Question

In Exercises \(19-26,\) find the center of mass of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\) Note: these are the same lamina as in Exercises \(11-18\). \(R\) is the rectangle with corners (1,-3),(1,2),(7,2) and (7,-3)\(; \delta(x, y)=5 \mathrm{gm} / \mathrm{cm}^{2}\)

Short answer

The center of mass is \((4, -0.5)\).

Step-by-step solution

Understand the Geometry of the Region

The region \( R \) is a rectangle with the vertices (1, -3), (1, 2), (7, 2), and (7, -3). This defines a rectangle on the plane where the width extends from \( x = 1 \) to \( x = 7 \) and the height from \( y = -3 \) to \( y = 2 \).

Calculate the Mass of the Lamina

The mass \( M \) of the lamina can be calculated using \( M = \int \int_R \delta(x, y) \, dA \). Since \( \delta(x, y) = 5 \) gm/cm² is constant, \( M = 5 \times \text{Area of }R \). The area of the rectangle is \((7 - 1)(2 - (-3)) = 30 \) cm², so \( M = 5 \times 30 = 150 \) grams.

Find the Coordinates of the Center of Mass

The coordinates of the center of mass \((\bar{x}, \bar{y})\) for a uniform density are given by the averages of the geometric centers of the region. Thus, \( \bar{x} = \frac{1+7}{2} = 4 \) and \( \bar{y} = \frac{-3+2}{2} = -0.5 \).

Calculate the Center of Mass

For the uniform density function, the center of mass is the same as the geometric center. Thus, \((\bar{x}, \bar{y}) = (4, -0.5)\).

Key concepts

lamina

In the realm of physics and mathematics, a lamina is a two-dimensional, flat surface that possesses mass. Imagine it as a thin sheet, similar to a piece of paper, but this sheet can have variable mass distribution.
The concept of a lamina is significant when determining the center of mass, especially in exercises related to mass distribution over a plane. To better understand, picture a rectangle drawn on the ground.
This rectangle, or lamina, can have mass distributed across it at different densities depending on a given function.

• Geometrical Aspect: Lamina usually defines an area like a rectangle, circle, or any plane figure.
• Mass Distribution: This is given by a density function, which tells us how mass is spread over each point in the region.

To solve problems involving lamina, just think of it as an object on which mathematics is used to understand physical properties like mass and center.

density function

The density function, in mathematics, tells us how much mass is present at each point on a two-dimensional object like a lamina. For this exercise, the density is constant across the rectangle and is noted as \( \delta(x, y) = 5 \mathrm{gm/cm}^2 \).
This means that for every square centimeter of the rectangle, the mass is 5 grams. This uniformity simplifies calculations considerably.
When mass is uniformly distributed, as in this case, finding the mass of the entire shape is straightforward.

• Constant Density: Here, since the density doesn't change, we multiply the density by the area of the lamina.
• Mass Calculation: For constant density, it follows the formula: \[ M = \int \int_R \delta(x, y) \, dA \]
• This is simplified to mass \( M = \delta \times \mathrm{Area \ of \ } R \).

Understanding density functions allows for solving many real-world physics problems, relating to weight distribution.

rectangle

The rectangle in this exercise serves as the plane region where mass is distributed and calculations are performed. It is defined by four vertices (1, -3), (1, 2), (7, 2), and (7, -3).Each of these points plays a crucial role in determining its length and width, parameters necessary for further calculations.
Here, the rectangle indeed forms a perfect shape due to perpendicular sides, lending straightforward computation of both area and perimeters necessary for determining different properties like mass.

• Width and Height: Width is the horizontal distance from \( x = 1 \) to \( x = 7 \), while height is vertical from \( y = -3 \) to \( y = 2 \).
• Area: Calculated by the product of width and height: \[ \mathrm{Area} = (7-1) \times (2-(-3)) = 6 \times 5 = 30 \text{ cm}^2 \]

Understanding these rectangular dimensions is vital for mastering related exercises dealing with planes and mass distributions on such regions.