Problem 19 - Thomas Calculus

Question

In Exercises \(13-24\) , find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region bounded by the \(x\) -axis and the curve \(y=\cos x\) \(-\pi / 2 \leq x \leq \pi / 2\)

Step-by-Step Solution

Answer

The center of mass is \((0, \frac{\pi}{8})\).

Step 1: Define the region and its boundaries

The region is bounded by the curve \(y = \cos x\), the \(x\)-axis, and vertical lines \(x = -\pi/2\) and \(x = \pi/2\). This is a symmetrical region around the \(y\)-axis from \(-\pi/2\) to \(\pi/2\).

Step 2: Set up formula for area A

The area \(A\) of the region is calculated using the integral of the curve from \(-\pi/2\) to \(\pi/2\). Since the density \(\delta\) is constant, it can be factored out. Calculate \(A\) as follows:\[ A = \int_{-\pi/2}^{\pi/2} \cos x \; dx \]

Step 3: Calculate the area A

Solve the integral to find the area:\[ A = \left[ \sin x \right]_{-\pi/2}^{\pi/2} = \sin(\pi/2) - \sin(-\pi/2) = 1 - (-1) = 2 \]

Step 4: Set up formula for finding the center of mass \(\bar{x}\)

The \(x\)-coordinate of the center of mass \(\bar{x}\) is found using symmetry and integration. Due to symmetry about the \(y\)-axis, \(\bar{x} = 0\).To verify, the calculation is:\[ \bar{x} = \frac{1}{A} \int_{-\pi/2}^{\pi/2} x \cos x \; dx \]

Step 5: Calculate \\(\bar{x}\\)

Calculate the integral:\[ \int_{-\pi/2}^{\pi/2} x \cos x \; dx \]By symmetry, this integral simplifies to zero since the region on the left and right of the \(y\)-axis cancels each other out, thus \(\bar{x} = 0\).

Step 6: Set up formula for finding the center of mass \(\bar{y}\)

The \(y\)-coordinate of the center of mass \(\bar{y}\) is calculated with:\[ \bar{y} = \frac{1}{A} \int_{-\pi/2}^{\pi/2} \frac{1}{2}(\cos x)^2 \; dx \]

Step 7: Calculate \\(\bar{y}\\)

Calculate the integral:\[ \int_{-\pi/2}^{\pi/2} \frac{1}{2} (\cos x)^2 \; dx = \frac{1}{2} \int_{-\pi/2}^{\pi/2} (\cos x)^2 \; dx \]Use the identity \((\cos x)^2 = \frac{1 + \cos(2x)}{2}\) to evaluate the integral:\[ \frac{1}{2} \left( \frac{x}{2} + \frac{1}{4} \sin(2x) \right) \bigg|_{-\pi/2}^{\pi/2} = \frac{1}{4}(\pi) = \frac{\pi}{4} \]Thus, \(\bar{y} = \frac{\pi}{4} \cdot \frac{1}{2} = \frac{\pi}{8}\).

Step 8: Assemble the center of mass

Based on calculations, the center of mass for the region is \((\bar{x}, \bar{y}) = (0, \frac{\pi}{8})\).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focusing on the concept of integration, which can be thought of as the reverse operation of differentiation. Integration is primarily used to compute areas under curves, among other things.
When finding the center of mass of a region defined by a curve, we use integral calculus to compute necessary quantities such as area and moment. The area under the curve gives us information about the size of the region while the moment informs us about how the mass is distributed.
For instance, to find the area of a region bounded by the curve \(y = \cos x\), the \(x\)-axis, and vertical lines at \(x = -\pi/2\) and \(x = \pi/2\), we calculate the integral of \(\cos x\) over this interval. This calculation is:

• \(\int_{-\pi/2}^{\pi/2} \cos x \, dx\)
• Evaluating the integral gives an area of 2.

This area is a fundamental part of the center of mass calculation.

Symmetry in Calculus

Symmetry plays an important role in simplifying calculations involving geometry and calculus. When a region is symmetrical about an axis, it can provide valuable information about the centroids and moments, often simplifying complex integrations.
In our exercise, the region bounded by the function \(y = \cos x\) from \(-\pi/2\) to \(\pi/2\) is symmetrical about the \(y\)-axis. This symmetry means that any calculus calculation concerning the center of mass can make use of these properties.
Due to symmetry, if the shape is balanced around the \(y\)-axis, the \(x\)-coordinate of the center of mass \(\bar{x}\) is zero. This is because the moments on either side of the \(y\)-axis cancel out, showing that the mass is evenly distributed.

Area under a Curve

The area under a curve is crucial in understanding the total size or extent of a specific region bounded by the curve. This is equivalent to the definite integral of a function between two points.
For a function \(y = \cos x\) from \(-\pi/2\) to \(\pi/2\), the process involves integrating \(\cos x\) over this interval. By evaluating the integral \(\int_{-\pi/2}^{\pi/2} \cos x \, dx\), we determine that the area is 2.
This resultant area gives us one component needed to calculate the center of mass. It's a crucial number because it sets a reference scale for other integral-related calculations like moments and densities.

Density

Density in a problem involving center of mass helps to understand how mass is distributed across the surface. In these exercises, it is often given as constant, simplifying the equations.
Given a constant density \(\delta\), any variable changes in shape or size do not affect the density value itself. Because the density is constant across the region bounded by \(y = \cos x\), calculations for the center of mass don't require adjustments for variable mass distribution.
Knowing that the density is uniform allows us to factor \(\delta\) out of integrals when calculating the center of mass. This makes expressions more manageable and highlights the geometric properties of the region.