Problem 8 - 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 first quadrant region bounded above by \(y=\frac{\sin (x)}{x}\) below by the \(x\) -axis, and on the sides by \(x=\pi / 6\) and \(x=\pi / 2\); about \(x=0\).

Step-by-Step Solution

Answer

The moment about the vertical axis \(x=0\) is \(\frac{\sqrt{3}}{2}\).

Step 1: Identify the Region

We are given the region \ \( \mathcal{R} \approx \{(x, y) \mid 0 \leq x \leq \pi/2, 0 \leq y \leq \frac{\sin(x)}{x} \} \) bounded by the curve \( y = \frac{\sin(x)}{x} \), the \(x\)-axis, and the lines \( x = \pi/6 \) and \( x = \pi/2 \).

Step 2: Set up the Moment Formula

To find the moment \( M \) about the vertical axis \( x = 0 \), we use the formula \( M = \int_{a}^{b} x \cdot \text{density} \cdot ( ext{height}) \, dx \). Here, since the density is uniform and equal to 1, it simplifies to \( M = \int_{\pi/6}^{\pi/2} x \cdot f(x) \, dx \), where \( f(x) = \frac{\sin(x)}{x} \).

Step 3: Substitute and Simplify the Integral

Substitute \( f(x) \) in the integral: \ \( M = \int_{\pi/6}^{\pi/2} x \cdot \frac{\sin(x)}{x} \, dx \). \ Simplifying, we have \( M = \int_{\pi/6}^{\pi/2} \sin(x) \, dx \).

Step 4: Calculate the Integral

Now compute the integral: \ \( M = \int_{\pi/6}^{\pi/2} \sin(x) \, dx = [-\cos(x)]_{\pi/6}^{\pi/2} \).

Step 5: Evaluate the Integral

Evaluate \(-\cos(x)\) from \(\pi/6\) to \(\pi/2\): \ \( M = [-\cos(\pi/2) + \cos(\pi/6)] = [0 + \frac{\sqrt{3}}{2}] = \frac{\sqrt{3}}{2} \).

Key Concepts

Moment About a Vertical Axis

The concept of a moment about a vertical axis is integral to understanding how forces or mass distributions influence certain points in space. In this exercise, we consider the **moment about the vertical axis at** \( x = 0 \) for the region \( \mathcal{R} \). The moment can be understood as the rotational equivalent of linear motion, describing the tendency of a mass to rotate around an axis. To find the moment about a vertical axis, we use the formula:

• \( M = \int_{a}^{b} x \cdot \text{density} \cdot (\text{height}) \, dx \)

Here, \( x \) represents the distance from the vertical axis, and \( \text{height} \) is given by the function \( f(x) = \frac{\sin(x)}{x} \). Notably, calculating the moment involves multiplying the distance from the axis by the mass distribution across the stipulated boundaries.

Uniform Mass Density

Uniform mass density implies that the mass is evenly distributed across the region \( \mathcal{R} \). This simplifies calculations as the density can be factored out when computing the moment.Since mass density is uniform and assumed to be 1 in this exercise, it simplifies the formula for calculating the moment. We can focus on the integration part without additional complexities arising from variable density.

• With density = 1, the moment formula becomes smaller and simpler: \( M = \int_{a}^{b} x \cdot f(x) \, dx \).

This simplification is critical for calculating the moment efficiently, making it an ideal scenario to learn about rotational dynamics in physics and engineering.

Integration

Integration is a crucial mathematical tool used to find the sum of infinite, small quantities when dealing with continuous variables, particularly with functions defined over an interval.To find the moment \( M \), we need to integrate:

• \( M = \int_{\pi/6}^{\pi/2} x \cdot f(x) \, dx \)

After substituting \( f(x) = \frac{\sin(x)}{x} \), the integral becomes simpler:

• \( M = \int_{\pi/6}^{\pi/2} \sin(x) \, dx \)

The integration process essentially computes the area under the curve \( \sin(x) \) from \( x = \pi/6 \) to \( x = \pi/2 \), giving us the measure of the moment for this region.

Trigonometric Functions

Trigonometric functions, like \( \sin(x) \), play a fundamental role in this moment calculation.Their periodic nature helps describe oscillations, waves, and other regular, repeating phenomena occurring in physics.

1. Here, \( f(x) = \frac{\sin(x)}{x} \) crafts the upper boundary of the region \( \mathcal{R} \).
2. This function is modified by multiplying with \( x \) in the integral, creating a different shape for the area we're calculating.

Understanding \( \sin(x) \) and how it varies between \( 0 \) and \( \pi/2 \) aids in grasping how the integral reflects these changes in the moment. Thus, trigonometric identities and evaluations become essential when computing integrals involving \( \sin(x) \).