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UIUC MATH 241 - Lecture040214

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Copyright © Cengage Learning. All rights reserved. 15 Multiple IntegralsCopyright © Cengage Learning. All rights reserved. 15.5 Applications of Double Integrals3 3 Applications of Double Integrals Learning objectives: § compute double integrals in applications: physics (mass, electric charge, moments, center of mass, moments of inertia, radii of gyration) § compute double integrals in applications: probability with two random variables (probability, expected values)4 4 Density and Mass5 5 Density and Mass We can consider a lamina with variable density. Suppose the lamina occupies a region D of the xy-plane and its density (in units of mass per unit area) at a point (x, y) in D is given by ρ (x, y), where ρ is a continuous function on D.6 6 Density and Mass This means that where Δm and ΔA are the mass and area of a small rectangle that contains (x, y) and the limit is taken as the dimensions of the rectangle approach 0. (See Figure 1.) Figure 17 7 To find the total mass m of the lamina we divide D into subrectangles Rij of the same size (as in Figure 2) and consider ρ (x, y) to be 0 outside D. If we choose a point in Rij, then the mass of the part of the lamina that occupies Rij is approximately ρ ΔA, where ΔA is the area of Rij. If we add all such masses, we get an approximation to the total mass: Density and Mass Figure 28 8 Density and Mass If we now increase the number of subrectangles, we obtain the total mass m of the lamina as the limiting value of the approximations:9 9 Density and Mass Physicists also consider other types of density that can be treated in the same manner. For example, if an electric charge is distributed over a region D and the charge density (in units of charge per unit area) is given by σ(x, y) at a point (x, y) in D, then the total charge Q is given by10 10 Moments and Centers of Mass11 11 Moments and Centers of Mass We have found the center of mass of a lamina with constant density; here we consider a lamina with variable density. Suppose the lamina occupies a region D and has density function ρ (x, y). Recall that we defined the moment of a particle about an axis as the product of its mass and its directed distance from the axis. We divide D into small rectangles.12 12 Moments and Centers of Mass Then the mass of Rij is approximately ρ ΔA, so we can approximate the moment of Rij with respect to the x-axis by If we now add these quantities and take the limit as the number of subrectangles becomes large, we obtain the moment of the entire lamina about the x-axis:13 13 Moments and Centers of Mass Similarly, the moment about the y-axis is As before, we define the center of mass so that and The physical significance is that the lamina behaves as if its entire mass is concentrated at its center of mass.14 14 Moments and Centers of Mass Thus the lamina balances horizontally when supported at its center of mass (see Figure 4). Figure 415 15 Moment of Inertia16 16 Moment of Inertia Other applications of double integrals (see Section 15.5 of your Stewart text): § moment of inertia (also called the second moment) of a lamina about the x- or y-axis or about the origin § radius of gyration of a lamina with respect to the x- or y-axis § Probability theory (recall worksheet on integrating normal distribution in two-variables).17 17 Copyright © Cengage Learning. All rights reserved. 15.10 Change of Variables in Double Integrals18 18 Change of Variables in Double Integrals Learning objective: § simplify double integrals by a change of variables (and then evaluate) § (a similar technique for triple integrals will be discussed later)19 19 Change of Variables in Double Integrals In one-dimensional calculus we often use a change of variable (a substitution) to simplify an integral. By reversing the roles of x and u, we can write where x = g(u) and a = g(c), b = g(d). Another way of writing Formula 1 is as follows:20 20 Change of Variables in Double Integrals A change of variables can also be useful in double integrals. We have already seen one example of this: conversion to polar coordinates. The new variables r and θ are related to the old variables x and y by the equations x = r cos θ y = r sin θ and the change of variables formula can be written as f (x, y) dA = f (r cos θ, r sin θ) r dr dθ where S is the region in the rθ -plane that corresponds to the region R in the xy-plane.21 21 Change of Variables in Double Integrals More generally, we consider a change of variables that is given by a transformation T from the uv-plane to the xy-plane: T(u, v) = (x, y) where x and y are related to u and v by the equations x = g (u, v) y = h (u, v) or, as we sometimes write, x = x (u, v) y = y (u, v)22 22 Change of Variables in Double Integrals We usually assume that T is a C1 transformation, which means that g and h have continuous first-order partial derivatives. A transformation T is really just a function whose domain and range are both subsets of If T(u1, v1) = (x1, y1), then the point (x1, y1) is called the image of the point (u1, v1). If no two points have the same image, T is called one-to-one.23 23 Change of Variables in Double Integrals Figure 1 shows the effect of a transformation T on a region S in the uv-plane. T transforms S into a region R in the xy-plane called the image of S, consisting of the images of all points in S. Figure 124 24 Change of Variables in Double Integrals If T is a one-to-one transformation, then it has an inverse transformation T –1 from the xy-plane to the uv-plane and it may be possible to solve Equations 3 for u and v in terms of x and y : u = G (x, y) v = H (x, y) You have seen the first examples of such transformations on the last worksheet. Here is another example:25 25 Example A transformation is defined by the equations x = u2 – v2 y = 2uv Find the image of the square S = {(u, v) | 0 ≤ u ≤ 1 , 0 ≤ v ≤ 1}. Solution: The transformation maps the boundary of S into the boundary of the image. So we begin by finding the images of the sides of S.26 26 Example – Solution The first side, S1, is given by v = 0 (0 ≤ u ≤ 1). (See Figure 2.) Figure 2 continued27 27 Example – Solution From the given equations we have x = u2, y = 0, and so 0 ≤ x ≤ 1. Thus S1 is mapped into the line segment from (0, 0) to


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UIUC MATH 241 - Lecture040214

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