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The first four line integrals comprise the line integral around the geometric set S 2 , and the last four comprise the line integral around the geometric set S 1 . On the other hand, the first and eighth line integrals here cancel out, for one is just the reverse of the other.Hence, the sum total of these eight line integrals, integrals 2–7, is just the line integral around the boundary C S of S . Therefore

C S ω = C S 1 ω + C S 2 ω

as desired.

We give next an outline of the proof for a general partition S 1 , ... , S n of S . Let S k be determined by the interval [ a k , b k ] and the two bounding functions u k and l k . Observe that, if the boundary C S k of S k intersects the boundary C S j of S j in a curve C , then the line integral of ω along C , when it is computed as part of integrating counterclockwise around S k , is the negative of the line integral along C , when it is computed as part of the line integral counterclockwise around S j . Indeed, the first line integral is the reverse of the second one. (A picture could be helpful.)Consequently, when we compute the sum of the line integrals of ω around the C S k 's, All terms cancel out except those line integrals that ar computed along parts of the boundaries of the S k 's that intersect no other S j . But such parts of the boundaries of the S k 's must coincide with parts of the boundary of S . Therefore, the sum of the line integrals of ω around the boundaries of the S k 's equals the line integral of ω around the boundary of S , and this is precisely what the theorem asserts.

Prove the analog of [link] for contour integrals: Let S 1 , ... , S n be a collection of closed geometric sets that constitute a partition of a geometric set S , and assume that the boundaries of all the S i 's, as well as the boundary of S , have finite length. Suppose f is a continuous complex-valued function on all the boundaries { C S k } as well as on the boundary C S . Then

C S f ( ζ ) d ζ = k = 1 n C S k f ( ζ ) d ζ .

We come now to the most remarkable theorem in the subject of integration over curves, Green's Theorem.Another fanfare, please!

Green

Let S be a piecewise smooth, closed, geometric set, let C S denote the closed curve that is the boundary of S , and assume that C S is of finite length. Suppose ω = P d x + Q d y is a continuous differential form on S that is smooth on the interior S 0 of S . Then

C S ω = C S P d x + Q d y = S t i a l Q t i a l x - t i a l P t i a l y .

REMARK The first thing to notice about this theorem is that it connects an integral around a (1-dimensional) curve with an integral over a (2-dimensional) set,suggesting a kind of connection between a 1-dimensional process and a 2-dimensional one. Such a connection seems to be unexpected, and it shouldtherefore have some important implications, as indeed Green's Theorem does.

The second thing to think about is the case when ω is an exact differential d f of a smooth function f of two real variables. In that case, Green's Theorem says

C S t i a l f t i a l x d x + t i a l f t i a l y d y = S ( f y x - f x y ) ,

which would be equal to 0 if f C 2 ( S ) , by [link] . Hence, the integral of d f around any such curve would be 0. If U is an open subset of R 2 , there may or may not be some other ω 's, called closed differential forms, having the property that their integral around every piecewise smooth curve of finite length in U is 0, and the study of these closed differential forms ω that are not exact differential forms d f has led to much interesting mathematics. It turns out that the structure of the open set U , e.g., how many “holes” there are in it, is what's important.Take a course in Algebraic Topology!

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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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