CHAPTER 4 Multiple Integrals 28. Double Integrals 28.1. The Volume Problem. Suppose one needs to determine the volume of a hill whose height f(r) as a function of position r = x, y in the base of the hill is known. For example, the hill must be leveled toh constri uct a highway. Its volume is required to estimate the number of truck loads needed to move the soil away. The following procedure can be used to estimate the volume. The base D of the hill is first partitioned into small pieces Dp of area ∆Ap, where p = 1, 2,...,N enumerates the pieces; that is, the union of all the pieces Dp is the region D. The partition elements should be small enough so that the height f(r) has no significant variation when r ranges over Dp. The volume of the portion of the hill above each partition element Dp is approximately ∆Vp f(rp)∆Ap, where rp is a point in Dp (see the left panel of Fig. 28.1). The≈ volume of the hill can therefore be estimated as N V f(r )∆A . ≈ p p p=1 X For practical purposes, the values f(rp) can be found, for example, from a contour map of f. The above approximation neglects variations of values of f within a partition element Dp. Therefore it is expected to become more accurate with decreasing the size of the partition elements (naturally, their number N has to increase). If Rp is the smallest radius of a disk that contains Dp, then put RN∗ = maxp Rp, which determines the size of the largest partition element. One says that the partition is refined if RN∗ is decreasing with increasing the number N of partition elements. Note that the reduction of the maximal area maxp ∆Ap versus the maximal size RN∗ may not be good enough to improve the accuracy of the estimate. If Dp looks like a narrow strip, its area is small, but the variations f along the strip may be significant and the accuracy of the approximation ∆Vp f(rp)∆Ap is poor. One can therefore expect that the exact value of the volume≈ is obtained in the limit N (28.1) V = lim f(rp)∆Ap . N→∞ (R∗ 0) p=1 N → X The volume V may be viewed as the volume of a solid bounded from above by the surface z = f(x, y), which is the graph of f, and by the portion D of 431 432 4. MULTIPLE INTEGRALS y z Rjk zp z = f(x, y) d yk D RD yk 1 y − rp D c ∆Ap x x a xj 1 xj − b Figure 28.1. Left: The volume of a solid region bounded from above by the graph z = f(x, y) and from below by a portion D of the xy plane is approximated by the sum of volumes ∆Vp = zp∆Ap of columns with the base area ∆Ap and the height zp = f(rp) where rp is a sample point within the base and p enumerates the columns. Right: A rectangular partition of a region D is obtained by embedding D into a rectangle RD. Then the rectangle RD is partitioned into smaller rectangles Rkj. the xy plane. Naturally, it is not expected to depend on the way the region D is partitioned, neither should it depend on the choice of sample points rp in each partition element. The limit (28.1) resembles the limit of a Riemann sum for a single- variable function f(x) on an interval [a, b] used to determine the area under the graph of f. Indeed, if xk, k = 0, 1,...,N, x0 = a<x1 < < xN 1 < ··· − xN = b is the partition of [a, b], then ∆Ap is the analog of ∆xj = xj xj 1, − − j = 1, 2,...,N, the number RN∗ is the analog of ∆N = maxj ∆xj , and the values f(rp) are analogous to f(xj∗), where xj∗ isin[xj 1,xj]. The area under the graph is then − N b A = lim f(x∗)∆xj = f(x) dx. →∞ j N a (∆N 0) j=1 Z → X So, the limit (28.1) seems to define an integral over a two-dimensional region D (i.e., with respect to both variables x and y used to label points in D). This observation leads to the concept of a double integral. However, the qualitative construction used to analyze the volume problem still lacks the level of rigor used to define the single-variable integration. For example, how does one choose the “shape” of the partition elements Dp, or how does one calculate their areas? These kinds of questions were not even present in the single-variable case and have to be addressed. 28.DOUBLEINTEGRALS 433 28.2. Preliminaries. The closure of a set D in a Euclidean space is the set obtained from D by adding all its limit points to D. The closure of D is denoted D¯. For example, let D be the open disk x2 + y2 < 1. Every point of D is a limit point and every point of the circle x2 + y2 = 1 is also a limit point. Therefore the closure D¯ is the (closed) disk x2 + y2 1. ≤ Definition 28.1. (A Region in a Euclidean Space) An open connected set in a Euclidean space is called an open region. The closure of an open connected set is called a closed region. A set D is a region in a Euclidean space if there is an open region G that is contained in D while the closure of G contains D. The whole idea of introducing the notion of a region is to give a name to sets in a plane that have a non-zero area and to sets in space that have a non-zero volume. Note that a region in a plane always contains an open set and this open set has a disk that lies in it. As any disk has a non-zero area, a region is expected to have a non-zero area. In particular, the volume problem considered above makes sense if D is a region. But in order to make the notion of the area (or volume) of a region precise, some additional conditions on the boundary of the region have to be imposed. If all points of an open region D are removed from its closure D¯, then the obtained set is called the boundary of D. In other words, The boundary of an open region D is the difference of its closure D¯ and D itself. For example, if D is the open disk x2 + y2 < 1, then its closure is the closed disk x2 + y2 1, and the difference between the two sets is the circle x2 + y2 = 1 which is≤ the boundary of D. Now recall that a point of a set is an interior point of the set if there is an open ball of sufficiently small radius that contains the point and lies in the set. So, The boundary of a closed region D is obtained from D by removing all inte- rior points of D. Clearly, if D is an open region, then the interior of D¯ is D. Let G be a region. Then by definition there exists an open region D that it lies in G, while G is contained in the closure D¯: D G D.¯ ⊂ ⊂ Then D is nothing but the collection of all interior points of G. It follows that the boundary of G coincides with the boundary of D or the boundary of D¯ (since D and D¯ have the same boundary). Thus, the difference between an open region and a region is that the region may contain its boundary or a part of it, while an open region contains no point of its boundary. 434 4. MULTIPLE INTEGRALS For example, let G be the set in the xy plane defined by the conditions x2 + y2 < 1 if y 0 and x2 + y2 1 if y < 0. So, G is the disk of radius 1. The upper part≥ of its boundary circle≤ (y 0) does not belong to G, while its lower part lies in G. The largest open set≥ D that is contained in G is the open disk x2 + y2 < 1. It is an open region. Its closure D¯ is the closed disk x2 + y2 1. Evidently, D G D¯. So, the boundary of G is the circle x2 + y2 =≤ 1. Note that G contains⊂ ⊂ a part of its boundary. Definition 28.2. (Smooth Boundary of a Region) The boundary of a region is called smooth if in a neighborhood of every point it coincides with a level set of a function that has continuous partial derivatives and whose gradient does not vanish. The boundary is called piecewise smooth if it consists of finitely many smooth pieces. Let D be a region in the plane. The boundary of D is smooth if in a neighborhood of each point (x0, y0) of the boundary there is a function g of two variables such that the boundary is the level set g(x, y) = g(x0, y0), where the function g has continuous partial derivatives, and ∇g = 0. Recall from Section 24.2 that under these conditions on g, the level set6 is a smooth curve. Similarly, a smooth boundary of a region in space is a smooth surface. For example, the disk x2 + y2 < 1 has the boundary x2 + y2 = 1 which is the level curve of the function g(x, y) = x2 + y2.