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Chapter 15 – Multiple Integrals 15.5 Applications of Double Integrals 1 Objectives: Understand the physical applications of double integrals Dr. Erickson
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Applications of Double Integrals In this section, we explore physical applications—such as computing: ◦ Mass ◦ Electric charge ◦ Center of mass ◦ Moment of inertia 15.5 Applications of Double Integrals2Dr. Erickson
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Density and Mass In Section 8.3, we used single integrals to compute moments and the center of mass of a thin plate or lamina with constant density. 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. Also, let its density (in units of mass per unit area) at a point (x, y) in D be given by ρ(x, y), where ρ is a continuous function on D. 15.5 Applications of Double Integrals3Dr. Erickson
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Mass This means that: where: ◦ Δm and ΔA are the mass and area of a small rectangle that contains (x, y). ◦ The limit is taken as the dimensions of the rectangle approach 0. 15.5 Applications of Double Integrals4Dr. Erickson
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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: 15.5 Applications of Double Integrals5Dr. Erickson
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Density and Mass Physicists also consider other types of density that can be treated in the same manner. For example, 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. 15.5 Applications of Double Integrals6Dr. Erickson
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Total Charge Then, the total charge Q is given by: 15.5 Applications of Double Integrals7Dr. Erickson
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Example 1 Electric charge is distributed over the disk x 2 + y 2 4 so that the charge density at (x, y) is (x, y) is (x, y) = x + y + x 2 + y 2 (measured in coulombs per square meter). Find the total charge on the disk. 15.5 Applications of Double Integrals8Dr. Erickson
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Moments and Centers of Mass In Section 8.3, we 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 from Chapter 8 that we defined the moment of a particle about an axis as the product of its mass and its directed distance from the axis. 15.5 Applications of Double Integrals9Dr. Erickson
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Moments and Center of Mass We divide D into small rectangles as earlier. Then, the mass of R ij is approximately: ρ(x ij *, y ij *) ∆A So, we can approximate the moment of R ij with respect to the x-axis by: [ρ(x ij *, y ij *) ∆A] y ij * 15.5 Applications of Double Integrals10Dr. Erickson
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Moment about the x-axis If we now add these quantities and take the limit as the number of sub rectangles becomes large, we obtain the moment of the entire lamina about the x-axis: 15.5 Applications of Double Integrals11Dr. Erickson
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Moment about the y-axis Similarly, the moment about the y-axis is: 15.5 Applications of Double Integrals12Dr. Erickson
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Center of Mass 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. Thus, the lamina balances horizontally when supported at its center of mass. 15.5 Applications of Double Integrals13Dr. Erickson
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Center of Mass The coordinates of the center of mass of a lamina occupying the region D and having density function ρ(x, y) are: where the mass m is given by: 15.5 Applications of Double Integrals14Dr. Erickson
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Example 2 Find the mass and center of mass of the lamina that occupies the region D and has the given density function . 15.5 Applications of Double Integrals15Dr. Erickson
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Moment of Inertia The moment of inertia (also called the second moment) of a particle of mass m about an axis is defined to be mr 2, where r is the distance from the particle to the axis. ◦ We extend this concept to a lamina with density function ρ(x, y) and occupying a region D by proceeding as we did for ordinary moments. 15.5 Applications of Double Integrals16Dr. Erickson
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Moment of Inertia (x-axis) The result is the moment of inertia of the lamina about the x-axis: 15.5 Applications of Double Integrals17Dr. Erickson
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Moment of Inertia (y-axis) Similarly, the moment of inertia about the y-axis is given by: 15.5 Applications of Double Integrals18Dr. Erickson
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Moment of Inertia (Origin) It is also of interest to consider the moment of inertia about the origin (also called the polar moment of inertia): ◦ Note that I 0 = I x + I y. 15.5 Applications of Double Integrals19Dr. Erickson
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Example 3 Find the moments of inertia I x, I y, I o for the lamina of the problem below. 15.5 Applications of Double Integrals20Dr. Erickson
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Example 4 A lamina occupies the part of the disk x 2 + y 2 1 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis. 15.5 Applications of Double Integrals21Dr. Erickson
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Radius of Gyration Dr. Erickson15.5 Applications of Double Integrals22
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Radius of Gyration Dr. Erickson15.5 Applications of Double Integrals23
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Example 5 A lamina with constant density (x, y) = occupies the given region. Find the moments of inertia I x, I y, I o and the radii of gyration. The part of the disk x 2 + y 2 ≤ a 2 in the first quadrant. Dr. Erickson15.5 Applications of Double Integrals24
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Probability and Expected Values Please read through these pages in your text book. Dr. Erickson15.5 Applications of Double Integrals25
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Dr. Erickson15.5 Applications of Double Integrals26
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More Examples The video examples below are from section 15.5 in your textbook. Please watch them on your own time for extra instruction. Each video is about 2 minutes in length. ◦ Example 2 Example 2 ◦ Example 3 Example 3 ◦ Example 4 Example 4 15.5 Applications of Double Integrals27Dr. Erickson
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Demonstrations Feel free to explore these demonstrations below. ◦ Center of Mass of a Polygon Center of Mass of a Polygon ◦ Moment of Inertia Moment of Inertia 15.5 Applications of Double Integrals28Dr. Erickson
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