15 Multiple Integrals Copyright Cengage Learning All rights reserved 15 5 Applications of Double Integrals Copyright Cengage Learning All rights reserved 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 3 Density and Mass 4 Density and Mass We were able to use single integrals to compute moments and the center of mass of a thin plate or lamina with constant density But now equipped with the double integral 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 x y where is a continuous function on D 5 Density and Mass This means that where m and A 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 1 6 Density and Mass 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 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 A where A is the area of Rij Figure 2 If we add all such masses we get an approximation to the total mass 7 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 8 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 x y at a point x y in D then the total charge Q is given by 9 Moments and Centers of Mass 10 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 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 11 Moments and Centers of Mass Then the mass of Rij is approximately A 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 12 Moments and Centers of Mass Similarly the moment about the y axis is As before we define the center of mass and so that The physical significance is that the lamina behaves as if its entire mass is concentrated at its center of mass 13 Moments and Centers of Mass Thus the lamina balances horizontally when supported at its center of mass see Figure 4 Figure 4 14 Moment of Inertia 15 Moment of Inertia Other applications of double integrals in physics 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 xor y axis 16 Probability 17 Probability We have considered the probability density function f of a continuous random variable X This means that f x 0 for all x and the probability that X lies between a and b is found by integrating f from a to b 18 Probability Now we consider a pair of continuous random variables X and Y such as the lifetimes of two components of a machine or the height and weight of an adult female chosen at random The joint density function of X and Y is a function f of two variables such that the probability that X Y lies in a region D is 19 Probability In particular if the region is a rectangle the probability that X lies between a and b and Y lies between c and d is See Figure 7 Figure 7 The probability that X lies between a and b and Y lies between c and d is the volume that lies above the rectangle D a b c d and below the graph of the joint density function 20 Probability Because probabilities aren t negative and are measured on a scale from 0 to 1 the joint density function has the following properties f x y 0 The double integral over is an improper integral defined as the limit of double integrals over expanding circles or squares and we can write 21 Probability Suppose X is a random variable with probability density function f1 x and Y is a random variable with density function f2 y Then X and Y are called independent random variables if their joint density function is the product of their individual density functions f x y f1 x f2 y 22 Expected Values 23 Expected Values Recall that if X is a random variable with probability density function f then its mean is Now if X and Y are random variables with joint density function f we define the X mean and Y mean also called the expected values of X and Y to be 24 Expected Values Notice how closely the expressions for 1 and 2 in resemble the moments Mx and My of a lamina with density function in Equations 3 and 4 In fact we can think of probability as being like continuously distributed mass We calculate probability the way we calculate mass by integrating a density function And because the total probability mass is 1 the expressions for and in show that we can think of the expected values of X and Y 1 and 2 as the coordinates of the center of mass of the probability distribution 25 Expected Values In the next example we deal with normal distributions A single random variable is normally distributed if its probability density function is of the form where is the mean and is the standard deviation For example for a class of this size about 1000 students exam scores and course total scores tend to be normally distributed 26 Example A factory produces cylindrically shaped roller bearings that are sold as having diameter 4 0 cm and length 6 0 cm In fact the diameters X are normally distributed with mean 4 0 cm and standard deviation 0 01 cm while the lengths Y are normally distributed with mean 6 0 cm and standard deviation 0 01 cm Assuming that X and Y are independent write the joint density function and graph it Find the probability that a bearing randomly chosen …
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