# 14.1. Multiple Integration. Iterated Integrals and Area in the Plane. Objectives. Iterated Integrals. Iterated Integrals

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1 14 Multiple Integration 14.1 Iterated Integrals and Area in the Plane Copyright Cengage Learning. All rights reserved. Copyright Cengage Learning. All rights reserved. Objectives! Evaluate an iterated integral.! Use an iterated integral to find the area of a plane region. Iterated Integrals 3 4 Iterated Integrals Example 1 Integrating with Respect to y Evaluate Note that the variable of integration cannot appear in either limit of integration. Solution: Considering x to be constant and integrating with respect to y produces For instance, it makes no sense to write 5 6

2 Example 2 The Integral of an Integral Evaluate Solution: Using the result of Example 1, you have Iterated Integrals The integral in Example 2 is an iterated integral. The brackets used in Example 2 are normally not written. Instead, iterated integrals are usually written simply as The inside limits of integration can be variable with respect to the outer variable of integration. However, the outside limits of integration must be constant with respect to both variables of integration. 7 8 Iterated Integrals For instance, in Example 2, the outside limits indicate that x lies in the interval 1! x! 2 and the inside limits indicate that y lies in the interval 1! y! x. Together, these two intervals determine the region of integration R of the iterated integral, as shown in Figure Area of a Plane Region Figure Area of a Plane Region Consider the plane region R bounded by a! x! b and g 1 (x)! y! g 2 (x), as shown in Figure Area of a Plane Region Specifically, if you consider x to be fixed and let y vary from g 1 (x) to g 2 (x), you can write The area of R is given by the definite integral Combining these two integrals, you can write the area of the region R as an iterated integral Using the Fundamental Theorem of Calculus, you can rewrite the integrand g 2 (x) g 1 (x) as a definite integral. Figure

3 Area of a Plane Region Area of a Plane Region Placing a representative rectangle in the region R helps determine both the order and the limits of integration. A vertical rectangle implies the order dy dx, with the inside limits corresponding to the upper and lower bounds of the rectangle, as shown in Figure This type of region is called vertically simple, because the outside limits of integration represent the vertical lines x = a and x = b. Figure Similarly, a horizontal rectangle implies the order dx dy, with the inside limits determined by the left and right bounds of the rectangle, as shown in Figure This type of region is called horizontally simple, because the outside limits represent the horizontal lines y = c and y = d. Figure Area of a Plane Region Example 3 The Area of a Rectangular Region Use an iterated integral to represent the area of the rectangle shown in Figure Solution: The region shown in Figure 14.4 is both vertically simple and horizontally simple, so you can use either order of integration. By choosing the order dy dx, you obtain the following. Figure Example 3 Solution 14.2 Double Integrals and Volume 17 Copyright Cengage Learning. All rights reserved. 18

4 Objectives! Use a double integral to represent the volume of a solid region.! Use properties of double integrals.! Evaluate a double integral as an iterated integral. Double Integrals and Volume of a Solid Region! Find the average value of a function over a region Double Integrals and Volume of a Solid Region You know that a definite integral over an interval uses a limit process to assign measures to quantities such as area, volume, arc length, and mass. In this section, you will use a similar process to define the double integral of a function of two variables over a region in the plane. Double Integrals and Volume of a Solid Region Consider a continuous function f such that f(x, y) " 0 for all (x, y) in a region R in the xy-plane. The goal is to find the volume of the solid region lying between the surface given by z = f(x, y) Surface lying above the xy-plane and the xy-plane, as shown in Figure Figure Double Integrals and Volume of a Solid Region You can begin by superimposing a rectangular grid over the region, as shown in Figure Double Integrals and Volume of a Solid Region Next, choose a point (x i, y i ) in each rectangle and form the rectangular prism whose height is f(x i, y i ), as shown in Figure Figure 14.9 The rectangles lying entirely within R form an inner partition #, whose norm # is defined as the length of the longest diagonal of the n rectangles. 23 Because the area of the ith rectangle is #A i Area of ith rectangle it follows that the volume of the ith prism is f(x i, y i ) #A i Volume of ith prism Figure

