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Textbook by: Lawrence Baggett. E-mail the author

# Integration Around Closed Curves, and Green's Theorem

Module by: Lawrence Baggett. E-mail the author

Summary: Two problems are immediately apparent concerning integrating around a closed curve. First, where do we start on the curve, which point is the initial point? And second, which way to we go around the curve? Recall tha if φ:[a,b]Cφ:[a,b]C is a parameterization of C,C, then ψ:[a,b]C,ψ:[a,b]C, defined by ψ(t)=φ(a+b-t),ψ(t)=φ(a+b-t), is a parameterization of CC that is the reverse of φ,φ, i.e., it goes around the curve in the other direction. If we are integrating with respect to arc length, this reverse direction won't make a difference, but, for contour integrals and line integrals, integrating in the reverse direction will introduce a minus sign.

Thus far, we have discussed integration over curves joining two distinct points z1z1 and z2.z2. Very important in analysis is the concept of integrating around a closed curve, i.e., one that starts and ends at the same point. There is nothing really new here; the formulas for all three kinds of integrals we have defined will look the same, in the sense that they all are described interms of some parameterization φ.φ. A parameterization φ:[a,b]Cφ:[a,b]C of a closed curve CC is just like the parameterization for a curve joining two points, except that the two points φ(a)φ(a) and φ(b)φ(b) are equal.

Two problems are immediately apparent concerning integrating around a closed curve. First, where do we start on the curve, which point is the initial point? And second, which way to we go around the curve? Recall tha if φ:[a,b]Cφ:[a,b]C is a parameterization of C,C, then ψ:[a,b]C,ψ:[a,b]C, defined by ψ(t)=φ(a+b-t),ψ(t)=φ(a+b-t), is a parameterization of CC that is the reverse of φ,φ, i.e., it goes around the curve in the other direction. If we are integrating with respect to arc length, this reverse direction won't make a difference, but, for contour integrals and line integrals, integrating in the reverse direction will introduce a minus sign.

The first question mentioned above is not so difficult to handle. It doesn't really matter where we start on a closed curve; the parameterization can easily be shifted.

## Exercise 1

Let φ[a,b]R2φ[a,b]R2 be a piecewise smooth function that is 1-1 except that φ(a)=φ(b).φ(a)=φ(b). For each 0<c<b-a,0<c<b-a, define φ^:[a+c,b+c]:R2φ^:[a+c,b+c]:R2 by φ^(t)=φ(t)φ^(t)=φ(t) for a+ctb,a+ctb, and φ^(t)=φ(t-b+aφ^(t)=φ(t-b+a for btb+c.btb+c.

1. Show that φ^φ^ is a piecewise smooth function, and that the range CC of φφ coincides with the range of φ^.φ^.
2. Let ff be an integrable (with respect to arc length) function on C.C. Show that
abf(φ(t))|φ'(t)|dt=a+cb+cf(φ^(t))|φ^'(t)|dt.abf(φ(t))|φ'(t)|dt=a+cb+cf(φ^(t))|φ^'(t)|dt.
(1)
That is, the integral Cf(s)dsCf(s)ds of ff with respect to arc length around the closed curve CC is independent of where we start.
3. Let ff be a continuous complex-valued function on C.C. Show that
abf(φ(t))φ'(t)dt=a+cb+cf(φ^(t))φ^'(t)dt.abf(φ(t))φ'(t)dt=a+cb+cf(φ^(t))φ^'(t)dt.
(2)
That is, the contour integral Cf(ζ)dζCf(ζ)dζ of ff around the closed curve CC is independent of where we start.
4. Let ω=Pdx+Qdyω=Pdx+Qdy be a differential form on C.C. Prove that
abP(φ(t))x'(t)+Q(φ(t))y'(t)dt=a+cb+cP(φ^(t))x^'(t)+Q(φ^(t))y^'(t)dt.abP(φ(t))x'(t)+Q(φ(t))y'(t)dt=a+cb+cP(φ^(t))x^'(t)+Q(φ^(t))y^'(t)dt.
(3)
That is, the line integral CωCω of ωω around CC is independent of where we start.

