Problem 272 - Calculus Volume 1

Question

For the following exercises, compute the center of mass \((\overline{x}, \overline{y}) .\) Use symmetry to help locate the center of mass whenever possible. $$ \begin{array}{l}{ \rho=2 \text { for the region bounded by } y=\cos (x)} \\\ {y=-\cos (x), \quad x=-\frac{\pi}{2}, \text { and } x=\frac{\pi}{2}}\end{array} $$

Step-by-Step Solution

Answer

The center of mass is \((0, \frac{\pi}{4})\).

Step 1: Understand the Region

First, note that the region of interest is bound by the curves \(y = \cos(x)\) and \(y = -\cos(x)\), between \(x = -\frac{\pi}{2}\) and \(x = \frac{\pi}{2}\). This region is symmetric about the y-axis because it spans equal distances to the left and right from the y-axis, and the curves are mirror images across the x-axis.

Step 2: Find Area of the Region

The area \(A\) of the region can be found by integrating the top curve minus the bottom curve over the given interval. So,\[ A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos(x) - (-\cos(x)))\, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2\cos(x)\, dx.\]We can simplify to:\[ A = 2 \cdot \left[ \sin(x) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 2 \cdot (\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})) = 2 \cdot (1 - (-1)) = 4. \]

Step 3: Calculate \(\overline{x}\) using Symmetry

Due to symmetry about the y-axis, the center of mass \(\overline{x}\) is 0 without computations because the region is symmetric around this axis.

Step 4: Calculate \(\overline{y}\)

For \(\overline{y}\), we use the formula: \[ \overline{y} = \frac{1}{A} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{\cos^2(x) - (-\cos^2(x))}{2}\right) \cdot 2 \cos(x)\, dx. \]This simplifies using trigonometric identities to:\[ \overline{y} = \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2\cos^2(x) \,dx = \frac{1}{4} \cdot 2 \cdot \frac{\pi}{2}, \text{ as } \frac{\pi}{2} \cdot \cos^2(x) \text{ averages to } \frac{1}{2}.\]Thus, \[ \overline{y} = \frac{\pi}{4}. \]

Step 5: State the Center of Mass

The center of mass coordinates for the region are \((\overline{x}, \overline{y}) = (0, \frac{\pi}{4})\). This result incorporates both the symmetry and the calculations of mass distribution in the y-direction.

Key Concepts

Symmetry in Calculus

Symmetry plays a crucial role in simplifying calculus problems, especially when determining the center of mass. In our exercise, the region is symmetric about the y-axis because the curves are mirror images across the x-axis. This symmetry means that the distribution of the region is identical on both the left and right sides of the y-axis.
Such symmetry allows us to immediately conclude that the x-coordinate of the center of mass, denoted as \(\overline{x}\), must be zero. This is because the total area on either side of the y-axis is equivalent, effectively balancing each other out without requiring further integration to calculate \(\overline{x}\). Utilizing symmetry reduces the complexity of solving problems and minimizes unnecessary computations. Always look for symmetry in calculus as a strategy to make problem-solving more efficient!

Integration Techniques

Integration is an essential technique in calculus for finding areas, volumes, and centers of mass. In this exercise, the integration technique used involves calculating the area of the region bounded by two curves: \(y = \cos(x)\) and \(y = -\cos(x)\).
To find the area, we integrate the difference between the upper and lower curves. This involves a definite integral: \[ A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos(x) - (-\cos(x))) \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2\cos(x) \, dx. \]This integration simplifies to finding the antiderivative of \(2\cos(x)\), which is \(2\sin(x)\), and evaluating it between the specified limits.
Upon evaluation, the calculated area is 4. Understanding integration, including definite integrals, is key in solving such problems where you determine bounded areas.

Trigonometric Functions

Trigonometric functions like \(\cos(x)\) play a pivotal role in calculus problems that involve periodic phenomena or symmetric regions. In our exercise, \(\cos(x)\) and \(-\cos(x)\) form the boundaries of the region of interest. These functions inherently possess properties such as periodicity and symmetry that are extremely useful for problem-solving.
A trigonometric identity applied here is that the integral of \(\cos(x)\) over its full period (from \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\)) averages out, thereby simplifying the computations needed to find areas and centroids. Moreover, using identities like \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\) helps in simplifying the expression for when calculating the center of mass.
Trigonometric functions and their properties are key tools for handling more complex calculus problems, making it easier to perform intricate integrations and solve for specific calculations like center of mass.

Calculus Problem Solving

Effective calculus problem solving involves several steps and strategies. First, understanding the problem and identifying the relevant region or function is crucial. In this exercise, we needed to acknowledge the bounds set by the curves and realize the symmetry present.
After understanding the problem setup, correctly applying mathematical techniques and formulas becomes the focus. These include identifying suitable integration limits and choices, utilizing trigonometric identities, and leveraging symmetry to simplify calculations. Throughout this process, maintaining organizational clarity in stepwise reasoning ensures consistent results.
In this specific problem, arriving at the center of mass was achieved through methodically leveraging symmetry, performing appropriate integrations, and applying trigonometric insights. This structured approach is vital across calculus problem-solving tasks, providing a roadmap to tackle complex mathematical challenges efficiently.