5 Double Integrals and Volume of a Solid Region Example 1 Approximating the Volume of a Solid You can approximate the volume of the solid region by the Riemann sum of the volumes of all n prisms, Approximate the volume of the solid lying between the paraboloid as shown in Figure This approximation can be improved by tightening the mesh of the grid to form smaller and smaller rectangles. and the square region R given by 0! x! 1, 0! y! 1. Use a partition made up of squares whose sides have a length of Figure Example 1 Solution Example 1 Solution Begin by forming the specified partition of R. For this partition, it is convenient to choose the centers of the subregions as the points at which to evaluate f(x, y). Because the area of each square is approximate the volume by the sum you can This approximation is shown graphically in Figure The exact volume of the solid is. Figure Example 1 Solution You can obtain a better approximation by using a finer partition. For example, with a partition of squares with sides of length the approximation is Double Integrals and Volume of a Solid Region In Example 1, note that by using finer partitions, you obtain better approximations of the volume. This observation suggests that you could obtain the exact volume by taking a limit. That is, Volume 29 30

6 Double Integrals and Volume of a Solid Region The precise meaning of this limit is that the limit is equal to L if for every! > 0 there exists a " > 0 such that Double Integrals and Volume of a Solid Region Using the limit of a Riemann sum to define volume is a special case of using the limit to define a double integral. The general case, however, does not require that the function be positive or continuous. for all partitions # of the plane region R (that satisfy # < ") and for all possible choices of x i and y i in the ith region. 31 Having defined a double integral, you will see that a definite integral is occasionally referred to as a single integral. 32 Double Integrals and Volume of a Solid Region Sufficient conditions for the double integral of f on the region R to exist are that R can be written as a union of a finite number of nonoverlapping subregions (see Figure 14.14) that are vertically or horizontally simple and that f is continuous on the region R. Double Integrals and Volume of a Solid Region A double integral can be used to find the volume of a solid region that lies between the xy-plane and the surface given by z = f(x, y). Figure Properties of Double Integrals Double integrals share many properties of single integrals. Properties of Double Integrals 35 36

7 Evaluation of Double Integrals Consider the solid region bounded by the plane z = f(x, y) = 2 x 2y and the three coordinate planes, as shown in Figure Evaluation of Double Integrals 37 Figure Evaluation of Double Integrals Each vertical cross section taken parallel to the yz-plane is a triangular region whose base has a length of y = (2 x)/2 and whose height is z = 2 x. Evaluation of Double Integrals By the formula for the volume of a solid with known cross sections, the volume of the solid is This implies that for a fixed value of x, the area of the triangular cross section is 39 Figure This procedure works no matter how A(x) is obtained. In particular, you can find A(x) by integration, as shown in Figure Evaluation of Double Integrals That is, you consider x to be constant, and integrate z = 2 x 2y from 0 to (2 x)/2 to obtain Evaluation of Double Integrals To understand this procedure better, it helps to imagine the integration as two sweeping motions. For the inner integration, a vertical line sweeps out the area of a cross section. For the outer integration, the triangular cross section sweeps out the volume, as shown in Figure Combining these results, you have the iterated integral Figure

8 Evaluation of Double Integrals Example 2 Evaluating a Double Integral as an Iterated Integral Evaluate where R is the region given by 0! x! 1, 0! y! Solution: Because the region R is a square, it is both vertically and horizontally simple, and you can use either order of integration. Choose dy dx by placing a vertical representative rectangle in the region, as shown in Figure Figure Example 2 Solution This produces the following. Average Value of a Function Average Value of a Function For a function f in one variable, the average value of f on [a, b] is Given a function f in two variables, you can find the average value of f over the region R as shown in the following definition. Example 6 Finding the Average Value of a Function Find the average value of f(x, y) = over the region R, where R is a rectangle with vertices (0, 0), (4, 0), (4, 3), and (0, 3). Solution: The area of the rectangular region R is A = 12 (see Figure 14.23). 47 Figure

9 Example 6 Solution The average value is given by 14.3 Change of Variables: Polar Coordinates 49 Copyright Cengage Learning. All rights reserved. 50 Objective! Write and evaluate double integrals in polar coordinates. Double Integrals in Polar Coordinates Double Integrals in Polar Coordinates The polar coordinates (r,! ) of a point are related to the rectangular coordinates (x, y) of the point as follows. Example 1 Using Polar Coordinates to Describe a Region Use polar coordinates to describe each region shown in Figure Figure

