25 25 Copyright © Cengage Learning. All rights reserved. 15 Multiple Integrals26 26 Copyright © Cengage Learning. All rights reserved. 15.1 Double Integrals over Rectangles27 27 Review of the Definite Integrals28 28 Review of the Definite Integrals First lets recall the basic facts concerning definite integrals of functions of a single variable. If f (x) is defined for a ≤ x ≤ b, we start by dividing the interval [a, b] into n subintervals [xi – 1, xi] of equal width Δx = (b – a)/n and we choose sample points in these subintervals. Then we form the Riemann sum and take the limit of such sums as n → to obtain the definite integral of f from a to b:29 29 Review of the Definite Integrals In the special case where f (x) ≥ 0, the Riemann sum can be interpreted as the sum of the areas of the approximating rectangles in Figure 1, and represents the area under the curve y = f (x) from a to b. Figure 130 30 Volumes and Double Integrals31 31 In a similar manner we consider a function f of two variables defined on a closed rectangle R = [a, b] × [c, d ] = {(x, y) ∈ | a ≤ x ≤ b, c ≤ y ≤ d } and we first suppose that f (x, y) ≥ 0. The graph of f is a surface with equation z = f (x, y). Let S be the solid that lies above R and under the graph of f, that is, S = {(x, y, z) ∈ | 0 ≤ z ≤ f (x, y), (x, y) ∈ R} (See Figure 2.) Volumes and Double Integrals Figure 232 32 Volumes and Double Integrals Our goal is to find the volume of S. The first step is to divide the rectangle R into subrectangles. We accomplish this by dividing the interval [a, b] into m subintervals [xi – 1, xi] of equal width Δx = (b – a)/m and dividing [c, d ] into n subintervals [yj – 1, yj] of equal width Δy = (d – c)/n.33 33 Volumes and Double Integrals By drawing lines parallel to the coordinate axes through the endpoints of these subintervals, as in Figure 3, we form the subrectangles Rij = [xi – 1, xi] × [yj– 1, yj] = {(x, y) | xi – 1 ≤ x ≤ xi , yj– 1 ≤ y ≤ yj } each with area ΔA = Δx Δy. Figure 3 Dividing R into subrectangles34 34 Volumes and Double Integrals If we choose a sample point in each Rij, then we can approximate the part of S that lies above each Rij by a thin rectangular box (or column ) with base Rij and height as shown in Figure 4. The volume of this box is the height of the box times the area of the base rectangle: Figure 435 35 Volumes and Double Integrals If we follow this procedure for all the rectangles and add the volumes of the corresponding boxes, we get an approximation to the total volume of S: (See Figure 5.) This double sum means that for each subrectangle we evaluate f at the chosen point and multiply by the area of the subrectangle, and then we add the results. Figure 536 36 Volumes and Double Integrals Our intuition tells us that the approximation given in (3) becomes better as m and n become larger and so we would expect that We use the expression in Equation 4 to define the volume of the solid S that lies under the graph of f and above the rectangle R.37 37 Volumes and Double Integrals Limits of the type that appear in Equation 4 occur frequently, not just in finding volumes but in a variety of other situations even when f is not a positive function. So we make the following definition.38 38 Volumes and Double Integrals The precise meaning of the limit in Definition 5 is that for every number ε > 0 there is an integer N such that for all integers m and n greater than N and for any choice of sample points in Rij. A function f is called integrable if the limit in Definition 5 exists.39 39 Volumes and Double Integrals It is shown in courses on advanced calculus that all continuous functions are integrable. In fact, the double integral of f exists provided that f is not too discontinuous. In particular, if f is bounded [that is, there is a constant M such that | f (x, y) | ≤ M for all (x, y) in R], and f is continuous there, except on a finite number of smooth curves, then f is integrable over R.40 40 Volumes and Double Integrals The sample point can be chosen to be any point in the subrectangle Rij, but if we choose it to be the upper right-hand corner of Rij [namely (xi, yj), see Figure 3], then the expression for the double integral looks simpler: Figure 3 Dividing R into subrectangles41 41 Volumes and Double Integrals By comparing Definitions 4 and 5, we see that a volume can be written as a double integral:42 42 Volumes and Double Integrals The sum in Definition 5, is called a double Riemann sum and is used as an approximation to the value of the double integral. [Notice how similar it is to the Riemann sum in for a function of a single variable.]43 43 Volumes and Double Integrals If f happens to be a positive function, then the double Riemann sum represents the sum of volumes of columns, as in Figure 5, and is an approximation to the volume under the graph of f. Figure 544 44 Example 1 Estimate the volume of the solid that lies above the square R = [0, 2] × [0, 2] and below the elliptic paraboloid z = 16 – x2 – 2y2. Divide R into four equal squares and choose the sample point to be the upper right corner of each square Rij. Sketch the solid and the approximating rectangular boxes.45 45 Example 1 – Solution The squares are shown in Figure 6. The paraboloid is the graph of f (x, y) = 16 – x2 – 2y2 and the area of each square is ΔA = 1. Figure 646 46 Example 1 – Solution Approximating the volume by the Riemann sum with m = n = 2, we have = f (1, 1) ΔA + f (1, 2) ΔA + f (2, 1) ΔA + f (2, 2) ΔA = 13(1) + 7(1) + 10(1) + 4(1) = 34 contd47 47 Example 1 – Solution This is the volume of the approximating rectangular boxes shown in Figure 7. Figure 7 contd48 48 The Midpoint Rule49 49 The Midpoint Rule The methods that we used for approximating single integrals (the Midpoint Rule, the Trapezoidal Rule, Simpsons Rule) all have counterparts for double integrals. Here we consider only the Midpoint Rule for double integrals. This means that we use a double Riemann sum to approximate the double integral, where the sample point in Rij is chosen to be the center …
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