The question of which way we proceed around a closed curve is one that leads to quite intricate and difficult mathematics, at least when we consider totaly general smooth curves. For our purposes it wil, suffice to study a special kind of closed curve, i.e., curves that are the boundaries of piecewise smooth geometric sets. Indeed, the intricate part of the general situation has a lot to do with determining which is the “inside” of the closed curve and which is the “outside,” a question that is easily settled in the case of a geometric set. Simple pictures make this general question seem silly, but precise proofs that there is a definite inside and a definite outside are difficult, and eluded mathematicians for centuries, culminating in the famous Jordan Curve Theorem, which asserts exactly what our intuition predicts:

## Theorem 1: Jordan Curve Theorem

The complement of a closed curve is the union of two disjoint components, one bounded and one unbounded.

We define the bounded component to be the inside of the curve and the unbounded component to be the outside.

We adopt the following convention for how we integrate around the boundary of a piecewise smooth geometric set S.S. That is, the curve CSCS will consist of four parts: the lower boundary (graph of the lower bounding function ll), the righthand boundary (a portion of the vertical line x=bx=b), the upper boundary (the graph of the upper bounding function uu), and finally the lefthand side (a portion of the vertical line x=ax=a). By integrating around such a curve CS,CS, we will always mean proceeding counterclockwise around the curves. Specifically, we move from left to right along the lower boundary, from bottom to top along the righthand boundary, from right to left across the upper boundary, and from top to bottom along the lefthand boundary. Of course, as shown in the exercise above, it doesn't matter where we start.

## Exercise 2

Let SS be the closed piecewise smooth geometric set that is determined by the interval [a,b][a,b] and the two piecewise smooth bounding functions uu and l.l. Assume that the boundary CSCS of SS has finite length. Suppose the graph of uu intersects the lines x=ax=a and x=bx=b at the points (a,c)(a,c) and (b,d),(b,d), and suppose that the graph of ll intersects those lines at the points (a,e)(a,e) and (b,f).(b,f). Find a parameterization φ:[a',b']CSφ:[a',b']CS of the curve CS.CS.

HINT: Try using the interval [a,b+d-f+b-a+c-e][a,b+d-f+b-a+c-e] as the domain [a',b'][a',b'] of φ.φ.

The next theorem, though simple to state and use, contains in its proof a combinatorial idea that is truly central to all that follows in this chapter. In its simplest form, it is just the realization that the line integral in one direction along a curve is the negative of the line integral in the opposite direction.

## Theorem 2

Let S1,...,SnS1,...,Sn be a collection of closed geometric sets that constitute a partition of a geometric set S,S, and assume that the boundaries of all the SiSi's, as well as the boundary of S,S, have finite length. Suppose ωω is a continuous differential form on all the boundaries {CSk}.{CSk}. Then

C S ω = k = 1 n C S k ω . C S ω = k = 1 n C S k ω .
(4)

### Proof

We give a careful proof for a special case, and then outline the general argument. Suppose then that SS is a piecewise smooth geometric set, determined by the interval [a,b][a,b] and the two bounding functions uu and l,l, and assume that the boundary CSCS has finite length. Suppose m(x)m(x) is a piecewise smooth function on [a,b],[a,b], satisfying ab|m'|<,ab|m'|<, and assume that l(x)<m(x)<u(x)l(x)<m(x)<u(x) for all x(a,b).x(a,b). Let S1S1 be the geometric set determined by the interval [a,b][a,b] and the two bounding functions mm and l,l, and let S2S2 be the geometric set determined by the interval [a,b][a,b] and the two bounding functions uu and m.m. We note first that the two geometric sets S1S1 and S2S2 comprise a partition of the geometric set S,S, so that this is indeed a pspecial case of the theorem.

Next, consider the following eight line integrals: First, integrate from left to write along the graph of m,m, second, up the line x=bx=b from (b,m(b))(b,m(b)) to (b,u(b)),(b,u(b)), third, integrate from right to left across the graph of u,u, fourth, integrate down the line x=ax=a from (a,u(a))(a,u(a)) to (a,m(a)),(a,m(a)), fifth, continue down the line x=ax=a from (a,m(a))(a,m(a)) to (a,l(a)),(a,l(a)), sixth, integrate from left to right across the graph of l,l, seventh, integrate up the line x=bx=b from (b,l(b))(b,l(b)) to (b,m(b)),(b,m(b)), and finally, integfrate from right to left across the graph of m.m.