10 Example 1 Solution a. The region R is a quarter circle of radius 2. It can be described in polar coordinates as Double Integrals in Polar Coordinates The regions in Example 1 are special cases of polar sectors as shown in Figure b. The region R consists of all points between concentric circles of radii 1 and 3. It can be described in polar coordinates as 55 Figure Double Integrals in Polar Coordinates To define a double integral of a continuous function z = f(x, y) in polar coordinates, consider a region R bounded by the graphs of r = g 1 (! ) and r = g 2 (! ) and the lines! = \$ and! = ". Instead of partitioning R into small rectangles, use a partition of small polar sectors. Double Integrals in Polar Coordinates The polar sectors R i lying entirely within R form an inner polar partition #, whose norm # is the length of the longest diagonal of the n polar sectors. Consider a specific polar sector R i, as shown in Figure On R, superimpose a polar grid made of rays and circular arcs, as shown in Figure Figure Figure Double Integrals in Polar Coordinates It can be shown that the area of R i is Double Integrals in Polar Coordinates The region R corresponds to a horizontally simple region S in the r! -plane, as shown in Figure where and. This implies that the volume of the solid of height above R i is approximately and you have The sum on the right can be interpreted as a Riemann sum for 59 Figure

11 Double Integrals in Polar Coordinates The polar sectors R i correspond to rectangles S i, and the area of S i is So, the right-hand side of the equation corresponds to the double integral Double Integrals in Polar Coordinates This suggests the following theorem 14.3 From this, you can write Double Integrals in Polar Coordinates The region R is restricted to two basic types, r-simple regions and! -simple regions, as shown in Figure Example 2 Evaluating a Double Polar Integral Let R be the annular region lying between the two circles and Evaluate the integral Solution: The polar boundaries are and as shown in Figure Figure Figure Example 2 Solution Example 2 Solution Furthermore, and So, you have 65 66

12 14.4 Center of Mass and Moments of Inertia Objectives! Find the mass of a planar lamina using a double integral.! Find the center of mass of a planar lamina using double integrals.! Find moments of inertia using double integrals. Copyright Cengage Learning. All rights reserved Mass If the lamina corresponding to the region R, as shown in Figure 14.34, has a constant density #, Mass Figure then the mass of the lamina is given by Mass A lamina is assumed to have a constant density. But now you will extend the definition of the term lamina to include thin plates of variable density. Double integrals can be used to find the mass of a lamina of variable density, where the density at (x, y) is given by the density function #. Example 1 Finding the Mass of a Planar Lamina Find the mass of the triangular lamina with vertices (0, 0), (0, 3), and (2, 3), given that the density at (x, y) is #(x, y) = 2x + y. Solution: As shown in Figure 14.35, region R has the boundaries x = 0, y = 3, and y = 3x/2 (or x = 2y/3). 71 Figure

13 Example 1 Solution Therefore, the mass of the lamina is Moments and Center of Mass Moments and Center of Mass For a lamina of variable density, moments of mass are defined in a manner similar to that used for the uniform density case. Moments and Center of Mass Assume that the mass of R i is concentrated at one of its interior points (x i, y i ). The moment of mass of R i with respect to the x-axis can be approximated by For a partition # of a lamina corresponding to a plane region R, consider the i th rectangle R i of one area #A i, as shown in Figure Similarly, the moment of mass with respect to the y-axis can be approximated by Figure Moments and Center of Mass By forming the Riemann sum of all such products and taking the limits as the norm of # approaches 0, you obtain the following definitions of moments of mass with respect to the x- and y-axes. Moments and Center of Mass For some planar laminas with a constant density #, you can determine the center of mass (or one of its coordinates) using symmetry rather than using integration

14 Moments and Center of Mass Example 3 Finding the Center of Mass For instance, consider the laminas of constant density shown in Figure Using symmetry, you can see that and for the second lamina. for the first lamina Find the center of mass of the lamina corresponding to the parabolic region 0 % y % 4 x 2 Parabolic region where the density at the point (x, y) is proportional to the distance between (x, y) and the x-axis, as shown in Figure Figure Figure Example 3 Solution Example 3 Solution Because the lamina is symmetric with respect to the y-axis and #(x, y) = ky the center of mass lies on the y-axis. So, To find, first find the mass of the lamina. Next, find the moment about the x-axis Example 3 Solution So, and the center of mass is Moments of Inertia 83 84

