Problem 9 - Calculus

Question

Find the mass \(M\), the moments \(M_{y}\) and \(M_{x}\), and the center of \(\operatorname{mass}(\bar{x}, \bar{y})\). $$ \rho=7 \text { in the square } 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1 $$

Step-by-Step Solution

Answer

Mass \( M = 7 \); Moments \( M_y = \frac{7}{2} \), \( M_x = \frac{7}{2} \); Center of mass \( (\bar{x}, \bar{y}) = \left( \frac{1}{2}, \frac{1}{2} \right) \).

Step 1: Understand the Problem

We have a square region in the plane defined by \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \) with a uniform density \( \rho = 7 \). Our task is to find the total mass \( M \), the moments \( M_y \) and \( M_x \), and the center of mass \( (\bar{x}, \bar{y}) \).

Step 2: Calculate the Mass (M)

To find the mass \( M \), integrate the constant density over the area of the square. The mass \( M \) is given by the integral \( M = \int_{0}^{1} \int_{0}^{1} \rho \, dx \, dy \). Since \( \rho = 7 \), we have:\[ M = \int_{0}^{1} \int_{0}^{1} 7 \, dx \, dy = 7 \cdot \int_{0}^{1} \int_{0}^{1} \, dx \, dy = 7 \cdot 1 \cdot 1 = 7. \]

Step 3: Calculate the Moment My

The moment about the \( y \)-axis, \( M_y \), is given by \( M_y = \int_{0}^{1} \int_{0}^{1} x \rho \, dx \, dy \). Substituting \( \rho = 7 \), we get:\[ M_y = \int_{0}^{1} \int_{0}^{1} x \cdot 7 \, dx \, dy = 7 \cdot \int_{0}^{1} x \, dx = 7 \cdot \left[ \frac{x^2}{2} \right]_{0}^{1} = 7 \cdot \frac{1}{2} = \frac{7}{2}. \]

Step 4: Calculate the Moment Mx

The moment about the \( x \)-axis, \( M_x \), is given by \( M_x = \int_{0}^{1} \int_{0}^{1} y \rho \, dx \, dy \). Since the density \( \rho = 7 \), it results in a similar calculation:\[ M_x = \int_{0}^{1} \int_{0}^{1} y \cdot 7 \, dy \, dx = 7 \cdot \int_{0}^{1} y \, dy = 7 \cdot \left[ \frac{y^2}{2} \right]_{0}^{1} = 7 \cdot \frac{1}{2} = \frac{7}{2}. \]

Step 5: Find the Center of Mass (\(\bar{x}, \bar{y}\))

The coordinates of the center of mass are given by:\( \bar{x} = \frac{M_y}{M} \) and \( \bar{y} = \frac{M_x}{M} \).Substituting the values:\[ \bar{x} = \frac{\frac{7}{2}}{7} = \frac{1}{2}, \quad \bar{y} = \frac{\frac{7}{2}}{7} = \frac{1}{2}. \]Therefore, \( (\bar{x}, \bar{y}) = \left( \frac{1}{2}, \frac{1}{2} \right) \).

Key Concepts

Moments and Centers of Mass

In the study of calculus, particularly when dealing with physical systems, it's essential to understand moments and centers of mass. Think of moments like balancing points. If you imagine a seesaw, the moment helps in finding that spot where the seesaw balances perfectly. The center of mass then is exactly where this balance happens for a uniform object.
For a given region, the moment about an axis is derived by multiplying the distance of each infinitesimal part from the axis with its density, and integrating over the entire area. This gives us the total effect of all the tiny "weights" in the region. We use

• \( M_y \) for the moment about the y-axis, and
• \( M_x \) for the moment about the x-axis.

These combined help us identify the center of mass, denoted as \( (\bar{x}, \bar{y}) \). This point is critical because it helps in understanding where an object will balance in space, or how it moves under forces.

Integration in Calculus

Integration is a fundamental concept in calculus. It allows us to accumulate quantities over a region, like finding the area under a curve or, in this case, finding total mass or moments. Imagine it as summing up infinite tiny pieces to get a whole picture.
In our square region exercise, we used double integration because we are dealing with an area—a two-dimensional space. Integration was carried out over the x and y dimensions to calculate characteristics like mass and moments.

• First, we integrated across one dimension, and then the other.
• The integration across both x and y allows us to consider every tiny part of the area, ensuring nothing is missed.

This tool is essential when working with continuous distributions and varying densities, though our case had a uniform density, making the problem simpler. Understanding integration helps in exploring much intricate concepts in physics, engineering, and even probability.

Uniform Density

Density tells us how much "stuff" is packed into a given space. If you have a loaf of bread with uniform slices, every slice has the same amount of bread. That's what uniform density means—every part has the same mass per unit volume.
In this exercise, we worked with a uniform density given by \( \rho = 7 \). This means that every point within the specified region has the same density. This simplifies calculations significantly.

• The mass \( M \) for a uniform region can be calculated by multiplying the area by the density.
• This uniformity also means that the center of mass is simply the geometric center of the shape.

Uniform density is a common scenario in exercises to introduce the concept of mass in calculus without the complexity of varying densities. As you advance in calculus, you may encounter problems with non-uniform density, adding layers of complexity to the calculation of mass and moments.