Mean value theorem for the double integral; vector forms of Green’s theorem - Integration - Two-Dimensional Calculus

Two-Dimensional Calculus (2011)

Chapter 4. Integration

27. Mean value theorem for the double integral; vector forms of Green’s theorem

The notion of the mean value (or average value) of a variable quantity is useful in many connections. As an example involving functions of one variable, the average speed of a car is defined to be the distance covered divided by the time elapsed. Thus, if s(t) represents distance covered at time t, then the instantaneous velocity at time t is f(t) = s'(t), and the average velocity between time t = a and time t =b is given by:

Image

Thus if the function f(t) represents instantaneous velocity, then the average velocity over any interval is equal to the integral of f divided by the integral of the function 1 over the interval. Whether the function f(t) represents velocity or not, the right-hand side of (27.1) is taken to define the average value of f(t) over the interval atb.

For an example involving functions of two variables, we may take a plate cut out in the shape of a figure F, having variable density f(x,y). The mass of the plate is then equal to ∫∫F fdA. We may define the average density of the plate to be the value μ such that if the figure F had constant density μ, the total mass would be the same :

Image

or

Image

Thus the average density equals the total mass divided by the area of F. In a similar fashion, the average height of a surface z = f(x, y) would be defined to be that constant height, which would yield the same volume. We would then obtain the same value μ, defined by (27.2), interpreted as the total volume divided by the area of the base.

Regardless of the interpretation of the function f(x, y), its average value is defined in the same way.

Definition 27.1The mean value or average value of a function f(x, y) over a figure F is the number μ defined by (27.2).

Thus, the mean value of f is that constant whose integral over F yields the same value as the integral of f.

Example 27.1

The mean value of the function f(x, y) = x over a figure F is precisely the x coordinate of the centroid of F. Similarly for f(x, y) = y. Thus the centroid of F may be described as the point whose coordinates are the average values over F of the x and y coordinates, respectively.

In general, the mean value of a function need not be a value that the function actually assumes. Thus, a function that is equal to 2 on one half of a figure, and 4 on the other half, has a mean value of 3, although it is not actually equal to 3 at any point. For a continuous function, however, this cannot happen. We have the following basic result.

Theorem 27.1 Mean-Value Theorem If f(x, y) is continuous on a figure F, then there is a point (x0, y0) in F such that

Image

PROOF.Let M be the maximum of f(x, y) on F, and let (x1, y1) be a point such that f(xl, y1) = M. Similarly, let m be the minimum of f(x, y) on F, and suppose f(x2, y2) = m. Then

Image

Dividing through by ∫∫F1 dA, and defining μ by (27.2), yields

Image

Let C be a curve in F joining (x1, y1) to (x2, y2)· Then fc(t) is a continuous function on an interval and takes on the values m and M, hence every value between m and M, and in particular the value μ. At the corresponding point (x0, y0) on the curve C, f(x0, y0) = μ, which proves the theorem. image

Remark It is crucial to the above proof, and to the correctness of the theorem, that the definition of a figure was in terms of a domain, which is a connected set.

We next give an important application of the mean-value theorem.

Theorem 27.2Let f(x, y) be continuous in a domain D, and let (x0, y0) be any point of D. Let F1, F2,… be a sequence of figures lying in D, which all contain (x0, y0) and whose diameters tend to zero. Let A(Fk) be the area of Fk; then

Image

Remark In many applications, each Fk is a circular disk centered at (x0, y0).

