15 Multiple Integrals Copyright Cengage Learning All rights reserved 25 15 1 Double Integrals over Rectangles Copyright Cengage Learning All rights reserved 26 Review of the Definite Integrals 27 Review of the Definite Integrals First let s 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 x 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 definite integral of f from a to b to obtain the 28 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 1 29 Volumes and Double Integrals 30 Volumes and Double Integrals 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 See Figure 2 Figure 2 0 z f x y x y R 31 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 x b a m and dividing c d into n subintervals yj 1 yj of equal width y d c n 32 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 A x y Dividing R into subrectangles Figure 3 33 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 4 34 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 5 35 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 36 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 37 Volumes and Double Integrals The precise meaning of the limit in Definition 5 is that for every number 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 38 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 39 Volumes and Double Integrals Dividing R into subrectangles Figure 3 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 40 Volumes and Double Integrals By comparing Definitions 4 and 5 we see that a volume can be written as a double integral 41 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 42 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 5 43 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 44 Example 1 Solution The squares are shown in Figure 6 Figure 6 The paraboloid is the graph of f x y 16 x2 2y2 and the area of each square is A 1 45 Example 1 Solution cont d Approximating the volume by the Riemann sum with m n 2 we have f 1 1 A f 1 2 A f 2 1 A f 2 2 A 13 1 7 1 10 1 4 1 34 46 Example 1 Solution cont d This is the volume of the approximating rectangular boxes shown in Figure 7 Figure 7 47 The Midpoint Rule 48 The Midpoint Rule The methods that we used for approximating single integrals the Midpoint Rule the Trapezoidal Rule Simpson s 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 of Rij In other words is the midpoint of xi 1 xi and is the midpoint of yj 1 yj 49 The Midpoint Rule 50 Example 3 Use the Midpoint Rule with m n 2 to estimate the value of the integral R x 3y2 dA where R x y 0 x 2 1 y 2 Solution In using the Midpoint Rule with m n 2 we …
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