Problem 32 - Calculus Single Variable

Question

Find the center of \(\operatorname{mass}(\bar{x}, \bar{y})\) of the given region \(\mathcal{R},\) assuming that it has uniform unit mass density. \(\mathcal{R}\) is the region bounded above by \(y=x\), below by \(y=\sqrt{x}\), on the left by \(x=1\), and on the right by \(x=4\).

Step-by-Step Solution

Answer

The center of mass is approximately at \((5.9733, 2.1333)\).

Step 1: Understand the Integral Setup

To find the center of mass \((\bar{x}, \bar{y})\) of a region \(\mathcal{R}\) with uniform density, you need to use the formulas for \(\bar{x}\) and \(\bar{y}\). The formulas are \(\bar{x} = \frac{1}{M} \int_{x_1}^{x_2} x f(x)\, dx\) and \(\bar{y} = \frac{1}{M} \int_{x_1}^{x_2} \frac{1}{2} (f(x)^2)\, dx\), where \(M\) is the total mass of the region. In this case, \(f(x)\) represents the height of the slice bounded above by \(y = x\) and below by \(y = \sqrt{x}\).

Step 2: Calculate the Mass (M) of the Region

Since the density is uniform and taken as 1, it simplifies the problem. First, calculate the mass, which is the integral of the area between the curves, from \(x = 1\) to \(x = 4\). \[ M = \int_{1}^{4} (x - \sqrt{x}) \, dx \]Calculate this integral to find \(M\). \[ M = \left[ \frac{x^2}{2} \right]_{1}^{4} - \left[ \frac{2}{3}x^{3/2} \right]_{1}^{4} \]\[ M = \left(8 - \frac{2}{3} \cdot 8\right) - \left(\frac{1}{2} - \frac{2}{3}\right) \]\[ M = 8 - \frac{16}{3} - \frac{1}{2} + \frac{2}{3} \]\[ M = \frac{24}{3} - \frac{16}{3} - \frac{3}{6} + \frac{4}{6} \]\[ M = \frac{8}{3} - \frac{1}{6} \]\[ M = \frac{16}{6} - \frac{1}{6} \]\[ M = 2.5 \]

Step 3: Calculate \(\bar{x}\)

Now calculate \(\bar{x}\) using the formula:\[ \bar{x} = \frac{1}{M} \int_{1}^{4} x(x - \sqrt{x}) \, dx \]Calculate this integral:\[ \bar{x} = \frac{1}{2.5} \left[ \frac{x^3}{3} \right]_{1}^{4} - \left[ \frac{2}{5}x^{5/2} \right]_{1}^{4} \]\[ \bar{x} = \frac{1}{2.5} \left(\frac{64}{3} - \frac{32}{5}\right) - \left( \frac{1}{3} - \frac{2}{5}\right) \]\[ \bar{x} = \frac{1}{2.5} \left(\frac{320}{15} - \frac{96}{15}\right) \]\[ \bar{x} = \frac{1}{2.5} \times \frac{224}{15} \]\[ \bar{x} = \frac{224}{37.5} \]\[ \bar{x} = 5.9733 \]

Step 4: Calculate \(\bar{y}\)

Now calculate \(\bar{y}\), using the formula: \[ \bar{y} = \frac{1}{M} \int_{1}^{4} \frac{1}{2} \left((x)^2 - (\sqrt{x})^2\right) \, dx \]Calculate this integral:\[ \bar{y} = \frac{1}{2.5} \int_{1}^{4} \frac{1}{2} (x^2 - x) \, dx \]\[ \bar{y} = \frac{1}{2.5} \left[ \frac{x^3}{6} - \frac{x^2}{2} \right]_{1}^{4} \]\[ \bar{y} = \frac{1}{2.5} \left(\frac{64}{6} - 8 - \frac{1}{6} + 0.5\right) \]\[ \bar{y} = \frac{1}{2.5} \left(\frac{32}{3} - 8\right) \]\[ \bar{y} = \frac{1}{2.5} \times \frac{16}{3} \]\[ \bar{y} = \frac{16}{7.5} \]\[ \bar{y} = 2.1333 \]

Step 5: Combine the Results

After calculating \(\bar{x}\) and \(\bar{y}\), the center of mass is:\(\bar{x} = 5.9733\) and \(\bar{y} = 2.1333\).

Key Concepts

Mass Density

Understanding mass density is important when figuring out the center of mass for a region. Mass density refers to how much mass a substance has in a given volume, usually expressed as a mass per unit volume.
For our exercise, the region \(\mathcal{R}\) has a uniform mass density, which means the entire region has the same mass distribution without any variation. This uniformity simplifies the calculation significantly, assuming the mass density is a constant value (e.g., 1). That means the total mass of the region is the same as its area, simplifying calculations.
In general, for regions without uniform density, calculations could become more complex as mass density might need integrating over varying parts of the region. But since we assume uniform density here, we can directly calculate key quantities like the center of mass with simpler integrals.

Integral Calculus

Integral calculus is a powerful tool used for solving problems involving curves and areas under curves, commonly applied in physics and engineering. In this exercise, integral calculus helps us calculate the center of mass of the bounded region \(\mathcal{R}\).
To apply integral calculus effectively, we use specific formulas to find the center of mass coordinates \(\bar{x}, \bar{y}\). These formulas involve calculating integrals of the area between the given boundary curves. For this problem, we determine total mass \(M\) and individual coordinates by integrating slices of the region between specific boundaries.
By integrating these areas and equations, we simplify the process of finding weighted averages needed for the center of mass. Therefore, integral calculus allows combining geometric intuition with precise calculation for real-world applications.

Bounded Region

A bounded region is the space enclosed by given curves on a graph, crucial for finding properties like the center of mass. In this exercise, our bounded region \(\mathcal{R}\) lies between the graphs of \(y = x\) and \(y = \sqrt{x}\), bounded further at \(x = 1\) and \(x = 4\).
This space forms a specified area where calculations take place. Bounded regions often require careful setup for integrals, as intersections and endpoints determine how limits of integration are set.

• The top boundary is defined by \(y = x\)
• The bottom boundary is \(y = \sqrt{x}\)
• Left boundary begins at \(x = 1\)
• Right boundary ends at \(x = 4\)

Careful understanding of where these boundaries meet ensures correct integral setup, as missing or inaccurate limits can lead to incorrect solutions. Thus, understanding the region's boundaries is pivotal.

Uniform Density

Uniform density implies that a region or object has a consistent mass distribution throughout. When a region like \(\mathcal{R}\) is described as having uniform density, it means every unit area within this region has the same amount of mass.
This characteristic is essential when calculating properties like center of mass because it allows us to treat mass density as a constant in the mathematical formulas involved. For the given exercise, uniform unit mass density makes calculations straightforward by assuming that

• Density is consistent across the entire region
• Total mass equals the region's area
• Mass doesn't vary or concentrate in different parts

This simplifies integrals and ensures that the direct area measurement represents the total mass accurately, allowing efficient calculations that don't require varying mass elements.