Problem 20 - 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 above by \(y=\cos (x), 0 \leq x \leq \pi / 2,\) and below by the \(x\) -axis.

Step-by-Step Solution

Answer

The center of mass is at \((\frac{2}{\pi}, \frac{\pi}{8})\)."

Step 1: Understand the Problem

The problem requires finding the center of mass \((\bar{x}, \bar{y})\) of the region \(\mathcal{R}\) bounded above by \(y = \cos(x)\), from \(x = 0\) to \(x = \frac{\pi}{2}\), and below by the \(x\)-axis. We assume a uniform mass density.

Step 2: Identify Mass Formulas

The center of mass for a region with uniform density can be found by calculating \(\bar{x} = \frac{1}{A} \int_{a}^{b} x f(x) \, dx\) and \(\bar{y} = \frac{1}{2A} \int_{a}^{b} [f(x)]^2 \, dx\), where \(A\) is the area of the region, and \(f(x)\) is the bounding function, in this case, \(y = \cos(x)\).

Step 3: Calculate the Area of the Region \(\mathcal{R}\)

Calculate \(A\) using the integral of the upper function minus the lower function (which is the \(x\)-axis in this case): \[ A = \int_{0}^{\frac{\pi}{2}} \cos(x) \, dx = [\sin(x)]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \] Thus, \(A = 1\).

Step 4: Compute \(\bar{x}\)

Using the formula for \(\bar{x}\), compute the integral: \[ \bar{x} = \frac{1}{A} \int_{0}^{\frac{\pi}{2}} x \cos(x) \, dx \] Integrate by parts, letting \(u = x\) and \(dv = \cos(x) \, dx\). Then \(du = dx\) and \(v = \sin(x)\). Hence, \[ \bar{x} = \int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx = x \sin(x) + \cos(x) \vert_0^{\frac{\pi}{2}} \] which after evaluation gives: \[ \bar{x} = \left[ \frac{\pi}{2} \cdot 1 + 0 - 0 \right] = \frac{\pi}{2} \cdot 0 + 1 - 1 = \frac{\pi}{2} \cdot 0 + 1 = 1 \] Therefore, \(\bar{x} = \frac{2}{\pi}\).

Step 5: Compute \(\bar{y}\)

Using the formula for \(\bar{y}\), calculate the integral: \[ \bar{y} = \frac{1}{2A} \int_{0}^{\frac{\pi}{2}} [\cos(x)]^2 \, dx \] Use the identity \(\cos^2(x) = \frac{1}{2}(1 + \cos(2x))\): \[ \int_0^{\frac{\pi}{2}} \cos^2(x) \, dx = \int_0^{\frac{\pi}{2}} \frac{1}{2}(1 + \cos(2x)) \, dx = \frac{1}{2} \left[ x + \frac{1}{2}\sin(2x) \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left[ \frac{\pi}{2} + 0 \right] = \frac{\pi}{4} \] Thus, \(\bar{y} = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}\).

Step 6: State the Solution

The center of mass for the region \(\mathcal{R}\) is \((\bar{x}, \bar{y}) = \left(\frac{2}{\pi}, \frac{\pi}{8}\right)\).

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in mathematics that allows us to find areas under curves, accumulate values over regions, and solve problems related to rates of change. In the context of finding the center of mass, integrals help us average the position of mass elements over a specific region. This involves computing definite integrals, which calculate the signed area under a curve between two bounds.For the exercise involving the region under the curve defined by the cosine function, integral calculus is used to determine both the area () and the moment of each coordinate. These are essential for finding coordinates (x, y) of the center of mass:- The area is found through a thorough integral involving the function y = cos(x) over the limits from 0 to \(\frac{\pi}{2}\).- The integrals make use of calculus techniques like integration by parts, which is crucial for calculating the specific moments for coordinates.

Mass Density

Mass density (\(\rho\)) is a measure of mass per unit area in situations dealing with two-dimensional regions. In this exercise, however, it is assumed that the mass density is uniform throughout the region . This simplification means that the center of mass can be calculated without needing to consider variations in density. When mass density is uniform, the center of mass can be computed by finding the average position, weighted by area. Because every point in the region has the same level of density, the calculations rely solely on the shape and boundaries of the region.Uniform mass density ensures that the center of mass (x, y) depends only on the geometry of the region rather than any concentration of mass at a particular point. Thus, solving for more complicated mass distributions is avoided, leading to a more straightforward integration process.

Area of a Region

To find the center of mass, one must first calculate the area (A) of the region in question. The area of a region bounded by curves is typically found using integration. In this particular problem, the region is defined between the curve of \(y = cos(x)\) and the x-axis from \(x = 0\) to \(x = \frac{\pi}{2}\).For the function y = cos(x), the integral of it with respect to x over the specified interval determines the area below the cosine curve and above the x-axis:

• The calculated integral of cos(x) from 0 to \(\frac{\pi}{2}\) gives the area of 1.
• This integral represents the total area over the specified domain, which acts as the normalization factor for calculating the average position in defined in center of mass formulas.

Trigonometric Integration

Trigonometric integration involves integrating functions that include trigonometric functions such as cosine, sine, or tangent. These integrals often require specific techniques or substitutions for simplification. For the problem with the region bounded by cos(x), trigonometric identities are employed to solve the integrals required for determining both \(\bar{x}\) and \(\bar{y}\).- To compute \(\bar{y}\), we use the identity \[\cos^2(x) = \frac{1}{2}(1 + \cos(2x))\]. This simplifies the integral into manageable parts that can easily be evaluated over the given limits.- For the integration needed to determine \(\bar{x}\), it involves an integration by parts setup, where one must choose functions for u and dv that make the integral solvable.Trigonometric integration is crucial in translating the area under periodic functions into numeric values needed for the mass center calculation. This process simplifies complex expressions, ensuring precise and efficient calculations.