Problem 28 - 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 mass density. \(\mathcal{R}\) is the region bounded by \(y=x+1 / x, x=1 / 2, x=2,\) and the \(x\) -axis.

Step-by-Step Solution

Answer

The center of mass is given by \( (\bar{x}, \bar{y}) = \left( \frac{33}{8(\frac{15}{8} + \ln 4)}, \frac{57}{16(\frac{15}{8} + \ln 4)} \right) \).

Step 1: Identify the Boundaries

The region \( \mathcal{R} \) is bounded by the curve \( y = x + \frac{1}{x} \), the vertical lines \( x = \frac{1}{2} \) and \( x = 2 \), and the \( x \)-axis. These boundaries help define the limits of integration.

Step 2: Set up Mass Calculation

Since the mass density is uniform, let it be \( \rho = 1 \). The mass \( M \) of region \( \mathcal{R} \) can be found using the integral \( M = \int_{1/2}^{2} \left( x + \frac{1}{x} \right) \, dx \).

Step 3: Calculate the Mass

Evaluate the integral: \[ M = \int_{1/2}^{2} \left( x + \frac{1}{x} \right) \, dx = \int_{1/2}^{2} x \, dx + \int_{1/2}^{2} \frac{1}{x} \, dx. \] Calculate each term separately.

Step 4: Simplify the Integrals

First integral: \( \int_{1/2}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1/2}^{2} = \frac{4}{2} - \frac{(1/2)^2}{2} = 2 - \frac{1}{8} = \frac{15}{8} \).Second integral: \( \int_{1/2}^{2} \frac{1}{x} \, dx = \left[ \ln |x| \right]_{1/2}^{2} = \ln 2 - \ln \frac{1}{2} = \ln 4 \).Thus, \( M = \frac{15}{8} + \ln 4 \).

Step 5: Determine \( \bar{x} \) Coordinate

The \( \bar{x} \) coordinate of the centroid is given by: \( \bar{x} = \frac{1}{M} \int_{1/2}^{2} x \left( x + \frac{1}{x} \right) \, dx = \frac{1}{M} \int_{1/2}^{2} (x^2 + 1) \, dx \).

Step 6: Calculate the \( \bar{x} \) Integral

Calculate \( \int_{1/2}^{2} (x^2 + 1) \, dx = \int_{1/2}^{2} x^2 \, dx + \int_{1/2}^{2} 1 \, dx \).First part: \( \int_{1/2}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{1/2}^{2} = \frac{8}{3} - \frac{1}{24} = \frac{63}{24} \).Second part: \( \int_{1/2}^{2} 1 \, dx = \left. x \right|_{1/2}^{2} = \frac{3}{2} \).Thus, this integral is \( \frac{63}{24} + \frac{3}{2} \).

Step 7: Simplify and Find \( \bar{x} \)

Combine and simplify: \( \int_{1/2}^{2} (x^2 + 1) \, dx = \frac{63}{24} + \frac{36}{24} = \frac{99}{24} = \frac{33}{8} \).Then, \( \bar{x} = \frac{1}{M} \cdot \frac{33}{8} = \frac{33}{8(\frac{15}{8} + \ln 4)} \).

Step 8: Determine \( \bar{y} \) Coordinate

The \( \bar{y} \) coordinate of the centroid is given by: \( \bar{y} = \frac{1}{2M} \int_{1/2}^{2} \left( x + \frac{1}{x} \right)^2 \, dx \).

Step 9: Calculate the \( \bar{y} \) Integral

Develop the integral: \( \int_{1/2}^{2} \left( x + \frac{1}{x} \right)^2 \, dx = \int_{1/2}^{2} \left( x^2 + 2 + \frac{1}{x^2} \right) \, dx \).