15 Moments of Inertia The moments of M x and M y used in determining the center of mass of a lamina are sometimes called the first moments about the x- and y-axes. In each case, the moment is the product of a mass times a distance. Moments of Inertia You will now look at another type of moment the second moment, or the moment of inertia of a lamina about a line. In the same way that mass is a measure of the tendency of matter to resist a change in straight-line motion, the moment of inertia about a line is a measure of the tendency of matter to resist a change in rotational motion. 85 For example, if a particle of mass m is a distance d from a fixed line, its moment of inertia about the line is defined as I = md 2 = (mass)(distance) Moments of Inertia As with moments of mass, you can generalize this concept to obtain the moments of inertia about the x- and y-axes of a lamina of variable density. These second moments are denoted by I x and I y, and in each case the moment is the product of a mass times the square of a distance. Example 4 Finding the Moment of Inertia Find the moment of inertia about the x-axis of the lamina in Example 3. Solution: From the definition of moment of inertia, you have The sum of the moments I x and I y is called the polar moment of inertia and is denoted by I Moments of Inertia The moment of inertia I of a revolving lamina can be used to measure its kinetic energy. For example, suppose a planar lamina is revolving about a line with an angular speed of \$ radians per second, as shown in Figure Moments of Inertia The kinetic energy E of the revolving lamina is On the other hand, the kinetic energy E of a mass m moving in a straight line at a velocity v is So, the kinetic energy of a mass moving in a straight line is proportional to its mass, but the kinetic energy of a mass revolving about an axis is proportional to its moment of inertia. Figure

16 Moments of Inertia The radius of gyration of a revolving mass m with moment of inertia I is defined as If the entire mass were located at a distance from its axis of revolution, it would have the same moment of inertia and, consequently, the same kinetic energy. For instance, the radius of gyration of the lamina in Example 4 about the x-axis is given by 14.5 Surface Area 91 Copyright Cengage Learning. All rights reserved. 92 Objective! Use a double integral to find the area of a surface. Surface Area Surface Area You know about the solid region lying between a surface and a closed and bounded region R in the xy-plane, as shown in Figure For example, you know how to find the extrema of f on R. Surface Area In this section, you will learn how to find the upper surface area of the solid. To begin, consider a surface S given by z = f(x, y) defined over a region R. Assume that R is closed and bounded and that f has continuous first partial derivatives. Figure

17 Surface Area To find the surface area, construct an inner partition of R consisting of n rectangles, where the area of the ith rectangle R i is #A i = #x i #y i, as shown in Figure In each R i let (x i, y i ) be the point that is closest to the origin. Surface Area The area of the portion of the tangent plane that lies directly above R i is approximately equal to the area of the surface lying directly above R i. That is, #T i \$ #S i. So, the surface area of S is given by At the point (x i, y i, z i ) = (x i, y i, f(x i, y i )) on the surface S, construct a tangent plane T i. Figure To find the area of the parallelogram #T i, note that its sides are given by the vectors and u = #x i i + f x (x i, y i ) #x i k v = #y i j + f y (x i, y i ) #y i k. 98 Surface Area Surface Area The area of #T i is given by, where So, the area of #T i is and Surface area of Surface Area As an aid to remembering the double integral for surface area, it is helpful to note its similarity to the integral for arc length. Example 1 The Surface Area of a Plane Region Find the surface area of the portion of the plane z = 2 x y that lies above the circle x 2 + y 2! 1 in the first quadrant, as shown in Figure Figure

18 Example 1 Solution Example 1 Solution Because f x (x, y) = %1 and f y (x, y) = %1, the surface area is given by Note that the last integral is simply region R. times the area of the R is a quarter circle of radius 1, with an area of or. So, the area of S is Triple Integrals and Applications Objectives! Use a triple integral to find the volume of a solid region.! Find the center of mass and moments of inertia of a solid region. Copyright Cengage Learning. All rights reserved Triple Integrals The procedure used to define a triple integral follows that used for double integrals. Triple Integrals Consider a function f of three variables that is continuous over a bounded solid region Q. 107 Then, encompass Q with a network of boxes and form the inner partition consisting of all boxes lying entirely within Q, as shown in Figure Figure

