Two-Dimensional Calculus (2011)
Chapter 4. Integration
25. Green’s theorem for a rectangle
In Sect. 22 we have seen how the fundamental theorem of calculus generalizes to line integrals. Our next objective is to obtain an analog of the fundamental theorem in the case of double integrals.
As our point of departure; we take the second form of the fundamental theorem, Eq. (22.2) which we may write in the form
The left-hand side of this equation consists of the integral over an interval of the derivative of a function of one variable. The corresponding expression in two variables consists of a double integral over a rectangle of a partial derivative of a function of two variables. Specifically, consider the rectangle
and let q(x, y)∈ in a domain that includes the rectangle R. We wish to evaluate
Applying Th. 24.3, we first integrate along each horizontal line y = t obtaining a function of t
For a fixed value of t, the partial derivative qx(x, t) reduces to the ordinary derivative with respect to x of the function f(x) = q(x, t), and by Eq. (25.1),
Then
In exactly the same fashion, applying Th. 24.4 to the partial derivative ∂p/∂y, where p(x, y) ∈ in some domain including R, we find
Equations (25.3) and (25.4) constitute, in primitive form, the two-variable generalization of Eq. (25.1). A more useful version is obtained by observing that the right-hand sides of (25.3) and (25.4) are in fact line integrals over parts of the boundary of the rectangle R. In fact, the boundary of R can be described as the piecewise smooth curve C consisting of the four line segments ( Fig. 25.1)
FIGURE 25.1 The curve C defined by Eqs. (25.5)
Note that these segments are oriented so that when traversed in succession they form the closed curve C, beginning and ending at (a, c).
Now the right-hand side of (25.3) consists of two line integrals of the form ∫ q dy taken over the two vertical sides of R, each side being traversed from bottom to top. Since side C4, as described in (25.5), is traversed from top to bottom, it follows from Lemma 21.2 that the value of the line integral has the opposite sign. Thus, (25.3) can be written as
since dy/dt ≡ 0 on the horizontal sides C1 and C3 we have
It follows that
In the same way, we have
and observing the orientation of the sides C1 and C3 we may write (25.4) in the form
Equations (25.7) and (25.8) together yield the desired two-variable generalization of (25.1). We may state the result as follows.
Theorem 25.1 Green’s Theorem for a Rectangle Let p(x, y), q(x, y) ∈ in a domain that includes the rectangle R defined by (25.2). Let the boundary of R be the closed curve C described by (25.5). Then
and hence
Remark The two equations in (25.9) are the direct generalizations of Eq. (25.1). In a rough way, these equations state the following: to evaluate the double integral over a rectangle of a partial derivative of a function, it is sufficient to know the values of the function on the boundary. Clearly, Eq. (25.1) yields an analogous statement for ordinary derivatives, replacing “rectangle” by “interval” and “boundary” by “endpoints.” For most applications, however, it is not the individual equations in (25.9) that occur, but the combined form (25.10).
Example 25.1
Evaluate ∫c x dy, where C is described by (25.5).
Method 1. By direct computation, using the definition of the line integral,
Method 2.Applying (25.10), with p = 0,q = x we obtain
We are now ready to take up again the thread of our discussion at the end of Sect. 22. We prove the following result, which is a partial converse of Lemma 19.1.
Theorem 25.2 Let p(x, y), q(x, y) be a vector field in a disk D:(x − x 1 + (y− y1). < r2. If
throughout D, then there exists a function f(x, y) ∈ in D such that
PROOF. We have seen in Lemmas 22.3 and 22.4 how to construct a pair of functions g(x, y), h(x, y) in D such that gx = p and hy = q. We shall show
FIGURE 25.2 Notation for the proof of Th. 25.2
that if (25.11) holds, then the functions g(x, y) and h(x, y) are in fact equal, and denoting their common value by f(x, y), we have the desired result (25.12).
We recall first the definition of g and h. Let (x, y) be any point of D, and consider the line segments (25.5), where (a, c) = (x1, y1) and (b, d) = (x, y) ( Fig. 25.2). Then from Lemma 22.4,
Similarly from Lemma 22.3,
But the rectangle R defined by (25.2) is included in the disk D, and we may apply Th. 25.1. The integrand on the right-side of (25.10) is identically zero by (25.11), and hence
Thus, for an arbitrary point (x, y) in D, h(x, y) = g(X Y) and the theorem is proved.
Let us now collect and summarize our various results for conservative vector fields.
Let v(x, y) = p(x, y), q(x, y) ∈ in a domain D. Consider the following properties.
1. There exists f(x, y) ∈ in D such that v = ∇f.
2. ∫c p dx + q dy is independent of path.
3. ∫c p dx + q dy = 0, for all closed curves C in D.
4. py = qx throughout D.
