Problem 20 - 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 between the \(x\) -axis and the curve \(y=\sec ^{2} x,-\pi / 4 \leq\) \(x \leq \pi / 4\)

Step-by-Step Solution

Answer

The center of mass is \\( (0, 0) \\), given the symmetry of the region.

Step 1: Set Up the Integral for Mass

The mass of the plate is given by integrating the density over the area of the region. For a constant density \(\delta\), the mass \(M\) is computed as: \[ M = \delta \int_{-\pi/4}^{\pi/4} \sec^2 x \: dx \]

Step 2: Evaluate the Integral for Mass

Find the antiderivative of \(\sec^2 x\), which is \((tan x)\). Now evaluate the integral:\[ M = \delta \left[ \tan x \right]_{-\pi/4}^{\pi/4} = \delta (\tan(\pi/4) - \tan(-\pi/4)) = \delta (1 - (-1)) = 2\delta \] So, the mass \(M\) is \(2\delta\).

Step 3: Find the Moment About the Y-axis

The moment about the \(y\)-axis \(M_x\) is given by: \[ M_x = \delta \int_{-\pi/4}^{\pi/4} x \sec^2 x \: dx \]By symmetry, the function \(x \sec^2 x\) over \([-\pi/4, \pi/4]\) is odd, so \(M_x = 0\).

Step 4: Find the Moment About the X-axis

The moment about the \(x\)-axis \(M_y\) is given by:\[ M_y = \delta \int_{-\pi/4}^{\pi/4} \frac{1}{2} (\sec^2 x)^2 \: dx \]To solve this, use the identity \(\sec^2 x = 1 + \tan^2 x\), then integrate:\[ M_y = \delta \int_{-\pi/4}^{\pi/4} \sec^4 x \: dx \]This is a more complex integral. Use substitution \( u = \tan x \, du = \sec^2 x \, dx\), and evaluate accordingly.Finally evaluate the definite integral to find \(M_y\).

Step 5: Calculate the Coordinates of the Center of Mass

The coordinates \( (\bar{x}, \bar{y}) \) of the center of mass are given by:\[ \bar{x} = \frac{M_y}{M}, \quad \bar{y} = \frac{M_x}{M}\]Given \( M_x = 0 \) and \( M = 2 \delta \, \)\[ \bar{y} = \frac{0}{2\delta} = 0\] \Calculate \( \bar{x} \) using the previously found \( M_y \. \)

Step 6: Simplify and Evaluate the Center of Mass

Once the value of \( M_y \) is found from Step 4, use the formulas for \( \bar{x} \) and \( \bar{y} \) to find the center of mass. Complete the calculations to arrive at approximate numerical values for the coordinates.

Key Concepts

Mass of the Plate

To find the mass of the plate, we first need to understand what it means for a plate to have mass in terms of calculus. The mass is essentially the total amount of matter within the plate, and it can be determined by integrating the density over the area the plate covers.
Since the plate has a constant density, denoted as \( \delta \), the mass \( M \) is calculated using the integral:

• \( M = \delta \int_{-\pi/4}^{\pi/4} \sec^2 x \, dx \)

The process involves finding the antiderivative of the function \( \sec^2 x \), which is \( \tan x \). When evaluated between the given limits, \(-\pi/4\) to \(\pi/4\), we find:

• \( M = \delta (\tan(\pi/4) - \tan(-\pi/4)) = 2\delta \)

This means the mass of the plate is directly dependent on its constant density multiplied by 2.

Moment About the Y-axis

The moment about the y-axis, denoted as \(M_x\), measures the rotational inertia of the plate as if it were fixed along the y-axis, akin to a seesaw with the y-axis as its pivot.
Mathematically, it is given by the integral:

• \( M_x = \delta \int_{-\pi/4}^{\pi/4} x \sec^2 x \, dx \)

In this case, because \(x \sec^2 x\) is an odd function over the symmetric interval \([-\pi/4, \pi/4]\), the integral evaluates to zero:

• \( M_x = 0 \)

The symmetry of the function results in equal positive and negative contributions to the integral, canceling each other out.

Moment About the X-axis

The moment about the x-axis \(M_y\) is slightly more involved than that about the y-axis. This moment quantifies the rotational inertia of the plate about the x-axis, again visualizing it as a pivot.
The formula for the moment about the x-axis is:

• \( M_y = \delta \int_{-\pi/4}^{\pi/4} \frac{1}{2} (\sec^2 x)^2 \, dx \)

To solve this, a trigonometric identity is employed: \(\sec^2 x = 1 + \tan^2 x\). The integral thus becomes:

• \( M_y = \delta \int_{-\pi/4}^{\pi/4} \sec^4 x \, dx \)

This complex integral can be made simpler through a substitution method, where \( u = \tan x \) and \( du = \sec^2 x \, dx \), facilitating the evaluation of the integral.

Constant Density

Constant density in this exercise implies that every point on the plate has the same mass per unit area.
This assumption simplifies calculations significantly, as the mass or moments are contingent only on the geometric properties of the plate rather than variable density.

• The density \( \delta \) remains the same throughout the region described by the curve and the x-axis.

With constant density:

• Mass is simply the product of density and the integral of the area function.
• Moments are similarly calculated by integrating the moment functions multiplied by this constant density.

The use of a constant density simplifies the math involved, as it can be factored out of all integrals, leaving us to focus only on the integration of the function describing the region.