Problem 11 - APEX Calculus

Question

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

Step-by-Step Solution

Answer

The mass of the lamina is 150 gm.

Step 1: Understand the Problem

We are given a rectangle with corners (1,-3), (1,2), (7,2), and (7,-3). We are also given a constant density function \( \delta(x, y) = 5 \, \text{gm/cm}^2 \). Our task is to find the mass of this rectangular region.

Step 2: Determine the Area of Region R

Since \( R \) is a rectangle, we can find the area by determining the length and width of the rectangle. The width is from \( x = 1 \) to \( x = 7 \), which is 6 units. The height is from \( y = -3 \) to \( y = 2 \), which is 5 units. Therefore, the area \( A \) is \( 6 \times 5 = 30 \, \text{cm}^2 \).

Step 3: Calculate the Mass Using the Density Function

The mass \( M \) of the lamina can be calculated as the product of its area and its constant density. Given the density \( \delta(x, y) = 5 \, \text{gm/cm}^2 \), the mass is given by \( M = \text{Area} \times \text{Density} = 30 \, \text{cm}^2 \times 5 \, \text{gm/cm}^2 = 150 \, \text{gm} \).

Key Concepts

Density Function

A density function describes how mass is distributed over a region. It assigns a mass per unit area or volume, commonly expressed in units such as grams per square centimeter (gm/cm²) or grams per cubic centimeter (gm/cm³). In the context of a lamina like the one in this exercise, the function is constant, meaning it doesn’t change across the lamina.

• In this particular exercise, the density function is given as \( \delta(x, y) = 5 \, \text{gm/cm}^2 \).
• This allows for straightforward calculations, as the density doesn't vary with position \((x, y)\).

Knowing the density helps determine the overall mass of objects when combined with their physical size, or area, for flat regions like rectangles.

Rectangular Region

A rectangular region is simply a flat, two-dimensional shape confined by four sides, with opposite sides being equal and parallel. Calculating properties of rectangular regions is straightforward due to their simple geometric attributes. In this exercise, the rectangle is defined using coordinates:

• The corners of the rectangle are given as (1,-3), (1,2), (7,2), and (7,-3).
• Knowing these points allows us to determine the shape's width and height.

To visualize this region, plot these points on a coordinate plane to form the shape. Rectangles are defined by right angles, and the sides should fit neatly between the given points. Understanding how to define and sketch these regions is crucial for spatial reasoning in geometry.

Calculation of Area

Calculating the area of the rectangle involves multiplying its length by its width. For any rectangle, this process is straightforward and foundational in geometry. Key Steps:

• Determine the width: It's the horizontal distance between points sharing the same y-coordinate. From this problem, this distance is \( x = 7 - x = 1 = 6 \) units.
• Determine the height: It's the vertical distance between points sharing the same x-coordinate. Here, it's \( y = 2 - y = -3 = 5 \) units.
• Area \( A \) is thus \( 6 \times 5 = 30 \, \text{cm}^2 \).

Applying these basics allows you to handle even more complex shapes in the future. Recognizing these measurements is fundamental when moving to calculate mass.

Constant Density

Constant density indicates that the distribution of mass does not change over the entire region of the shape. Here, the constant density is set at \( 5 \, \text{gm/cm}^2 \) across the entire lamina. Benefits of Constant Density:

• The mass of any portion of the lamina can be easily calculated by multiplying the area by this fixed density value.
• Simplifies the integration needed to find mass since density remains unchanged throughout the region.

Calculating mass with constant density means you apply the density equally across the entire shape, making it straightforward to compute the total mass using the equation \( M = A \times \delta \). In this case, for a 30 cm² lamina, it results in \( 150 \, \text{gm} \). This exemplifies why knowing the density, especially a constant one, is important for practical applications in physics and engineering.