem to convert curl integrals to double integrals?

*print*Print*list*Cite

### 1 Answer

Let's define the parameters of Green's Theorem so we don't have any ambiguity:

`vecF(x,y) = Aveci + Bvecj`

`oint_C Adx + Bdy = int int_R (delA)/(delx) - (delB)/(dely) dxdy`

`vecF` is defined as a vector field of class `C^1` (their partial derivatives up to at least an order of 1 exist, otherwise the double integral wouldn't make sense!) throughout `R`. `R` is the region enclosed by the curves that define `C`.

1) Curve and region are defined in `RR^2`

This is the main "special case" specification for Green's Theorem with respect to Stoke's Theorem. It just means we don't deal with a 3-D input.

2) Curve `C` is a set of piecewise smooth, simple, closed curves whose internal region is defined by `R` .

This is a pretty loaded qualification (it's actually 4, in a way!). First off, C must be able to be defined by a finite number of piecewise functions that are differentiable over a closed interval (with the exception at the intervals' endpoints, at which derivatives from the left and right must exist, but not necessarily be equal). The simple aspect of the curve means that it will not intersect itself, except to ensure that it is closed. For example, the flowers you learned to make when you first started using polar coordinates would be disqualified! Finally, as I alluded to just now, `C` must enclose `R`, meaning I can draw no path from a point `SinR` (the region bounded by the curve) to a point `T` outside `R` without intersecting `C`.

2) `(delA)/(delx)` and `(delB)/(dely)` are both continuous in the region R.

This is why `vecF` must be of class `C^1`. `A` and `B` must both have continuous partial derivatives, or the double integral will not be defined! For example, if `A` has any holes in `R`, some of the `(x,y)` values will not exist! This would indicate that in the closed region containing those "holes," the partial derivatives would not exist either. Keep in mind, that example does not extend to holes that would have already been defined by a piecewise `C`. By this i mean that if we had a "doughnut" where the hole is defined by `C` , we're fine. Otherwise, the theorem wouldn't apply!

3) `C` is positively oriented

This condition is to ensure that negative and positive line up on both sides of the equation. Because the curl, integral, and what-have-you are based on positive (right-handed) rotation, you want the curve to be positively oriented, too. This means that as you go through a trip around any of the curves defined by `C`, `R` will always be on the left.

There you have all of the conditions necessary for Green's Theorem! I hope all of that makes sense. If you think about it conceptually, though, each of these conditions is definitely understandable. Green's theorem is, in a way, equating the overall rotation of a vector field at the perimeter to the sum of rotations at an infinitesimal scale. In order for this to work, you would need to know that you could quantify all of the rotation going on in a 2-D space, and that you could sum each infinitesimal bit!

I hope that helps!

**Side Note**: The link below seems to imply that `C` can only be a single curve. This is not true, as `C` can be used to identify any number of "holes" in `R`. Another way to think of this is that you can separate a region with "holes" into a number of regions without holes and recognize that summing the opposing orientations at the boundaries would cancel them out, allowing you to solve for regions defined by multiple simple curves.

**Sources:**