Problem 24 - 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 above by the curve \(y=1 / x^{3},\) below by the curve \(y=-1 / x^{3},\) and on the left and right by the lines \(x=1\) and \(x=a>1 .\) Also, find \(\lim _{a \rightarrow \infty} \overline{x}\) .

Step-by-Step Solution

Answer

The center of mass \( \overline{x} \) approaches 2 as \( a \to \infty \).

Step 1: Determine the boundaries of the region

The region is bounded by four lines or curves:- Above: \( y = \frac{1}{x^3} \)- Below: \( y = -\frac{1}{x^3} \)- Left: \( x = 1 \)- Right: \( x = a \) (where \( a > 1 \))The region forms a symmetrical strip along the x-axis due to the symmetry of the functions above and below it.

Step 2: Set up expressions for area and mass

Since the plate has a constant density \( \delta \), we need to calculate the mass (total area due to constant density). The height of the rectangle at each \( x \) is given by \( 2y = \frac{2}{x^3} \), and its width is \( dx \). This makes the infinitesimal area \( dA = \frac{2}{x^3} dx \). The total area (mass) \( A \) is determined by integrating over the interval \([1, a]\): \[ A = \int_{1}^{a} \frac{2}{x^3} \, dx \]

Step 3: Integrate to find the total area (mass)

Evaluate the integral for the total area:\[ A = \int_{1}^{a} \frac{2}{x^3} \, dx = 2 \left[ \frac{1}{-2x^2} \right]_{1}^{a} \]\[ = 2 \left( \frac{1}{2} - \frac{1}{2a^2} \right) = 1 - \frac{1}{a^2} \]

Step 4: Integrate to find the first moment about y-axis

The first moment about the y-axis (with respect to \( x \)) is given by:\[ M_x = \int_{1}^{a} x \cdot \frac{2}{x^3} \, dx = \int_{1}^{a} \frac{2}{x^2} \, dx\]Evaluate this integral:\[ M_x = 2 \left[ \frac{-1}{x} \right]_{1}^{a} = 2 \left( 1 - \frac{1}{a} \right) = 2 - \frac{2}{a} \]

Step 5: Calculate the center of mass \( \overline{x} \)

The center of mass along the x-axis \( \overline{x} \) is calculated using:\[ \overline{x} = \frac{M_x}{A} = \frac{2 - \frac{2}{a}}{1 - \frac{1}{a^2}} \]

Step 6: Simplify \( \lim_{a \rightarrow \infty} \overline{x} \)

To find \( \lim_{a \rightarrow \infty} \overline{x} \), consider the expressions as \( a \to \infty \):\[ \overline{x} \approx \frac{2 - 0}{1 - 0} = 2 \]Thus, the limit is \( \overline{x} = 2 \) as \( a \) approaches infinity.

Key Concepts

Integral Calculus

Integral calculus is a powerful tool for finding the area under curves and solving problems involving accumulation of quantities. In this exercise, we used integral calculus to determine the center of mass for a symmetrical region defined by specific curves and lines. Let’s break down the essentials:

To start, the given region is bounded by curves and lines: above by the curve \( y = \frac{1}{x^3} \), below by \( y = -\frac{1}{x^3} \), to the left by \( x = 1 \), and to the right by \( x = a \).
This setup creates a symmetrical region along the x-axis, which we explore using integration.

In the process, to calculate the area (and thus the mass of the plate), we set up an integral:

\[A = \int_{1}^{a} \frac{2}{x^3} \, dx\]

This integral computes the total area covered by the strip, which is crucial for determining the mass since the density is constant.
Solving this integral involves identifying the antiderivative of \( \frac{2}{x^3} \), resulting in:

• \( A = 1 - \frac{1}{a^2} \)

This expression tells us how the area changes concerning \( a \). Mastering this component of integral calculus allows us to apply these principles to various practical problems.

Symmetrical Regions

Understanding symmetrical regions is key when finding the center of mass. In this exercise, the region is defined by curves symmetrical about the x-axis: above by \( y = \frac{1}{x^3} \) and below by \( y = -\frac{1}{x^3} \).
This symmetry simplifies our computations and enables straightforward application of integral calculus principles. Let's explore why:

Symmetry implies that certain properties, such as the center of mass, distribute evenly on either side of the axis. Therefore, we only need to find the center of mass along the x-axis, \( \overline{x} \), as the vertical component will cancel out. This reduces the complexity of calculations.

To calculate \( \overline{x} \), we integrate to find the first moment \( M_x \) about the y-axis:

• \( M_x = \int_{1}^{a} x \cdot \frac{2}{x^3} \, dx = 2 - \frac{2}{a} \)

The result reflects how the region's symmetry affects the first moment. It makes it easier to find \( \overline{x} \) using symmetry to balance computations around the axis.

Limits

Limits are crucial in calculus to determine the behavior of functions as inputs approach specific values. In this exercise, we are asked to find the limit of \( \overline{x} \) as the right boundary \( a \) approaches infinity.
This concept helps us understand how the location of the center of mass changes when the limiting process is applied.

In step 6, the expression for \( \overline{x} \) is:

\[\overline{x} = \frac{2 - \frac{2}{a}}{1 - \frac{1}{a^2}}\]

To find \( \lim_{a \rightarrow \infty} \overline{x} \), we consider how both the numerator and the denominator behave when \( a \) becomes very large. With increasing \( a \), both terms \( \frac{2}{a} \) and \( \frac{1}{a^2} \) approach zero. Thus, the limit simplifies to:

• \( \overline{x} \approx \frac{2 - 0}{1 - 0} = 2 \)

This indicates that as the strip grows infinitely long, the center of mass along the x-axis stabilizes to 2.
The limit helps us observe the long-term trend and stability of the function, providing insights into the behavior of symmetrical regions in integral calculus.