PROOF. Applying the mean value theorem to the figure Fk, we find that there is a point (xk, yk) in Fk such that f(xk, yk) = image Since the diameters of Fk tend to zero as k tends to infinity, it follows that (xk, yk)→ (x0, y0). By the continuity of f at (x0, y0), f(x0, y0) = limk→∞ f(xk, yk). This proves the theorem. image

Corollary Given a differentiable transformation

Image

in a domain D, let (x0, y0) be a point at which the Jacobian is different from zero. Let F1, F2,. . . be a sequence of figures lying in D, containing the point (x0, y0), whose diameters tend to zero. Let imagek be the image of Fk under G, and let A(Fk), A(image k) denote the areas of Fk and imagek; then

Image

PROOF. Since the Jacobian of G is different from zero at (x0, y0), it follows from the inverse mapping theorem (Theorem 17.3) that there is a neighborhood D' of (x0, y0) in which the mapping G has a differentiable inverse. Thus G restricted to D' is a diffeomorphism, and since the diameters of Fk tend to zero, from some point on they must all lie in D'. We may then apply Th. 26.2 to conclude that

Image

Consequently, setting f(x, y) = |∂(u, v)/∂(x, y)|, Eq. (27.3) reduces to (27.4), and the result is proved. image

RemarkThe property of the Jacobian expressed in Eq. (27.4) is the fundamental geometric interpretation of the Jacobian, which we have already mentioned at the end of Sect. 16.

We next give applications of Th. 27.2 to the theory of vector fields.

Let v(x, y) = imagep(x, y), q(x, y)imageimage in a domain D. We noted in Eqs.

(21.12) and (21.13a) that a line integral may be written in vector form as

Image

where vT is the tangential component of v along the curve C. Now if F is a figure lying in D such that ∂F consists of the single closed curve C, then Green’s theorem takes the form

Image

For v(x, y) the velocity field of a fluid flow, the right-side of (27.5) was interpreted as the total rate of flow around the curve C.

Choose any point (x0, y0) in D, and choose a sequence of curves Cn such that Cn = ∂Fn, where Fn is a figure lying in D and containing (x0, y0) and where the diameters of Fn tend to zero. For example, the Cn could be circles centered at (x0, y0). Then applying (27.3), with f(x, y) = qxpy, we find

Image

Thus, the quantity qx − py may be described as a measure of the “vorticity” of the flow at each point. As we mentioned at the end of Sect. 22, a vector field is called “irrotational” if qxpy ≡ 0.

Regardless of the physical interpretation, Eq. (27.6) has an important consequence. Namely, in view of (27.5), the numerators on the right-hand side of (27.6) may be described in terms of the vector field v, in a fashion that is completely independent of coordinates. Suppose, in fact, that D is a plane domain, and that without choosing any coordinate system in the plane, we assign to each point of D a vector v in the form of a directed line segment. If Cn is a circle of radius rn, then the tangential component vT of v is defined at each point of Cn and provides a function whose integral

Image

is on the one hand defined without any reference to a coordinate system, and on the other hand, once a coordinate system is chosen, is equal by (27.5) to image. Thus, the right-hand side of (27.6) consists of a sequence of numbers, which are independent of the coordinate system, and their limit, the value of qx − py at (x0, y0) must also be independent of the coordinates. One can show directly that under a rotation of coordinates the quantity qx − py remains invariant (see Ex. 20.19c). However, such a computation provides no insight as to why this particular quantity is invariant. It is always more illuminating to prove the invariance of a quantity by obtaining an expression for it that is intrinsic and independent of coordinates. Equation (27.6) , together with (27.5), does this for the quantity qx − py.

Example 27.2

Let p = x, q = y ( Fig. 27.1).

If C is a circle with center at the origin, then the vector v = imagex, yimage is perpendicular to the circle at the point (x, y) and hence vT = 0 at each point. Thus

Image

If C is a circle not centered at the origin, then vT is not zero at each point, but it is clear from the symmetry of the flow that the line through the origin and the center of the circle divides the circle into two halves such that at each pair of points on the circle which are symmetric about this line, the quantity vT has the same magnitude, but opposite sign ( Fig. 27.1). Thus the total clockwise flow equals the total counterclockwise flow, and again

Image

Image

FIGURE 27.1 The vector field imagex, yimage along various circles

If we now choose a sequence of circular disks Fn centered at a point (x0, y0), then the flow around ∂Fn is zero for each n, and therefore the right-hand side of (27.6) is equal to zero. That the left-hand side is also zero at every point is easily verified.