We have the following relations between these properties.
a. 1 ⇒ 4 for any domain D (Lemma 19.1).
b. 1,2, and 3 are equivalent for any domain D (Th. 22.1 and Lemma 22.5).
c. In the case that D is a disk, 4 ⇒ 1 (Th. 25.2).
d. In the case that D is a disk, all four properties are equivalent (combining relations a, b, c).
e. For certain domains D, the four properties are not equivalent (Example 22.3).
These facts are basic in the theory of vector fields. There is an important distinction between property 4 and the other three properties that may further clarify the situation. Property 4 is a local property, in the sense that its validity at each point of the domain depends only on the values of the functions p and q in a neighborhood of that point and is not affected by the values in the rest of the domain. The first three properties, on the other hand are globalproperties in the sense that they can only be verified if the functions p and q are known throughout the domain D. We may then add to our list of relations
f. 4 ⇒ 1 locally.
g. All four properties are locally equivalent.
The precise meaning of these statements is the following. Suppose that property 4 holds in D, and let (x1, y1) be any point of D. If D' is a disk centered at (x1, y1) and included in D ( Fig. 25.3), then by Th. 25.2, there exists a function f(x, y) defined in D' satisfying v = ∇f. Thus, such a function may be defined in some neighborhood of each point of D, but it may be impossible; to define a single function in all of D satisfying property 1 (see Example 25.1 below). This is the meaning of relation f. Similarly, relation g means the properties 2 and 3 also hold, provided that the curves C referred to in these properties are required to lie in the neighborhood D'.
Example 25.2
Let
in the domain D consisting of the whole plane except the origin. We have seen in the discussion of Example 22.3 that in this case property 4 holds, but properties 1, 2, 3 do not. On the other hand, we now know by Th. 25.2 that in
FIGURE 25.3 Localization to a neighborhood of (x1, y1)
some neighborhood of each point of D, there must exist a function f(x, y) satisfying
In order to find such a function, let us consider for example the second equation in (25.13) for a fixed value of x as a one-variable problem:
It follows that
where c(x0) denotes a constant depending on x0.This means that the function f(x, y) satisfying (25.13), whose local existence we know, must be of the form
where g(x) is some function of x. Furthermore, since f(x, y) ∈ , g(x) is differentiable, and it follows that
Comparing this with (25.13), we see that g'(x) must be zero, and hence g(x) is constant. We thus have a solution of (25.13) in the form
Geometrically, the function tan−1 (y/x) represents the polar angle θ between the positive x axis and the ray from the origin to the point (x, y) ( Fig. 25.4). The additive constant c may be interpreted geometrically by saying that the function f(x, y) represents the angle between the ray from the origin to (x, y) and any fixed ray through the origin.
FIGURE 25.4 Geometric interpretation of the function tan−1 y/x
Example 25.2 illustrates clearly the distinction between local and global potential functions alluded to in relation f above. Namely, given any point of D, that is to say, any point other than the origin, then in some neighborhood of that point one can choose a single-valued branch of tan−1 (y/x), and that will be a potential function of the vector field. Furthermore, up to an additive constant, it is the unique potential function in the neighborhood. On the other hand, it is impossible to find a function f(x, y)∈ in all of D that satisfies (25.13), since that would imply a continuous single-valued choice of the polar angle θ in the whole plane except the origin. But if we choose a determination of θ (or f) near some point, and then continue to observe its value as we traverse a circle going once counterclockwise around the origin, then the final value is 2π more than the starting value. We are thus faced with the choice of making θ multivalued or discontinuous, neither of which fits the definition of a potential function.
Let us conclude with the observation that this discussion explains the value 2π found by a direct computation in Example 22.3 for the line integral
where C is the circle x = cos t, y = sin t, 0 ≤ t ≤ 2π. Namely, in a neighborhood of any point on C, there is a well-defined potential function f = θ, and if we choose a short arc C1 of C, which lies in the neighborhood, then the line integral over C1 is equal to the change in the potential function (Fig. 25.5):
which is the angle subtended by that arc. If we divide the circle into a number of such arcs, the total integral around C is the sum of these angles, which is 2π.
FIGURE 25.5 Geometric interpretation of ∫C1 − y/ (x2 + y2) dx + x/(x2 + y2) dy
Exercises
25.1 Let C be the piecewise smooth curve consisting of the four successive line segments C1 C2, C3, C4 described in Eq. (25.5) (see Fig. 25.1). Compute each of the following line integrals in two ways, first by a direct computation, and then by Green’s theorem.
a. ∫C y dx
b. ∫C x2 dy
c. ∫C x dx + y dy
d. ∫C yex dx + (x + ex) dy
25.2 Obtain the answer to Ex. 25.1c by a third method, using a potential function for the vector field x, y .
25.3 Show that the answer to Ex. 25.1b is equal to twice the x moment of the rectangle R bounded by C. Write down the coordinates of the centroid of R, and check the answer to Ex. 25.1b.
