Problem 7 - Calculus Single Variable

Question

Find the moment of the given region \(\mathcal{R}\) about the given vertical axis. Assume that \(\mathcal{R}\) has uniform unit mass density. \(\mathcal{R}\) is the region bounded above by \(y=1 / x\), below by the \(x\) axis, and on the sides by the vertical lines \(x=1\) and \(x=2\); about \(x=-3\)

Step-by-Step Solution

Answer

The moment \( M = \ln(2) + 3\ln^2(2) \) about \( x = -3 \).

Step 1: Understanding the Moment

The moment of an area about a vertical axis is calculated using the integral formula: \[ M = ho imes ext{Area} imes ar{x} \]Where \( \rho \) is the mass density (which is 1), Area is the integral of the region, and \( \bar{x} \) is the average distance from the axis.

Step 2: Determine the Limits and Function

The region \( \mathcal{R} \) is bounded by \( y = \frac{1}{x} \), the x-axis, \( x=1 \), and \( x=2 \). The function \( y = \frac{1}{x} \) describes the upper boundary.

Step 3: Calculate the Area of the Region

Compute the integral of \( y = \frac{1}{x} \) from \( x=1 \) to \( x=2 \) to find the area:\[ \text{Area} = \int_{1}^{2} \frac{1}{x} \, dx \]This integral evaluates to:\[ \ln|x| \bigg|_1^2 = \ln(2) - \ln(1) = \ln(2) \]

Step 4: Calculate the Average Distance

Since we want the moment about \( x = -3 \), the distance \( d(x) \) from each point on \( \mathcal{R} \) to this line is \( |x + 3| \). The \( \bar{x} \) is given as:\[ \bar{x} = \frac{1}{\text{Area}} \cdot \int_{1}^{2} (x+3) \cdot \frac{1}{x} \, dx \]

Step 5: Evaluate the Integral for Moment

Calculate the integral for \( \bar{x} \):\[ \int_{1}^{2} \frac{x+3}{x} \, dx = \int_{1}^{2} \left(1 + \frac{3}{x}\right) \, dx \]This evaluates to:\[ (x + 3\ln|x|) \bigg|_1^2 = \left(2 + 3\ln(2)\right) - \left(1 + 3\ln(1)\right) = 1 + 3\ln(2) \]

Step 6: Compute the Moment

Finally, compute the moment \( M \) by multiplying the area by \( \bar{x} \):\[ M = \ln(2) \times (1 + 3\ln(2)) \]

Step 7: Simplify and Conclude

So, the moment of the region \( \mathcal{R} \) about the vertical line \( x = -3 \) is:\[ M = \ln(2) + 3\ln^2(2) \]

Key Concepts

Integral Calculus

Integral calculus is a part of calculus that deals with the concept of integration. It is essential for solving problems that involve finding the area under a curve or other accumulative processes. In the context of this problem, integral calculus helps us calculate the area of a region defined by a curve, such as

• Determining the area under or between curves.
• Finding the total accumulated change across a region.
• Calculating moment or torque by integrating along an axis.

In the solution, the integral of the function \( y = \frac{1}{x} \) from \( x=1 \) to \( x=2 \) is calculated to determine the area. This is achieved by evaluating the natural logarithm, \( \ln(x) \), between these limits, resulting in \( \ln(2) \). Integrals provide a reliable method to solve for quantities over a continuum.

Area Under a Curve

The concept of the area under a curve is integral to understanding calculus. In essence, when you integrate a function over a specific interval, you find the "accumulated area" between the curve and the axis.

• The area is said to be positive if the curve lies above the x-axis within the selected bounds.
• To find this area, we use definite integrals, applying limits to the integration function.
• In the given problem, the function \( y = \frac{1}{x} \) acts as the upper boundary when compared to the x-axis.

This integral provides the required area, which forms a crucial part of calculating the moment because it represents part of the distribution of mass across the region. Understanding this helps grasp how specific regions contribute more to overall calculations based on their positioning above the x-axis.

Vertical Axis

When calculating moments, particularly in areas defined with respect to vertical axes, we determine how mass distribution affects torque. In this exercise, we examine the area about \( x = -3 \), letting us understand how far each segment of the area is from this vertical line.

• The vertical axis can serve as a pivot line, measuring how much torque or moment is produced due to a region's position.
• In this problem, the calculation \( |x + 3| \) represents the distance from each point in \( \mathcal{R} \) to \( x = -3 \).
• The average distance or centroid, denoted as \( \bar{x} \), is used key into the moment calculation formula.

Effectively, the moment considers both the area of the region and its average distance from the vertical axis, culminating in how mass affects this line.

Uniform Mass Density

Uniform mass density implies every part of the region or object has the same mass per unit area or volume. In simpler terms, density is the measure of how "packed" the mass is in the space it occupies.

• When dealing with moments, uniform mass density helps simplify calculations because the density, \( \rho \), remains constant throughout the region.
• Given a density of 1 (unit mass density), our calculations in this problem focus on the geometric factors, like area and distance.
• This uniform distribution allows us to use the moment formula directly, without adjusting for variations in density.

In scenarios with non-uniform density, one would have to integrate the density function as well. However, for unit density, the focus rests on how effectively a uniform "sheet" of mass exerts force about a pivot point, such as our vertical line here.