19 Triple Integrals Triple Integrals The volume of the ith box is Taking the limit as leads to the following definition. The norm of the partition is the length of the longest diagonal of the n boxes in the partition. Choose a point (x i, y i, z i ) in each box and form the Riemann sum Triple Integrals Triple Integrals Some of the properties of double integrals can be restated in terms of triple integrals Example 1 Evaluating a Triple Iterated Integral Evaluate the triple iterated integral Example 1 Solution For the second integration, hold x constant and integrate with respect to y. Solution: For the first integration, hold x and y constant and integrate with respect to z. Finally, integrate with respect to x

20 Triple Integrals To find the limits for a particular order of integration, it is generally advisable first to determine the innermost limits, which may be functions of the outer two variables. Triple Integrals For instance, to evaluate Then, by projecting the solid Q onto the coordinate plane of the outer two variables, you can determine their limits of integration by the methods used for double integrals. first determine the limits for z, and then the integral has the form Triple Integrals By projecting the solid Q onto the xy-plane, you can determine the limits for x and y as you did for double integrals, as shown in Figure Center of Mass and Moments of Inertia Figure Center of Mass and Moments of Inertia Center of Mass and Moments of Inertia Consider a solid region Q whose density is given by the density function!. The center of mass of a solid region Q of mass m is given by, where and The quantities M yz, M xz, and M xy are called the first moments of the region Q about the yz-, xz-, and xy-planes, respectively. 119 The first moments for solid regions are taken about a plane, whereas the second moments for solids are taken about a line. 120

21 Center of Mass and Moments of Inertia The second moments (or moments of inertia) about the and x-, y-, and z-axes are as follows. Center of Mass and Moments of Inertia For problems requiring the calculation of all three moments, considerable effort can be saved by applying the additive property of triple integrals and writing where I xy, I xz, and I yz are as follows Example 5 Finding the Center of Mass of a Solid Region Find the center of mass of the unit cube shown in Figure 14.61, given that the density at the point (x, y, z) is proportional to the square of its distance from the origin. Example 5 Solution Because the density at (x, y, z) is proportional to the square of the distance between (0, 0, 0) and (x, y, z), you have You can use this density function to find the mass of the cube. Because of the symmetry of the region, any order of integration will produce an integral of comparable difficulty. Figure Example 5 Solution Example 5 Solution The first moment about the yz-plane is Note that x can be factored out of the two inner integrals, because it is constant with respect to y and z. After factoring, the two inner integrals are the same as for the mass m

22 Example 5 Solution Therefore, you have 14.7 Triple Integrals in Cylindrical and Spherical Coordinates So, Finally, from the nature of! and the symmetry of x, y, and z in this solid region, you have, and the center of mass is. 127 Copyright Cengage Learning. All rights reserved. 128 Objectives! Write and evaluate a triple integral in cylindrical coordinates.! Write and evaluate a triple integral in spherical coordinates. Triple Integrals in Cylindrical Coordinates Triple Integrals in Cylindrical Coordinates Triple Integrals in Cylindrical Coordinates The rectangular conversion equations for cylindrical coordinates are x = r cos " y = r sin " z = z. In this coordinate system, the simplest solid region is a cylindrical block determined by r 1! r! r 2, " 1! "! " 2, z 1! z! z 2 as shown in Figure To obtain the cylindrical coordinate form of a triple integral, suppose that Q is a solid region whose projection R onto the xy-plane can be described in polar coordinates. That is, and Q = {(x, y, z): (x, y) is in R, h 1 (x, y)! z! h 2 (x, y)} R = {(r, "): " 1! "! " 2, g 1 (")! r! g 2 (")}. Figure