We may note that this vector field has the potential function f(x, y) = image(x2 + y2), and consequently for every closed curve C, ∫c x dx + y dy = 0, even though this may not be obvious geometrically for arbitrary C.

Example 27.3

Let p = −y, q = x ( Fig. 27.2).

Let C be the circle x = r cos t, y = r sin t, 0 ≤ t ≤ 2π. Then imagex', y'image = image − r sin t, r cos timage , and the unit tangent is image = < —sin t, cos t >. Thus, at each point of C the vector v = image−y, x>image = imager sin t, r cos timage has the same direction as image, and the tangential component is

Image

whence

Image

Since the area inside C is πr2, if we choose for the figures Fn a sequence of disks centered at the origin, then each term on the right-hand side of (27.6) is equal to image, and the limit is also equal to 2. As for the left-hand side, we have qx = 1, Py = 1, qx − Py = 2. Thus Eq. (27.6) is verified for (x0, y0) = (0, 0).

Image

FIGURE 27.2 The vector field <—y, x> along various circles

For an arbitrary point (x0, y0) the left-hand side of (27.6) is still equal to 2. It is more difficult to compute directly the flow around a circle not centered at the origin, but it is clear geometrically that the value obtained must be a positive number. In fact, the circle is divided into two arcs by the pair of points whose tangent lines pass through the origin ( Fig. 27.2). On the arc further from the origin the flow is in the counterclockwise direction, while on the arc nearer the origin it is clockwise. But the further arc is the longer one, and the vector field has greater magnitude. It is therefore clear that the amount of counterclockwise flow is greater than the amount of flow in the clockwise direction ; that is, vT ds > 0. Using Green’s theorem, we find that if C bounds a disk F of radius r, then

Image

so that the flow around an arbitrary circle depends only on its radius, and not on the position of its center.

Example 27.4

Let p = y, q = 0 ( Fig. 27.3).

If we consider the vector field v = imagey, 0image just in the domain D:0 < y < 1, it may be pictured as representing a horizontal flow along a canal, where the velocity is lower near the bottom because of friction. Although this flow does not have the obvious vorticity of the previous example, we nevertheless find that

Image

at every point. This is explained by the fact that on the upper half of every circle, where the velocity is greater, the flow is clockwise, whereas on the lower half it is counterclockwise ( Fig. 27.3). Thus the total flow is clockwise, or ∫c vT ds < 0.

Image

FIGURE 27.3 A “horizontal” vector field

We may again find the precise value by Green’s theorem

Image

where F is a disk of radius r, and C = ∂F.

We turn next to the quantity px + qy, which we introduced in Sect. 19 as the divergence of the vector field v. We note that by Green’s theorem :

Image

and

Image

so that

Image

Suppose now that ∂F consists of a single smooth curve

Image

Then

Image

We observe that the integrand on the right-hand side is equal to the dot product of imagep, qimage andimage y'(t), −x'(t)image . Furthermore, the vector imagey'(t), − x'(t)image has length (y'(t)2 + x'(t)2)1/2 = ds/dt, and direction perpendicular to the tangent vector imagex'(t), y'(t)image . By the definition of ∂F, the curve is positively oriented with respect to F, so that the vector imagey'(t), x'(t)image is directed toward the interior of F. Thus, the vector imagey'(t), −x'(t)image has the opposite direction, and is equal to ds/dt N, where N is the unit normal directed toward the exterior of F ( Fig. 27.4). Equation (27.8) then takes the form

Image

where vN = v·N denotes the normal component of v in the direction of the exterior normal to F. We note again the important fact that the right-hand side of (27.9) is an intrinsic quantity, which may be defined without any reference to coordinates.

Combining (27.7) and (27.9), and introducing the notation

Image

Image

FIGURE 27.4 The exterior normal N

for the divergence of the vector field v, we obtain

Image

This equation is just one more form of Green’s theorem, emphasizing the interpretation of the pair of functions p, q as the components of a vector field v.