The first four line integrals comprise the line integral around the geometric set S2,S2, and the last four comprise the line integral around the geometric set S1.S1. 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 CSCS of S.S. Therefore

C S ω = C S 1 ω + C S 2 ω C S ω = C S 1 ω + C S 2 ω
(5)

as desired.

We give next an outline of the proof for a general partition S1,...,SnS1,...,Sn of S.S. Let SkSk be determined by the interval [ak,bk][ak,bk] and the two bounding functions ukuk and lk.lk. Observe that, if the boundary CSkCSk of SkSk intersects the boundary CSjCSj of SjSj in a curve C,C, then the line integral of ωω along C,C, when it is computed as part of integrating counterclockwise around Sk,Sk, is the negative of the line integral along C,C, when it is computed as part of the line integral counterclockwise around Sj.Sj. 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 ωω around the CSkCSk's, All terms cancel out except those line integrals that ar computed along parts of the boundaries of the SkSk's that intersect no other Sj.Sj. But such parts of the boundaries of the SkSk's must coincide with parts of the boundary of S.S. Therefore, the sum of the line integrals of ωω around the boundaries of the SkSk's equals the line integral of ωω around the boundary of S,S, and this is precisely what the theorem asserts.

## Exercise 3

Prove the analog of Theorem 2 for contour integrals: Let S1,...,SnS1,...,Sn be a collection of closed geometric sets that constitute a partition of a geometric set S,S, and assume that the boundaries of all the SiSi's, as well as the boundary of S,S, have finite length. Suppose ff is a continuous complex-valued function on all the boundaries {CSk}{CSk} as well as on the boundary CS.CS. Then

C S f ( ζ ) d ζ = k = 1 n C S k f ( ζ ) d ζ . C S f ( ζ ) d ζ = k = 1 n C S k f ( ζ ) d ζ .
(6)

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

## Theorem 3: Green

Let SS be a piecewise smooth, closed, geometric set, let CSCS denote the closed curve that is the boundary of S,S, and assume that CSCS is of finite length. Suppose ω=Pdx+Qdyω=Pdx+Qdy is a continuous differential form on SS that is smooth on the interior S0S0 of S.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 . 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 .
(7)

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 should therefore have some important implications, as indeed Green's Theorem does.

The second thing to think about is the case when ωω is an exact differential dfdf of a smooth function ff 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 ) , 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 ) ,
(8)

which would be equal to 0 if fC2(S),fC2(S), by (Reference). Hence, the integral of dfdf around any such curve would be 0. If UU is an open subset of R2,R2, there may or may not be some other ωω's, called closed differential forms, having the property that their integral around every piecewise smooth curve of finite length in UU is 0, and the study of these closed differential forms ωω that are not exact differential forms dfdf has led to much interesting mathematics. It turns out that the structure of the open set U,U, e.g., how many “holes” there are in it, is what's important. Take a course in Algebraic Topology!

The proof of Green's Theorem is tough, and we break it into several steps.

## Lemma 1

Suppose SS is the rectangle [a,b]×[c,d].[a,b]×[c,d]. Then Green's Theorem is true.

### Proof

We think of the closed curve CSCS bounding the rectangle as the union of four straight lines, C1,C2,C3C1,C2,C3 and C4,C4, and we parameterize them as follows: Let φ:[a,b]C1φ:[a,b]C1 be defined by φ(t)=(t,c);φ(t)=(t,c); let φ:[b,b+d-c]C2φ:[b,b+d-c]C2 be defined by φ(t)=(b,t-b+c);φ(t)=(b,t-b+c); let φ:[b+d-c,b+d-c+b-a]C3φ:[b+d-c,b+d-c+b-a]C3 be defined by φ(t)=(b+d-c+b-t,d);φ(t)=(b+d-c+b-t,d); and let φ:[b+d-c+b-a,b+d-c+b-a+d-c]C4φ:[b+d-c+b-a,b+d-c+b-a+d-c]C4 be defined by φ(t)=(a,b+d-c+b-a+d-t).φ(t)=(a,b+d-c+b-a+d-t). One can check directly to see that this φφ parameterizes the boundary of the rectangle S=[a,b]×[c,d].S=[a,b]×[c,d].

As usual, we write φ(t)=(x(t),y(t)).φ(t)=(x(t),y(t)). Now, we just compute, use the Fundamental Theorem of Calculus in the middle, and use part (d) of (Reference) at the end.