23 Triple Integrals in Cylindrical Coordinates If f is a continuous function on the solid Q, you can write the triple integral of f over Q as where the double integral over R is evaluated in polar coordinates. That is, R is a plane region that is either r-simple or "-simple. If R is r-simple, the iterated form of the triple integral in cylindrical form is Triple Integrals in Cylindrical Coordinates To visualize a particular order of integration, it helps to view the iterated integral in terms of three sweeping motions each adding another dimension to the solid. For instance, in the order dr d" dz, the first integration occurs in the r-direction as a point sweeps out a ray. Then, as " increases, the line sweeps out a sector Triple Integrals in Cylindrical Coordinates Finally, as z increases, the sector sweeps out a solid wedge, as shown in Figure Example 1 Finding Volume in Cylindrical Coordinates Find the volume of the solid region Q cut from the sphere x 2 + y 2 + z 2 = 4 by the cylinder r = 2 sin ", as shown in Figure Figure Figure Example 1 Solution Because x 2 + y 2 + z 2 = r 2 + z 2 = 4, the bounds on z are Example 1 Solution Let R be the circular projection of the solid onto the r"-plane. Then the bounds on R are 0! r! 2 sin " and 0! "! #. So, the volume of Q is

24 Triple Integrals in Spherical Coordinates Triple Integrals in Spherical Coordinates 139 The rectangular conversion equations for spherical coordinates are x =! sin % cos " y =! sin % sin " z =! cos %. In this coordinate system, the simplest region is a spherical block determined by {(!, ", %):! 1!!!! 2, " 1! "! " 2, % 1! %! % 2 } where! 1 " 0, " 2 " 1! 2#, and 0! % 1! % 2! #, as shown in Figure Figure Triple Integrals in Spherical Coordinates If (!, ", %) is a point in the interior of such a block, then the volume of the block can be approximated by #V \$! 2 sin % #! #% #" Using the usual process involving an inner partition, summation, and a limit, you can develop the following version of a triple integral in spherical coordinates for a continuous function f defined on the solid region Q. Triple Integrals in Spherical Coordinates Triple integrals in spherical coordinates are evaluated with iterated integrals. You can visualize a particular order of integration by viewing the iterated integral in terms of three sweeping motions each adding another dimension to the solid Triple Integrals in Spherical Coordinates For instance, the iterated integral is illustrated in Figure Example 4 Finding Volume in Spherical Coordinates Find the volume of the solid region Q bounded below by the upper nappe of the cone z 2 = x 2 + y 2 and above by the sphere x 2 + y 2 + z 2 = 9, as shown in Figure Figure Figure 14.70

25 Example 4 Solution In spherical coordinates, the equation of the sphere is! 2 = x 2 + y 2 + z 2 = 9 Furthermore, the sphere and cone intersect when (x 2 + y 2 ) + z 2 = (z 2 ) + z 2 = 9 and, because z =! cos %, it follows that Example 4 Solution Consequently, you can use the integration order d! d% d", where 0!!! 3, 0! %! #/4, and 0! "! 2#. The volume is Change of Variables: Jacobians Objectives! Understand the concept of a Jacobian.! Use a Jacobian to change variables in a double integral. Copyright Cengage Learning. All rights reserved Jacobians For the single integral Jacobians you can change variables by letting x = g(u), so that dx = g\$(u) du, and obtain where a = g(c) and b = g(d). 149 The change of variables process introduces an additional factor g\$(u) into the integrand. 150

26 Jacobians This also occurs in the case of double integrals Jacobians In defining the Jacobian, it is convenient to use the following determinant notation. where the change of variables x = g(u, v) and y = h(u, v) introduces a factor called the Jacobian of x and y with respect to u and v Example 1 The Jacobian for Rectangular-to-Polar Conversion Find the Jacobian for the change of variables defined by x = r cos " and y = r sin ". Solution: From the definition of the Jacobian, you obtain Jacobians Example 1 points out that the change of variables from rectangular to polar coordinates for a double integral can be written as 153 where S is the region in the r"-plane that corresponds to the region R in the xy-plane, as shown in Figure Figure Jacobians In general, a change of variables is given by a one-to-one transformation T from a region S in the uv-plane to a region R in the xy-plane, to be given by T(u, v) = (x, y) = (g(u, v), h(u, v)) where g and h have continuous first partial derivatives in the region S. Note that the point (u, v) lies in S and the point (x, y) lies in R. Change of Variables for Double Integrals

27 Change of Variables for Double Integrals Example 3 Using a Change of Variables to Simplify a Region Let R be the region bounded by the lines x 2y = 0, x 2y = 4, x + y = 4, and x + y = 1 as shown in Figure Evaluate the double integral Figure Example 3 Solution You can use the following change of variables. and The partial derivatives of x and y are Example 3 Solution which implies that the Jacobian is So, by Theorem 14.5, you obtain Example 3 Solution 161

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