At the end of Sect. 20, we showed that the divergence of a vector field is a quantity that does not depend on the choice of coordinates. We are now in a position to give a direct proof of this fact by obtaining an intrinsic expression for the divergence. We again choose any point (x0 .y0) and a sequence of figures F1, F2,…,which contain (x0, y0) and whose diameters tend to zero. We apply Eq. (27.3) with f(x, y) = div v, to obtain

Image

Again, the terms on the right-hand side are independent of coordinates, hence so is their limit.

Returning to the interpretation of v(x, y) as the velocity field of a fluid flow, we note that the quantity vN represents the component of the flow velocity perpendicular to C, so that the integral

Image

may be regarded as the “total rate of flow across C.” Equation (27.10) states that the total flow across ∂F is equal to the integral over F of the divergence of v. If we take a sequence of circles around a point, form the ratio of flow across these circles to area inside, and take the limit, then by (27.11) we obtain the values of the divergence at that point. Note in particular, that since N is the exterior normal to F, a positive value for (27.10) means flow out of F, and a negative value means flow into F. By (27.11), if the divergence is positive at a point, this means that there is flow away from that point, and if it is negative, there is flow toward the point. A divergence-free vector field, in which ux + vy ≡ 0, represents a flow that goes past each point, without any “sources” or “sinks.”

We now examine Examples 27.2–27.4 in terms of divergence.

Example 27.2a

Image

If C is a circle of radius r centered at the origin, then at each point (x, y) of C the vector v = imagex, yimage has length (x2 + y2)1/2 = r and is normal to C, directed toward the exterior. Thus v = rN, and vN = v·N = r. It follows that

Image

The area inside this circle is πr2 and the ratio

Image

Thus, if the figures Fk are chosen to be disks centered at the origin, then each term on the right-hand side of (27.11) is equal to 2. The limit is also 2, and we verify directly for the left-hand side of (27.11) that px + qy = 2.

The fact that the divergence is everywhere equal to 2 for this vector field means that the rate of flow out of any given figure F is equal to twice its area.

Image

It is clear geometrically from Fig. 27.1 that the amount of flow out of any disk is greater than the amount entering, so that one would expect the divergence to be positive.

Example 27.3a

Image

In this case the vector field is tangential to any circle C around the origin, so that vN = 0. This implies that the divergence at the origin must be zero, which may also be verified directly. In fact div v ≡ 0 in this case, in accordance with the fact that for an arbitrary closed curve C, by (27.9),

Image

as we have observed at the end of our discussion of Example 27.2.

Example 27.4a

Image

In this case it is clear that for an arbitrary circle, the flow into it equals the flow out of it, and it follows from (27.11) that the divergence must be everywhere zero. We may confirm this directly, since both px and qy are everywhere zero.

Exercises

27.1Find the average height of the surface z = xy over each of the following figures F.

a. F:0 ≤ x1, 0y ≤ 1

b. F:x ≥ 0, y ≥ 0, x2 + y2 ≤ 1

c. F:x ≥ 0, y ≥ 0, x2 + y2 ≤ 4

d. F:x ≥ 0, y ≥ 0, 1 ≤ x2 + y2≤ 4

e. F:x ≥ 0, y ≥ 0, x + ya, (a> 0)

f. F:x ≥ 0, y ≥ 0, x2/a2 + y2/b2 ≤ 1, (a > 0, b > 0)

g. F:0 ≤ xa, 0yb, (a > 0, b > 0)

27.2 a. What is the average height of the hemisphere image(Hint: recall the geometric interpretation of average height given in the text.)

b. What is the average density of the half disk x ≥ 0, x2 + y2a2, if the density at each point is equal to x3 ?