C S ω = C 1 ω + C 2 ω + C 3 ω + C 4 ω = C 1 P d x + Q d y + C 2 P d x + Q d y + C 3 P d x + Q d y + C 4 P d x + Q d y = a b P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b b + d - c P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b + d - c b + d - c + b - a P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b + d - c + b - a b + d - c + b - a + d - c P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t = a b P ( t , c ) d t + b b + d - c Q ( b , t - b + c ) d t + b + d - c b + d - c + b - a P ( b + d - c + b - t , d ) ( - 1 ) d t + b + d - c + b - a b + d - c + b - a + d - c Q ( a , b + d - c + b - a + d - t ) ( - 1 ) d t = a b P ( t , c ) d t + c d Q ( b , t ) d t - a b P ( t , d ) d t - c d Q ( a , t ) d t = c d ( Q ( b , t ) - Q ( a , t ) ) d t - a b ( P ( t , d ) - P ( t , c ) ) d t = c d a b t i a l Q t i a l x ( s , t ) d s d t - a b c d t i a l P t i a l y ( t , s ) d s d t = S ( t i a l Q t i a l x - t i a l P t i a l y , C S ω = C 1 ω + C 2 ω + C 3 ω + C 4 ω = C 1 P d x + Q d y + C 2 P d x + Q d y + C 3 P d x + Q d y + C 4 P d x + Q d y = a b P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b b + d - c P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b + d - c b + d - c + b - a P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b + d - c + b - a b + d - c + b - a + d - c P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t = a b P ( t , c ) d t + b b + d - c Q ( b , t - b + c ) d t + b + d - c b + d - c + b - a P ( b + d - c + b - t , d ) ( - 1 ) d t + b + d - c + b - a b + d - c + b - a + d - c Q ( a , b + d - c + b - a + d - t ) ( - 1 ) d t = a b P ( t , c ) d t + c d Q ( b , t ) d t - a b P ( t , d ) d t - c d Q ( a , t ) d t = c d ( Q ( b , t ) - Q ( a , t ) ) d t - a b ( P ( t , d ) - P ( t , c ) ) d t = c d a b t i a l Q t i a l x ( s , t ) d s d t - a b c d t i a l P t i a l y ( t , s ) d s d t = S ( t i a l Q t i a l x - t i a l P t i a l y ,
(9)

proving the lemma.

## Lemma 2

Suppose SS is a right triangle whose vertices are of the form (a,c),(b,c)(a,c),(b,c) and (b,d).(b,d). Then Green's Theorem is true.

### Proof

We parameterize the boundary CSCS of this triangle as follows: For t[a,b],t[a,b], set φ(t)=(t,c);φ(t)=(t,c); for t[b,b+d-c],t[b,b+d-c], set φ(t)=(b,t+c-b);φ(t)=(b,t+c-b); and for t[b+d-c,b+d-c+b-a],t[b+d-c,b+d-c+b-a], set φ(t)=(b+d-c+b-t,b+d-c+d-t).φ(t)=(b+d-c+b-t,b+d-c+d-t). Again, one can check that this φφ parameterizes the boundary of the triangle S.S.

Write φ(t)=(x(t),y(t)).φ(t)=(x(t),y(t)). Again, using the Fundamental Theorem and (Reference), we have