25.4 Let C be the curve of Ex. 25.1, where (a, c) = (−1, −1), (b, d) = (1, 1). Use Green’s theorem to evaluate the following line integrals.
a.
b. ∫C (sinh −1 x + yexy) dx + (tanh y + xexy) dy
c. ∫C (xy2 − x2y4) dx + (x2y + 3x3y3) dy
(Hint: use Ex. 24.13 or 24.14a.)
25.5 Let F be the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. For each of the following functions f(x, y), express ∫∫F f dA as a line integral over the boundary of F, and evaluate that integral.
a. f(x, y) = xy
b. f(x, y) = 3x2y + y3
c. f(x, y) = ex sin π y − ey cos π x
25.6a. Evaluate by direct computation
where C is the curve of Ex. 25.4.
b. If Green’s theorem (Eq. (25.10)) is applied to the integral in part a, then since qx = py in this case, the value of the double integral would appear to be zero. This contradicts the result obtained by direct computation. What is the source of difficulty?
c. Find the answer to part a by using a local potential function on each of the line segments constituting the curve C.
25.7 Let C be the curve of Ex. 25.1. For any point (X, Y) not on C, define φ(X, Y) by
*a. Find φ(x, y) for a point (x, y) inside C. (Hint: find a local potential function on each side of C, and use it to evaluate the integral over that side.)
b. Find φ(x, y) for a point (x, y) outside C. (Hint: use either the method of part a, or Green’s theorem.)
*25.8 Let u(x, y) ∈ in a domain that includes a rectangle R. Let the boundary of R be the curve C consisting of the four line segments described in Eq.(25.5). At each point of C other than the corners, let n denote the exterior normal to R; that is, the unit vector perpendicular to C, pointing toward the exterior of R. Denote by ∂u/∂n, the directional derivative of u in the direction of n. Show that
25.9 Find a potential function f(x, y) for each of the following vector fields p, q , using the method given in the text to solve Eq. (25.13).
a.
b. p, q = x2 − y2,−2xy
c.
d. p,q = 2 sec2 x tan x tan y, sec2 x sec2y
25.10 Apply the method of Ex. 25.9 to obtain the answers to Ex. 22.8.
25.11 Let v(x, y) be a “horizontal” vector field in a domain D; v(x y) = p(x, y),0 , where p(x, y) ∈ .
a. Show that if D is a disk, then v is conservative if and only if p(x, y) depends only on x. (Hint: see Ex. 7.27.)
*b. Show that it is possible to find a domain D and a function p(x, y) ∈ in D such that p(x, y) cannot be expressed as a function of x alone, and nevertheless p(x, y), 0 is conservative in D. (Hint: see Ex. 7.29.)
25.12 Show that harmonic vector fields in a disk coincide with the gradient vector fields of harmonic functions.
25.13 Show that a vector field w = u(x, y), υ(x, y) in a disk D whose divergence is identically zero always has a stream function. (Hint: apply Th.25.2 to the vector field p, q = −υ, u , and use Ex. 19.12.)
25.14 Let F be the figure bounded by the lines x = 0, y = 0, x/a + y/b = 1. Let C be the boundary of F where C consists of the three successive line segments: C1, from (0, 0) to (a, 0), C2, from (a, 0) to (0, b), and C3, from (0, b) to (0, 0). Show that Green’s theorem holds in the following form.
a. Given q(x, y)∈ in a domain which includes the figure F, verify each of the following steps.
b. If p(x, y) ∈ in a domain that includes F show by a reasoning analogous to part a that ∫∫F py dA = −∫C p dx.
c. Deduce that
25.15 Let C be the curve of Ex. 25.14. Use Ex. 25.14c to evaluate
a. ∫c x dy
b. ∫c y dx
c. ∫c cos ex dx + ecos y dy
*25.16Let C and φ(x, Y) be defined as in Ex. 25.7.
a. Show that for any point (X, Y) not on C, the partial derivatives φX, φY, may be obtained by “differentiating under the integral sign,” and that the result in this case takes the following form:
(Hint: apply Lemma 7.2 to the integrals over each side of C.)
b. Using the fact that qx = py at each point of the curve C in part a, show that ∇ φ = 0 at each point (X, Y) not in C.
c. Use part b together with the answer to Ex. 25.6a, to obtain the answer to Ex. 25.7a.
25.17 a. Consider the curve C:x = cos kt, y = sin kt, 0 ≤ t ≤ 2π, where k is an integer, positive or negative. Evaluate
b. Let C be a closed piecewise smooth curve that does not pass through the origin. Explain why you would expect the integral in part a to equal 2πn for some integer n. (Hint: use the reasoning given in the text at the end of this section; divide the curve into short arcs, and describe geometrically the value of the integral over each of these arcs.)
