Problem 13 - Calculus Single Variable

Question

Find the moment of the given region \(\mathcal{R}\) about the \(x\) -axis. Assume that \(\mathcal{R}\) has uniform unit mass density. \(\mathcal{R}\) is the region bounded above by \(y=\sqrt{4-x^{2}}\) and below by the \(x\) -axis.

Step-by-Step Solution

Answer

The moment of \(\mathcal{R}\) about the \(x\)-axis is \(\frac{32}{3}\).

Step 1: Understanding the Problem

We are tasked to find the moment of the region \(\mathcal{R}\) about the \(x\)-axis where the region is the area under the curve \(y=\sqrt{4-x^{2}}\) and above the \(x\)-axis. This indicates that \(\mathcal{R}\) is a semicircle with radius 2.

Step 2: Setting Up the Moment Formula

The moment about the \(x\)-axis, \(M_x\), for a region with uniform density is given by \(M_x = \int_{a}^{b} \overline{y} \cdot f(x) \, dx\). Here, \(\overline{y}\) is the distance from the \(x\)-axis (\(y\) itself, since we calculate moment about the \(x\)-axis), and \(f(x)\) is the function \(y=\sqrt{4-x^{2}}\).

Step 3: Defining the Integration Limits

The limits of integration \(a\) and \(b\) are determined from the points where \(y=\sqrt{4-x^2}\) meets the \(x\)-axis, which are at \(x=-2\) and \(x=2\). Thus, the integration limits are from \(-2\) to \(2\).

Step 4: Writing the Integral for Moment

The expression for the moment \(M_x\) is given by:\[ M_x = \int_{-2}^{2} y \cdot \sqrt{4-x^2} \, dx = \int_{-2}^{2} y^2 \, dx, \] where \(y^2 \) represents \((\sqrt{4-x^2})^2 = 4-x^2\).

Step 5: Solving the Integral

Now we solve the integral: \[ M_x = \int_{-2}^{2} (4-x^2) \, dx. \] This can be broken down as: \[ \int_{-2}^{2} 4 \, dx - \int_{-2}^{2} x^2 \, dx. \]Calculating the separate integrals:1. \(\int_{-2}^{2} 4 \, dx = 4x \Big|_{-2}^{2} = 4(2) - 4(-2) = 8 + 8 = 16\)2. \(\int_{-2}^{2} x^2 \, dx = \frac{x^3}{3} \Big|_{-2}^{2} = \left( \frac{8}{3} - \left(-\frac{8}{3}\right) \right) = \frac{16}{3}\)Subtracting these results, we get: \[ M_x = 16 - \frac{16}{3} = \frac{32}{3}. \]

Step 6: Conclusion

The moment of the region \( \mathcal{R} \) about the \(x\)-axis is \( \frac{32}{3} \). This value represents how the mass is distributed with respect to the \(x\)-axis given a uniform density.

Key Concepts

Integration Techniques

Integration is a critical tool in calculus that helps in finding the accumulation of quantities like areas and volumes. To find the moment of inertia, like in the given exercise, we use an integrated approach. In our example, the moment is calculated as an integral of

• the function over a defined region,
• multiplied by its distance from the axis of rotation.

In most instances involving moment computations along an axis, the formula is \[ M_x = \int_{a}^{b} \overline{y} \cdot f(x) \, dx, \] where \( \overline{y} \) represents the distance from the axis and \( f(x) \) is the function that defines the curve.

Particularly, you will often break this integral into simpler parts for easier computation. Like in our problem, we separated it into two different integrals: one for a constant function and one for a polynomial function. By assigning limits and computing individually, it simplifies the solving process. This highlights the usefulness of understanding how to apply specific integration techniques based on the function's form.

Semicircles in Calculus

The semicircle is a common shape encountered in calculus problems, often involving integrals for calculating areas, moments, and centers of mass. In this exercise, the region \( \mathcal{R} \) is defined by the semicircle above the \( x \)-axis with the equation \( y=\sqrt{4-x^2} \). This equation derives from the Pythagorean theorem, representing half of the circle \( x^2 + y^2 = 4 \).

To solve problems involving semicircles, it's important to grasp several key steps:

• Understand the boundary as it represents half of a full circle with radius 2.
• Set up integration limits based on the points where the semicircle meets the \( x \)-axis.

In this case, these limits are from \( x = -2 \) to \( x = 2 \).
As with many physics and engineering problems, semicircles can be used to model situations where calculating a property like a moment about an axis is necessary. By understanding the geometry and characteristics, such as symmetry and radius, it becomes straightforward to apply calculus techniques to solve for desired values.

Uniform Mass Density

Uniform mass density is a significant concept in physics and calculus, indicating that mass is evenly distributed throughout a given region. This assumption streamlines calculations because the density does not vary over the area.

In the case of this exercise, treating the semicircle as having a uniform unit mass density means every part of \( \mathcal{R} \) has an equal contribution to the moment about the \( x \)-axis. This allows you to focus on integrating the function of the region without adjusting for variations in mass density.
When solving for moments or centers of mass, knowing that the mass density is uniform means:

• You can use the entire region's area in computations.
• The function defining the region determines the distribution contextually without extra variables for density.

This assumption simplifies calculations considerably, helping to obtain results more precisely and efficiently, especially in theoretical physics or engineering using calculus-based approaches.