C S ω = C S P d x + Q d y = a b P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b b + d - c P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b + d - c b + d - c + b - a P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t = a b P ( t , c ) d t + b b + d - c Q ( b , t + c - b ) d t + b + d - c b + d - c + b - a P ( b + d - c + b - t , b + d - c + d - t ) ( - 1 ) d t + b + d - c b + d - c + b - a Q ( b + d - c + b - t , b + d - c + d - t ) ( - 1 ) d t = a b P ( t , c ) d t + c d Q ( b , t ) d t - a b P ( s , ( d + s - b a - b ( c - d ) ) ) d s - c d Q ( b + s - d c - d ( a - b ) ) , s ) d s = c d ( Q ( b , s ) - Q ( ( b + s - d c - d ( a - b ) ) , s ) ) d s - a b ( P ( s , ( d + s - b a - b ( c - d ) ) ) - P ( s , c ) ) d s = c d b + s - d c - d ( a - b ) b t i a l Q t i a l x ( t , s ) d t d s - a b c d + s - b a - b ( c - d ) t i a l P t i a l y ( s , t ) d t d s = S ( t i a l Q t i a l x - t i a l P t i a l y , C S ω = C S P d x + Q d y = a b P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b b + d - c P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t + b + d - c b + d - c + b - a P ( φ ( t ) ) x ' ( t ) + Q ( φ ( t ) ) y ' ( t ) d t = a b P ( t , c ) d t + b b + d - c Q ( b , t + c - b ) d t + b + d - c b + d - c + b - a P ( b + d - c + b - t , b + d - c + d - t ) ( - 1 ) d t + b + d - c b + d - c + b - a Q ( b + d - c + b - t , b + d - c + d - t ) ( - 1 ) d t = a b P ( t , c ) d t + c d Q ( b , t ) d t - a b P ( s , ( d + s - b a - b ( c - d ) ) ) d s - c d Q ( b + s - d c - d ( a - b ) ) , s ) d s = c d ( Q ( b , s ) - Q ( ( b + s - d c - d ( a - b ) ) , s ) ) d s - a b ( P ( s , ( d + s - b a - b ( c - d ) ) ) - P ( s , c ) ) d s = c d b + s - d c - d ( a - b ) b t i a l Q t i a l x ( t , s ) d t d s - a b c d + s - b a - b ( c - d ) t i a l P t i a l y ( s , t ) d t d s = S ( t i a l Q t i a l x - t i a l P t i a l y ,
(10)

which proves Lemma 2.

## Lemma 3

Suppose S1,...,SnS1,...,Sn is a partition of the geometric set S,S, and that the boundary CSkCSk has finite length for all 1kn.1kn. If Green's Theorem holds for each geometric set Sk,Sk, then it holds for S.S.

### Proof

From Theorem 2 we have

C S ω = k = 1 n C S k ω , C S ω = k = 1 n C S k ω ,
(11)

and from (Reference) we have

S Q x - P y = k = 1 n S k Q x - P y . S Q x - P y = k = 1 n S k Q x - P y .
(12)

Since Green's Theorem holds for each k,k, we have that

C S k ω = S k Q x - P y , C S k ω = S k Q x - P y ,
(13)

and therefore

C S ω = S Q x - P y , C S ω = S Q x - P y ,
(14)

as desired.

## Exercise 4

1. Prove Green's Theorem for a right triangle with vertices of the form (a,c),(b,c),(a,c),(b,c), and (a,d).(a,d).
2. Prove Green's Theorem for a trapezoid having vertices of the form (a,c),(b,c),(a,c),(b,c),(b,d),(b,d), and (a,e),(a,e), where both dd and ee are greater than c.c. HINT: Represent this trapezoid as the union of a rectangle and a right triangle that share a border. Then use Lemma 3.
3. Prove Green's Theorem for SS any quadrilateral that has two vertical sides.
4. Prove Green's Theorem for any geometric set SS whose upper and lower bounding functions are piecewise linear functions. HINT: Show that SS can be thought of as a finite union of quadrilaterals, like those in part (c), each one sharing a vertical boundary with the next. Then, using induction and the previous exercise finish the argument.

We need one final lemma before we can complete the general proof of Green's Theorem. This one is where the analysis shows up; there are carefully chosen ϵϵ's and δδ's.

## Lemma 4

Suppose SS is contained in an open set UU and that ωω is smooth on all of U.U. Then Green's Theorem is true.

### Proof

Let the piecewise smooth geometric set SS be determined by the interval [a,b][a,b] and the two bounding functions uu and l.l. Using (Reference), choose an r>0r>0 such that the neighborhood Nr(S)U.Nr(S)U. Now let ϵ>0ϵ>0 be given, and choose deltadelta to satisfy the following conditions:

1. δ<r/2,δ<r/2, from which it follows that the open neighborhood Nδ(S)Nδ(S) is a subset of the compact set N¯r/2(S).N¯r/2(S). (See part (f) of (Reference).)
2. δ<ϵ/4M,δ<ϵ/4M, where MM is a common bound for all four continuous functions |P|,|Q|,|Py|,|P|,|Q|,|Py|, and |Qx||Qx| on the compact set N¯r/2(S).N¯r/2(S).
3. δ<ϵ/4M(b-a).δ<ϵ/4M(b-a).
4. δδ satisfies the conditions of (Reference).

