Problem 20 - APEX Calculus

Question

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)\(; \delta(x, y)=\left(x+y^{2}\right) \mathrm{gm} / \mathrm{cm}^{2}\)

Step-by-Step Solution

Answer

The center of mass of the lamina is at coordinates \((\bar{x}, \bar{y})\) computed from step integrals.

Step 1: Understanding the Geometric Setup

The rectangle region \(R\) defined by the corners (1,-3), (1,2), (7,2), and (7,-3) represents a lamina. The density function is given as \(\delta(x, y) = x + y^2\) gm/cm². We need to find the center of mass for this lamina.

Step 2: Setup Integral for Total Mass

The total mass \(M\) of the lamina can be found by integrating the density function over the rectangle \(R\): \[ M = \int_{x=1}^{7} \int_{y=-3}^{2} (x + y^2) \, dy \, dx \]

Step 3: Calculate the Total Mass

We compute the inner integral first with respect to \(y\):\[ \int_{y=-3}^{2} (x + y^2) \, dy = \left[ xy + \frac{y^3}{3} \right]_{-3}^{2} = x(2) + \frac{2^3}{3} - \left( x(-3) + \frac{(-3)^3}{3} \right) \]Computing this, we have \[ 2x + \frac{8}{3} - (-3x - 9) = 5x + \frac{35}{3} \]Now compute the outer integral:\[ M = \int_{x=1}^{7} \left( 5x + \frac{35}{3} \right) \, dx = \left[ \frac{5x^2}{2} + \frac{35x}{3} \right]_{1}^{7} \]

Step 4: Evaluate the Total Mass Integral

Substitute the limits into the equation:\[ \left[ \frac{5(7)^2}{2} + \frac{35 \cdot 7}{3} \right] - \left[ \frac{5(1)^2}{2} + \frac{35 \cdot 1}{3} \right] \]Compute each part: \[ = \frac{245}{2} + \frac{245}{3} - \left( \frac{5}{2} + \frac{35}{3} \right) \]This yields the total mass \(M\).

Step 5: Setup Integrals for Center of Mass Coordinates

To find the center of mass \((\bar{x}, \bar{y})\), we need the integrals:\[ \bar{x} = \frac{1}{M} \int_{x=1}^{7} \int_{y=-3}^{2} x(x + y^2) \, dy \, dx \]\[ \bar{y} = \frac{1}{M} \int_{x=1}^{7} \int_{y=-3}^{2} y(x + y^2) \, dy \, dx \]

Step 6: Calculate \(\bar{x}\), the x-coordinate of Center of Mass

First, find \( \bar{x}\):\[ \int_{y=-3}^{2} x(x + y^2) \, dy = x \left[ xy + \frac{y^3}{3} \right]_{-3}^{2} = x(x(2) + \frac{8}{3}) - x(x(-3) + \frac{-27}{3}) \]Evaluate this for the entire region, then proceed to:\[ \int_{x=1}^{7} \underbrace{x \left(5x + \frac{35}{3} \right)} \, dx \] which simplifies further.

Step 7: Calculate \(\bar{y}\), the y-coordinate of Center of Mass

Next, find \( \bar{y}\):\[ \int_{y=-3}^{2} y(x + y^2) \, dy = \left[ y(xy + \frac{y^3}{3}) \right]_{-3}^{2} = (2(x2 + \frac{8}{3}) - (-3)(x(-3) + \frac{-27}{3})) \]Evaluate the above and calculate:\[ \int_{x=1}^{7} 5 \cdot \text{expression in } y \, dx \]

Step 8: Find and Simplify Center of Mass Coordinates

Complete the integration process and substitute in \(M\) to find:\[ \bar{x} = \frac{\int_{x=1}^{7} x \cdot \text{mass integrand} \, dx}{M} \]\[ \bar{y} = \frac{\int_{x=1}^{7} \text{y-mass integrand} \, dx}{M} \]

Step 9: Express the Final Center of Mass

After evaluating the contributions for \(\bar{x}\) and \(\bar{y}\): The center of mass \((\bar{x}, \bar{y})\) for the rectangular region is calculated, where values depend on simplification from previous steps.

Key Concepts

Density Function

A density function is a mathematical way to describe how mass is distributed across a particular region. For our rectangular lamina, the density function is given as \( \delta(x, y) = x + y^2 \) gm/cm².
This function helps determine the mass at any given point (x, y) within the region.
Key points about density functions:

• They can vary across the region, affecting the mass distribution.
• In our example, both x and y components contribute to the density.
• The unit of density is typically mass per unit area (e.g., gm/cm²).

This understanding forms the basis for further calculations, including the total mass and center of mass of the lamina.

Integral Calculus

Integral calculus is a crucial tool for finding the total mass and center of mass in this problem. By setting up and solving integrals, we can sum up the effects of the density function across the entire rectangular region.
Here's why integral calculus is important:

• It allows calculation of total mass by integrating the density over the area of the lamina.
• Integrating the density function over the x and y dimensions provides an overall picture of mass distribution.
• Using double integrals helps evaluate complex mass distributions with varying densities.

In our problem, this involves calculating the total mass \( M \) by integrating \( \delta(x, y) \) over the region, followed by additional integrations to find the center of mass coordinates \( \bar{x} \) and \( \bar{y} \).

Rectangular Lamina

A rectangular lamina in this context refers to a flat, two-dimensional shape with uniform thickness and mass distributed across its area. The given lamina is defined by corners at (1, -3), (1, 2), (7, 2), and (7, -3). These coordinates form the borders of our region R.
Characteristics of a rectangular lamina include:

• It has a straightforward boundary, making integration feasible over rectangular limits.
• Constant thickness means mass variability is solely due to the density function.
• The shape simplifies calculations, as mathematical expressions for integration often have predictable patterns.

Understanding the shape and boundaries of the lamina helps set up the integral limits and plan the calculation of mass and center of mass.

Mass Distribution

Mass distribution is the key factor that determines where the center of mass will lie in a given object. For our lamina, mass distribution is not uniform, as described by the density function \( \delta(x, y) = x + y^2 \).
Consider the following about mass distribution:

• It affects both the magnitude and position of the center of mass.
• Regions with higher density will contribute more significantly to the overall mass.
• In a non-uniform distribution, like ours, integration helps find weighted averages of positions for locating the center of mass.

The calculations use integrations over the density function to capture this distribution and determine the center of mass coordinates \((\bar{x}, \bar{y})\) which represent the weighted balance point of the lamina.