27.3Find the average value of the function yexover the following figures F.

a. F:0 ≤ x1, 0y≤ 1

b. F:x2 + y2 ≤ 1

27.4Suppose that f(x, y) is of the form g(x)h(y). If F is a rectangle axb, cyd, show that the average value of f(x, y) over F is equal to the product of the average value of g(x) over axb and the average value of h(y) over cyd.

27.5Let F be a figure that is symmetric about the x axis and suppose f(x, y)satisfies f(x, −y)= −f(x, y). Show that the average value of f(x, y) over F is equal to zero.

27.6For each part of Ex. 27.1, verify the mean-value theorem (Th. 27.1) by finding a specific point (x0, y0) in the given figure F such that x0y0 equals the average value of xy over F.

27.7Show that if f(x, y) is a linear function,ax + by + c, then the mean value of f(x, y) over any figure F is equal to the value of f(x, y) at the centroid of F

27.8In the statement of Th. 27.2, let f(x, y) =(y + 2)2, (x0, y0)= (0, 0), and let Fk be the disk x2 + y2 ≤ 1/k2.Evaluate explicitly the expression image for each positive integer k, and verify Eq. (27.3).

*27.9Give a proof of Th. 27.2 without using the mean value theorem, as follows.

a. Show that the theorem holds if f(x0, y0) = 0.(Hint: given any ∈ > 0, by continuity of f(x, y) there exists a δ > 0 such that |f(x, y)| < ∈ for (x, y)in the disk D of radius δ about (x0, y0). For k sufficiently large, the figure Fk lies in D. By Ex. 24.27,image and hence the limit on the right- hand side of (27.3) is ≤∈. Since this is true for arbitrary small ∈, the limit must be zero.)

b. If f(x0, y0) ≠ 0, set g(x, y) = f(x, y) − f(x0, y0) and apply part a to g(x, y).

27.10Suppose, in the Corollary to Th. 27.2, that G is the transformation u = x2, v = y3.

a. If Fk is the squarex0−1/kxx0 + 1/k, y0 − 1/kyy0 + 1/k, find the area A(image k) of its image. (Assume that x0 > 0, y0 > 0 and that k is sufficiently large so that Fk lies in the first quadrant.)

b. Compute explicitly the ratio A(image k)/A(Fk) for each k, and verify Eq. (27.4).

27.11Let G be the transformation u = x cos y, v = x sin y. Let F be a rectangle x1xx2, y1y y2, where x1 > 0.

a. Sketch the image image of F under the transformation G, and find its area A(image ).

b. Verify Eq. (27.4) for this transformation, where each Fk is a rectangle containing (x0, y0).

27.12Let G:(x, y) → (u, v) and H:(u, v) → (z, w) be differentiable transformations. Suppose that G is regular at (x0, y0) and H is regular at (u0, v0) = G(x0, y0). In Ex. 17.9, it was shown by a geometrical argument how the sign of the Jacobian of the composed mapping HG at (x0, y0)was determined by the signs of the Jacobian of G at (x0, y0) and the Jacobian of H at (u0, v0). Show how the Corollary to Th. 27.2 may be used to give a geometric derivation of the relation between the absolute values of these Jacobians; namely,

Image

27.13The mean-value theorem for integrals in one variable states that if f(t) is continuous for atb, then there is a value c, a < c < b, such that

Image

a. Show that this is equivalent to the ordinary mean value theorem for derivatives. (Hint: set image

b. State this result in terms of the average value of f(t).

27.14Show that the equality of the mixed derivatives (Th. 11.1) is a consequence of the mean value theorem (Th. 27.1) and the fact that the value of an iterated integral does not depend on the order of integration (Corollary to Th. 24.4). (Hint: to prove fxy(x1, y1) = fyx(x1, y1), let F be a rectangle x1xx2, y1yy2, apply the mean value theorem to ∫∫F fxy dA and ∫∫F fyx dA, evaluate each of these integrals by a suitable iteration, and take the limit as x2x1, y2y1)

27.15Show that Eq. (27.10) holds for an arbitrary figure F, even though the integrand vN may have discontinuities at corners of ∂F. (At a corner of the boundary the normal N is not well-defined, but by the definition of a piecewise smooth curve, it tends to a limit from each side of the corner. Thus, the value of vN at a corner may be defined as either of these limits, since the value of an integral is independent of changes in the value of the integrand at a finite number of points.)

