Applications of Double Integrals

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Presentation transcript:

Applications of Double Integrals Section 16.5 Applications of Double Integrals

LAMINAS AND DENSITY A lamina is a flat sheet (or plate) that is so thin as to be considered two-dimensional. Suppose the lamina occupies a region D of the xy-plane and its density (in units of mass per area) at a point (x, y) in D is given by ρ(x, y), where ρ is a continuous function on D. 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.

MASS OF A LAMINA To find the mass of a lamina, we partition D into small rectangles Rij of the same size. Pick a point in Rij. The mass of Rij is approximately and the total mass of the lamina is approximately The actual mass is obtained by taking the limit of the above expression as both k and l approach zero. That is,

EXAMPLE Find the mass of the lamina bounded by the triangle with vertices (0, 1), (0, 3) and (2, 3) and whose density is given by ρ(x, y) = 2x + y, measured in g/cm2.

MOMENTS The moment of a point about an axis is the product of its mass and its distance from the axis. To find the moments of a lamina about the x- and y-axes, we partition D into small rectangles and assume the entire mass of each subrectangle is concentrated at an interior point. Then the moment of Rij about the x-axis is given by and the moment of Rk about the y-axis is given by

MOMENTS (CONCLUDED) The moment about the x-axis of the entire lamina is The moment about the y-axis of the entire lamina is

CENTER OF MASS The center of mass of a lamina is the “balance point.” That is, the place where you could balance the lamina on a “pencil point.” The coordinates (x, y) of the center of mass of a lamina occupying the region D and having density function ρ(x, y) is where the mass m is given by

EXAMPLES 1. Find the mass and center of mass of the triangular lamina bounded by the x-axis and the lines x = 1 and y = 2x if the density function is ρ(x, y) = 6x + 6y + 6. 2. Find the center of mass of the lamina in the shape of a quarter-circle of radius a whose density is proportional to the distance from the center of the circle.

MOMENTS 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 region D as we did for ordinary moments.

MOMENTS OF INERTIA CONCLUDED The moment of inertia about the x-axis is The moment of inertia about the y-axis is The moment of inertia about the origin (or polar moment) is

EXAMPLE Find the moments of inertia Ix, Iy, and I0 of the lamina bounded by y = 0, x = 0, and y = 4 − x2 where the density is given by ρ(x, y) = 2y.