*25.18 Let C be a piecewise smooth closed curve. For any point (X, Y) not on C let
Show that n(C; X, Y) has the following properties.
a. If D is any domain which does not meet the curve C, then n(C: X, Y) is constant on D. (Hint: see Ex. 25.16.)
b. n(C; X, Y) is an integer. (Hint: see Exs. 25.7a and 25.17b.)
c. n(C; X, Y) may be described as “the number of times the curve C goes around the point (X, Y) in the counterclockwise direction.”
Note: the number n (C: X, Y) is called the winding number of the curve C about the point (X, Y). Using the notation described in Ex. 21.25 we could write, for fixed (X, Y)
where (see Fig. 25.6)
Although θ is not single-valued, the derivatives θx, θy are single-valued, and the integral ∫c dθ is well-defined. Intuitively, it represents the total variation of the angle θ as the point (x, y) traverses the curve C. For an excellent discussion of winding numbers and their applications, see Part II of [7].
FIGURE 25.6 A curve C which winds twice about a point (X, Y)
Addendum to Section 25 Simple Connectivity
Remark It is quite clear that the proof of Th. 25.2 can be adapted for domains D that are more general than a disk. The widest class of domains for which the theorem holds is the class of simply-connected domains. Simple connectivity is a very important geometric concept, which may be applied not only to plane domains, but to surfaces in space, three-dimensional bodies, and to still more general situations. However, in the case of the plane there are a number of different, but equivalent characterizations of simple connectivity. The following definition is specially adapted to plane domains, making use of the winding number n(C; X, Y) defined in Ex. 25.18.
Definition 25.1 A plane domain D is simply-connected if the winding number n(C; X, Y) is equal to zero for every piecewise smooth closed curve C lying in D and for all points (X, Y) not in D. A plane domain which is not simply connected is called multiply-connected.
Intuitively, a simply-connected domain is one “without holes.” If a domain D has a hole in it, then choosing a point (X, Y) inside the hole and a curve C going around the hole gives n(C; X, Y) ≠ 0. For example, if D is the annulus 1 < x2 + y2 < 9, then choose (X, Y) = (0, 0) and C: x cos t, y = sin t, 0 ≤ t ≤ 2π, then n(C; X, Y) = 1 ; hence D is multiply-connected.
A fact that is useful to bear in mind is that a domain bounded by a single closed curve is simply-connected, whereas a domain bounded by two or more closed curves is multiply-connected ( Fig. 25.7).
FIGURE 25.7 Examples of simple connectivity and of multiple connectivity
Theorem 25.3 Given a domain D, the necessary and sufficient condition that every vector field p(x, y), q(x, y) in D satisfying py ≡ qx should have a potential function in D is that D be simply-connected.
PROOF I. Suppose that D is not simply-connected. Then there exists a closed curve C in D and a point (X, Y) not in D such that n(C: X, Y) ≠ 0. Since the fixed point (X, Y) is not in D, it follows that the functions
are functions in D. By direct computation py ≡ qx. However,
and by Th. 22.2, the vector field p, q cannot have a potential function in D.
II. Suppose that D is simply-connected. Let p, q be an arbitrary vector field in D satisfying py ≡ qx. To define a potential function f(x, y) we fix a point (x1, y1) in D, and then for an arbitrary point (x2, y2) we set
where C is a curve in D that starts at (x1, y1), ends at (x2, y2), and consists of a finite number of horizontal and vertical line segments. (The existence of such a curve is easily proved.) If we can show that the value of this integral depends only on the point (x2, y2) and not on the particular choice of path C, then by choosing first a path ending in a horizontal segment and then a path ending in a vertical segment, it follows by the reasoning of Lemmas 22.3 and 22.4 that the function f(x, y) satisfies fx = p, fy = q and is therefore a potential function. Suppose then that C1, C2 are two paths having the properties described. Let be the closed curve in D consisting of C1 followed by C2described in the opposite direction. We wish to show that
The idea of the proof is to show that the curve can be broken up into a number of line segments, some of which are described twice in opposite directions making the line integrals cancel out, and the rest of which may be grouped together to form the boundaries of rectangles lying in D ( Fig. 25.8). Applying Green’s theorem to each of these rectangles (using the hypothesis that qx − py ≡ 0 in D) and adding over all the rectangles gives the desired result . A complete proof that can be decomposed in the manner indicated requires a careful analysis, which is omitted here. We refer to the proof of Theorem 15 in Chapter 4, Section 4.3 of [2].
FIGURE 25.8 Decomposition of the curve into a collection of boundaries of rectangles
We note in conclusion that the above discussion is based on the treatment of simple connectivity given in the book of Ahlfors [2]. The equivalence of the definition given here with other characterizations of simply-connected plane domains is proved in Section 4.2 of Chapter 4 and in Section 1.5 of Chapter 8 of Ahlfors’ book.