Next, using (Reference), choose two piecewise linear functions pupu and plpl so that

1.  |u(x)-pu(x)|<δ/2|u(x)-pu(x)|<δ/2 for all x[a,b].x[a,b].
2.  |l(x)-pl(x)|<δ/2|l(x)-pl(x)|<δ/2 for all x[a,b].x[a,b].
3.  ab|u'(x)-pu'(x)|dx<δ.ab|u'(x)-pu'(x)|dx<δ.
4.  ab|l'(x)-pl'(x)|dx<δ.ab|l'(x)-pl'(x)|dx<δ.

Let S^S^ be the geometric set determined by the interval [a,b][a,b] and the two bounding functions u^u^ and l^,l^, where u^=pu+δ/2u^=pu+δ/2 and l^=pl-δ/2.l^=pl-δ/2. We know that both u^u^ and l^l^ are piecewise linear functions. We have to be a bit careful here, since for some xx's it could be that pu(x)<pl(x).pu(x)<pl(x). Hence, we could not simply use pupu and plpl themselves as bounding functions for S^.S^. We do know from (1) and (2) that u(x)<u^(x)u(x)<u^(x) and l(x)>l^(x),l(x)>l^(x), which implies that the geometric set SS is contained in the geometric set S^.S^. Also S^S^ is a subset of the neighborhood Nδ(s),Nδ(s), which in turn is a subset of the compact set N¯r/2(S).N¯r/2(S).

Now, by part (d) of the preceding exercise, we know that Green's Theorem holds for S^.S^. That is

C S ^ ω = S ^ ( Q x - P y ) . C S ^ ω = S ^ ( Q x - P y ) .
(15)

We will show that Green's Theorem holds for SS by showing two things: (i) |CSω-CS^ω|<4ϵ,|CSω-CS^ω|<4ϵ, and (ii) |S(Qx)-Py)-S^(Qx-Py)|<ϵ.|S(Qx)-Py)-S^(Qx-Py)|<ϵ. We would then have, by the usual adding and subtracting business, that

| C S ω - S ( Q x - P y ) | < 5 ϵ , | C S ω - S ( Q x - P y ) | < 5 ϵ ,
(16)

and, since ϵϵ is an arbitrary positive number, we would obtain

C S ω = S ( Q x - P y ) . C S ω = S ( Q x - P y ) .
(17)

Let us estabish (i) first. We have from (1) above that |u(x)-u^(x)|<δ|u(x)-u^(x)|<δ for all x[a,b],x[a,b], and from (3) that

a b | u ' ( x ) - u ^ ' ( x ) | d x = a b | u ' ( x ) - p u ' ( x ) | d x < δ . a b | u ' ( x ) - u ^ ' ( x ) | d x = a b | u ' ( x ) - p u ' ( x ) | d x < δ .
(18)

Hence, by (Reference),

graph ( u ) ω - graph ( u ^ ) ω | < ϵ . graph ( u ) ω - graph ( u ^ ) ω | < ϵ .
(19)

Similarly, using (2) and (4) above, we have that

| graph ( l ) ω - graph ( l ^ ) ω | < ϵ . | graph ( l ) ω - graph ( l ^ ) ω | < ϵ .
(20)

Also, the difference of the line integrals of ωω along the righthand boundaries of SS and S^S^ is less than ϵ.ϵ. Thus

| C ( b , l ( b ) ) ( b , u ( b ) ) ω - C ( b , l ^ ( b ) ) ( b , u ^ ( b ) ) ω | = | l ( b ) u ( b ) Q ( b , t ) d t - l ^ ( b ) u ^ ( b ) Q ( b , t ) d t | | u ( b ) u ^ ( b ) Q ( b , t ) d t | + | l ^ ( b ) l ( b ) Q ( b , t ) d t | M ( | l ( b ) - l ^ ( b ) | + | u ( b ) - u ^ ( b ) | ) M ( δ + δ ) = 2 M δ < ϵ . | C ( b , l ( b ) ) ( b , u ( b ) ) ω - C ( b , l ^ ( b ) ) ( b , u ^ ( b ) ) ω | = | l ( b ) u ( b ) Q ( b , t ) d t - l ^ ( b ) u ^ ( b ) Q ( b , t ) d t | | u ( b ) u ^ ( b ) Q ( b , t ) d t | + | l ^ ( b ) l ( b ) Q ( b , t ) d t | M ( | l ( b ) - l ^ ( b ) | + | u ( b ) - u ^ ( b ) | ) M ( δ + δ ) = 2 M δ < ϵ .
(21)

Of course, a similar calculation shows that

| C ( a , u ( a ) ) ( a , l ( a ) ) ω - C ( a , u ^ ( a ) ) ( a , l ^ ( a ) ) ω | < ϵ . | C ( a , u ( a ) ) ( a , l ( a ) ) ω - C ( a , u ^ ( a ) ) ( a , l ^ ( a ) ) ω | < ϵ .
(22)

These four line integral inequalities combine to give us that

| C S ω - C S ^ ω | < 4 ϵ , | C S ω - C S ^ ω | < 4 ϵ ,
(23)

establishing (i).

