16 MULTIPLE INTEGRALS
MULTIPLE INTEGRALS 16.5 Applications of Double Integrals • In this section, we will learn about: • The physical applications of double integrals.
APPLICATIONS OF DOUBLE INTEGRALS • We have already seen one application of double integrals: computing volumes. • Another geometric application is finding areas of surfaces. • This will be done in Section 17.6
APPLICATIONS OF DOUBLE INTEGRALS • In this section, we explore physical applications—such as computing: • Mass • Electric charge • Center of mass • Moment of inertia
APPLICATIONS OF DOUBLE INTEGRALS • We will see that these physical ideas are also important when applied to probability density functions of two random variables.
DENSITY AND MASS • In Section 9.3, 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.
DENSITY • Suppose the lamina occupies a region Dof 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 ρis a continuous function on D.
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. Fig. 16.5.1, p. 1016
MASS • To find the total mass m of the lamina, we: • Divide a rectangle Rcontaining D into subrectangles Rijof equal size. • Consider ρ(x, y) to be 0 outside D. Fig. 16.5.2, p. 1016
MASS • If we choose a point (xij*, yij*) in Rij , then the mass of the part of the lamina that occupies Rij is approximately ρ(xij*, yij*) ∆A where ∆A is the area of Rij.
MASS • If we add all such masses, we get an approximation to the total mass:
MASS Equation 1 • If we now increase the number of subrectangles, we obtain the total mass mof the lamina as the limiting value of the approximations:
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.
TOTAL CHARGE Equation 2 • Then, the total charge Q is given by:
TOTAL CHARGE Example 1 • Charge is distributed over the triangular region D so that the charge density at (x, y) is σ(x, y) = xy, measured in coulombs per square meter (C/m2). • Find the total charge. Fig. 16.5.3, p. 1017
TOTAL CHARGE Example 1 • From Equation 2 and the figure, we have: Fig. 16.5.3, p. 1017
TOTAL CHARGE Example 1 • The total charge is: C Fig. 16.5.3, p. 1017
MOMENTS AND CENTERS OF MASS • In Section 9.3, we found the center of mass of a lamina with constant density. • Here, we consider a lamina with variable density.
MOMENTS AND CENTERS OF MASS • 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.
MOMENTS AND CENTERS OF MASS • We divide D into small rectangles as earlier. • Then, the mass of Rij is approximately: ρ(xij*, yij*) ∆A • So,we can approximate the moment of Rij with respect to the x-axis by: [ρ(xij*, yij*) ∆A] yij*
MOMENT ABOUT X-AXIS Equation 3 • If we now add these quantities and take the limit as the number of subrectangles becomes large, we obtain the momentof the entire lamina about the x-axis:
MOMENT ABOUT Y-AXIS Equation 4 • Similarly, the moment about the y-axis is:
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.
CENTER OF MASS • Thus, the lamina balances horizontally when supported at its center of mass. Fig. 16.5.4, p. 1017
CENTER OF MASS Formulas 5 • 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:
CENTER OF MASS Example 2 • Find the mass and center of mass of a triangular lamina with vertices (0, 0), (1, 0), (0, 2) and if the density function is ρ(x, y) = 1 + 3x + y
CENTER OF MASS Example 2 • The triangle is shown. • Note that the equation of the upper boundary is: y = 2 – 2x Fig. 16.5.5, p. 1018
CENTER OF MASS Example 2 • The mass of the lamina is:
CENTER OF MASS Example 2 • Then, Formulas 5 give:
CENTER OF MASS Example 2
CENTER OF MASS Example 2 • The center of mass is at the point . Fig. 16.5.5, p. 1018
CENTER OF MASS Example 3 • The density at any point on a semicircular lamina is proportional to the distance from the center of the circle. • Find the center of mass of the lamina.
CENTER OF MASS Example 3 • Let’s place the lamina as the upper half of the circle x2 + y2 = a2. • Then, the distance from a point (x, y) to the center of the circle (the origin) is: Fig. 16.5.6, p. 1018
CENTER OF MASS Example 3 • Therefore, the density function is: where K is some constant.
CENTER OF MASS Example 3 • Both the density function and the shape of the lamina suggest that we convert to polar coordinates. • Then, and the region Dis given by: 0 ≤ r ≤ a, 0 ≤ θ ≤ π
CENTER OF MASS Example 3 • Thus, the mass of the lamina is:
CENTER OF MASS Example 3 • Both the lamina and the density function are symmetric with respect to the y-axis. • So, the center of mass must lie on the y-axis, that is, = 0 Fig. 16.5.6, p. 1018
CENTER OF MASS Example 3 • The y-coordinate is given by:
CENTER OF MASS Example 3 • Thus, the center of mass is located at the point (0, 3a/(2π)). Fig. 16.5.6, p. 1018
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 mr2, 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.
MOMENT OF INERTIA • Thus, we: • Divide D into small rectangles. • Approximate the moment of inertia of each subrectangle about the x-axis. • Take the limit of the sum as the number of subrectangles becomes large.
MOMENT OF INERTIA (X-AXIS) Equation 6 • The result is the moment of inertiaof the lamina about the x-axis:
MOMENT OF INERTIA (Y-AXIS) Equation 7 • Similarly, the moment of inertia about the y-axisis:
MOMENT OF INERTIA (ORIGIN) Equation 8 • It is also of interest to consider the moment of inertia about the origin (also called the polar moment of inertia): • Note that I0 = Ix + Iy.
MOMENTS OF INERTIA Example 4 • Find the moments of inertia Ix , Iy , and I0of a homogeneous disk Dwith: • Density ρ(x, y) = ρ • Center the origin • Radius a
MOMENTS OF INERTIA Example 4 • The boundary of D is the circle x2 + y2 = a2 • In polar coordinates, D is described by: • 0 ≤ θ≤ 2π, 0 ≤ r ≤ a
MOMENTS OF INERTIA Example 4 • Let’s compute I0 first:
MOMENTS OF INERTIA Example 4 • Instead of computing Ix and Iy directly, we use the facts that Ix + Iy =I0 andIx = Iy (from the symmetry of the problem). • Thus,
MOMENTS OF INERTIA • In Example 4, notice that the mass of the disk is: • m = density x area = ρ(πa2)
MOMENTS OF INERTIA • So, the moment of inertia of the disk about the origin (like a wheel about its axle) can be written as: • Thus, if we increase the mass or the radius of the disk, we thereby increase the moment of inertia.