27.16 Show that the identity

Image

is a consequence of Eq. (27.10) (see also Ex. 27.15).

27.17Show that the expression for Δu must be invariant under rotation of coordinates by obtaining an invariant description of this quantity at any point. (Hint: use Ex. 27.16 and find an expression analogous to Eq. (27.11).)

*27.18Using the methods developed in this section, it is possible to deduce the expression for the Laplacian in polar coordinates (Eq. (20.10)) without actually computing the way the second derivatives transform under change of coordinates. This may be carried out in the following steps. Let h(x, y) ∈image , and let g(r, θ) = h(r cos θ, r sin θ) be the same function expressed in polar coordinates. Let F be the figure defined in polar coordinates by θ1θθ2, r1rr2.

a. Show that

Image

where

Image

(Hint: see Ex. 21.14.)

b.Show that

Image

c.Show that

Image

(Hint: use Ex. 27.16.)

d.Using the mean-value theorem for integrals in one variable (Ex. 27.13), and then the mean-value theorem for derivatives, show that

Image

and

Image

where r3, r4 are between r1 and r2, and θ3, θ4 are between θ1 and θ2

e. Show that there is a point (x0, y0) in F such that

Image

(Hint: apply the mean value theorem to the left-hand side of part c, use the answer to Ex. 27.11a for A(F) in polar coordinates, and use part d above.)

f. Show that

Image

where x1 = r1 cos θ1, x2 = r2 cos θ2. (Hint: take the limit in part e as r2r1, θ2→ θ1.)

27.19Show that Green’s identity, Ex. 26.9a, is a consequence of Eq. (27.10). (Hint : apply Eq. (27.10) to the vector field w = uv.)

27.20Sketch each of the following vector fields v(x, y), and compute the integrals ∫c vT ds and ∫c vN ds, where C is a circle of radius r centered at a point (x0, y0), T is the unit tangent to C in the counterclockwise direction, and Nis the unit normal directed toward the exterior of C. Choosing a sequence of such circles Cn whose radii tend to zero, verify Eqs. (27.6) and (27.11).

a. v(x, y) = image1, 0image (uniform flow)

b. v(x, y) = imagex, −yimage

c. v(x, y) = imagey, −ximage

d. v(x, y) = image x, 0image

e. v(x, y) = imagex2, 0image

f. v(x, y) = imagey/(x2 + y2), x/(x2 + y2)image .

27.21Show that if v(x, y) is a harmonic vector field in a domainD, then ∫∂F vT ds = 0, ∂F vN ds = 0 for every figure F lying in D.

*27.22 The generalized mean value theorem for double integrals may be stated as follows :if f(x, y) and g(x, y) are continuous on a figure F, and if g(x, y) > 0 everywhere on F, then there is a point (x0, y0) in F such that

Image

a. Prove this theorem. (Hint : use the proof of Th. 27.1 as a model.)

b. Show by an example that the conclusion may be false if the hypothesis g(x, y) > 0 is omitted.

*27.23Let f(x, y) ∈ imagein a domain D, and suppose that the figure

Image

lies in D.

a. Show that

Image

where R2(x, y) is the remainder term in the Taylor expansion of f(x, y) about (x0,y0) through terms of second order (see Eqs. (12.8), (12.9), (12.10), and (12.11)).

b. Deduce that

Image

c. Show that

Image

27.24A mapping G defined in a domain D is said to be area-preserving if for every figure F in D, the image image of F has the same area as F. Show that a diffeomorphism G: (x, y) → (u, v) is area-preserving if and only if |∂(u, v)/∂(x, y)| ≡ 1.