Finally, to see (ii), we just compute

0 | S ^ ( Q y - P x ) - S ( Q y - P x ) | = | a b l ^ ( t ) u ^ ( t ) ( Q x ( ( t , s ) - P y ( t , s ) ) d s d t - a b l ( t ) u ( t ) ( Q x ( t , s ) - P y ( t , s ) ) d s d t | = | a b l ^ ( t ) l ( t ) ( Q x ( t , s ) - P y ( t , s ) ) d s d t + a b u ( t ) u ^ ( t ) ( Q x ( t , s ) - P y ( t , s ) ) d s d t | 2 M ( a b | l ( t ) - l ^ ( t ) | + | u ^ ( t ) - u ( t ) | d t 4 M δ ( b - a ) < ϵ . 0 | S ^ ( Q y - P x ) - S ( Q y - P x ) | = | a b l ^ ( t ) u ^ ( t ) ( Q x ( ( t , s ) - P y ( t , s ) ) d s d t - a b l ( t ) u ( t ) ( Q x ( t , s ) - P y ( t , s ) ) d s d t | = | a b l ^ ( t ) l ( t ) ( Q x ( t , s ) - P y ( t , s ) ) d s d t + a b u ( t ) u ^ ( t ) ( Q x ( t , s ) - P y ( t , s ) ) d s d t | 2 M ( a b | l ( t ) - l ^ ( t ) | + | u ^ ( t ) - u ( t ) | d t 4 M δ ( b - a ) < ϵ .
(24)

This establishes (ii), and the proof is complete.

At last, we can finish the proof of this remarkable result.

## Theorem 4: PROOF OF GREEN'S THEOREM

### Proof

As usual, let SS be determined by the interval [a,b][a,b] and the two bounding functions uu and l.l. Recall that u(x)-l(x)>0u(x)-l(x)>0 for all x(a,b).x(a,b). For each natural number n>2,n>2, let SnSn be the geometric set that is determined by the interval [a+1/n,b-1/n][a+1/n,b-1/n] and the two bounding functions unun and ln,ln, where un=u-(u-l)/nun=u-(u-l)/n restricted to the interval [a+1/n,b-1/n],[a+1/n,b-1/n], and ln=l+(u-l)/nln=l+(u-l)/n restricted to [a+1/n,b-1/n].[a+1/n,b-1/n]. Then each SnSn is a piecewise smooth geometric set, whose boundary has finite length, and each SnSn is contained in the open set S0S0 where by hypothesis ωω is smooth. Hence, by Lemma 4, Green's Theorem holds for each Sn.Sn. Now it should follow directly, by taking limits, that Green's Theorem holds for S.S. In fact, this is the case, and we leave the details to the exercise that follows.

## Exercise 5

Let S,ω,S,ω, and the SnSn's be as in the preceding proof.

1. Using (Reference), show that
CSω=limCSnω.CSω=limCSnω.
(25)
2. Let ff be a bounded integrable function on the geometric set S.S. Prove that
Sf=limSnf.Sf=limSnf.
(26)
3. Complete the proof to Green's Theorem; i.e., take limits.

REMARK Green's Theorem is primarily a theoretical result. It is rarely used to “compute” a line integral around a curve or an integral of a function over a geometric set. However, there is one amusing exception to this, and that is when the differential form ω=xdy.ω=xdy. For that kind of ω,ω, Green's Theorem says that the area of the geometric set SS can be computed as follows:

A ( S ) = S 1 = S t i a l Q t i a l x = C S x d y . A ( S ) = S 1 = S t i a l Q t i a l x = C S x d y .
(27)

This is certainly a different way of computing areas of sets from the methods we developed earlier. Try this way out on circles, ellipses, and the like.

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