Step 10: Evaluate Parts of the Integral

Calculate each part: \( \int_{1/2}^{2} x^2 \, dx = \frac{63}{24} \), \( \int_{1/2}^{2} 2 \, dx = 3 \), and \( \int_{1/2}^{2} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_{1/2}^{2} = -\frac{1}{2} + 2 = \frac{3}{2} \).So, \( \int_{1/2}^{2} \left( x + \frac{1}{x} \right)^2 \, dx = \frac{63}{24} + 3 + \frac{3}{2} = \frac{63}{24} + \frac{72}{24} + \frac{36}{24} = \frac{171}{24} = \frac{57}{8} \).

Step 11: Simplify and Find \( \bar{y} \)

Now calculate \( \bar{y} = \frac{1}{2M} \cdot \frac{57}{8} = \frac{57}{16 (\frac{15}{8} + \ln 4)} \).

Step 12: Present Final Answer

The coordinates of the center of mass \((\bar{x}, \bar{y})\) for the region \( \mathcal{R} \) are:\[ \bar{x} = \frac{33}{8(\frac{15}{8} + \ln 4)} \] and \[ \bar{y} = \frac{57}{16 (\frac{15}{8} + \ln 4)}. \]

Key Concepts

Mass Density

Mass density is a fundamental concept in physics and engineering, especially when evaluating the distribution of mass within a given region or object. It's defined as mass per unit volume, generally denoted by the symbol \( \rho \).
The mass density allows us to understand how mass is spread across a body or region if it's uniformly distributed. When we say an object has a uniform mass density, it means that the density \( \rho \) is constant throughout the entire object. This simplifies calculations, as seen in the exercise above, where we assume \( \rho = 1 \).
For this problem, understanding mass density helps us set up the integral calculations, as it directly influences the formulas used to find the total mass \( M \) and the centroid (\( \bar{x}, \bar{y} \)). When dealing with non-uniform densities, the problem becomes more complex as \( \rho \) would need to be included in the integration process.

Integrals

Integrals are a vital mathematical tool used to calculate areas, volumes, and other quantities under a curve. They help in summing up infinitely small quantities, which is exceedingly useful when dealing with continuous shapes and densities.
In this problem, integrals help us compute the mass of the region and the coordinates of the centroid. For instance, the integral \( \int_{1/2}^{2} \left( x + \frac{1}{x} \right) \, dx \) gives the mass \( M \) of the region by adding up the area under the curve \( y = x + \frac{1}{x} \) within the given bounds \( x = 1/2 \) to \( x = 2 \).
The centroid coordinates \( \bar{x} \) and \( \bar{y} \) are also found using integrals. These calculations involve breaking integrals into manageable parts, such as \( \int_{1/2}^{2} (x^2 + 1) \, dx \), and easily simplify using the properties of integrals.

Centroid

The centroid represents the center of mass of a geometric object and is essentially a point where the distribution of mass is equal in all directions. For a plane region with uniform mass density, finding the centroid involves integrating the area's moments over its boundaries.
\( \bar{x} \), the x-coordinate of the centroid, is determined by dividing the moment of the area with respect to the y-axis by the total mass \( M \). The formula is \( \bar{x} = \frac{1}{M} \int_{1/2}^{2} x \left( x + \frac{1}{x} \right) \, dx \). Similarly, \( \bar{y} \), the y-coordinate, requires dividing the moment with respect to the x-axis by \( 2M \), which accounts for the symmetry in the x-direction.
These calculations give us the exact point \( (\bar{x}, \bar{y}) \) where we can "balance" the mass of the region \( \mathcal{R} \), assuming uniform density.

Uniform Density

Uniform density means that every unit of volume in the object carries the same amount of mass. This simplification often turns a complex real-world problem into something solvable with basic mathematics as seen in our exercise.
Assuming a uniform density of \( \rho = 1 \) allows us to treat the mass as directly proportional to the area of region \( \mathcal{R} \). This way, we only need to integrate the function that describes the boundary conditions—without adding complexity from varying densities.
With uniform density, the expressions for centroid coordinates simplify too. The extra variable of density drops out, and we're left with straightforward area calculations. Thus, the beauty of assuming uniform density lies in the ease with which it allows us to handle integrals and get actionable results.