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Chapter 6: Problem 18

In Exercises \(18,\) 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 \(\lim _{a \rightarrow \infty} \overline{x}\)

Short Answer

Expert verified
The center of mass is \((2, 0)\). As \(a\) approaches infinity, \(\overline{x}\) approaches 2.

Step by step solution

01

Define the Region R

The region R is defined as bounded above by the curve \(y = \frac{1}{x^3}\), below by \(y = -\frac{1}{x^3}\), and vertically by the lines \(x = 1\) and \(x = a\). This region is symmetrical about the x-axis.
02

Determine the Limits of Integration

Since the region is symmetrical around the x-axis and vertically bounded between \(x = 1\) and \(x = a\), we will focus on the upper half of R and multiply by 2 later. The limits of integration for \(x\) are from \(1\) to \(a\), and for \(y\) it's from \(0\) to \(\frac{1}{x^3}\).
03

Expression for Total Mass M

The mass \(M\) of the region is given by the integral of the density function across the region. With constant density \(\delta\), the total mass \(M\) is:\[M = \delta \int_{1}^{a} \int_{0}^{\frac{1}{x^3}} dy \cdot dx\]The inner integral with respect to \(y\) gives \(\delta \frac{1}{x^3}\).
04

Evaluate Total Mass M

Computing the integral for mass:\[M = \delta \int_{1}^{a} \frac{1}{x^3} dx = \delta \left[-\frac{1}{2x^2}\right]_{1}^{a} = \delta \left(\frac{1}{2} - \frac{1}{2a^2}\right)\]
05

Find Center of Mass \(\overline{x}\)

To find \(\overline{x}\), use:\[\overline{x} = \frac{1}{M} \cdot \delta \int_{1}^{a} \int_{0}^{\frac{1}{x^3}} x \cdot dy \cdot dx\]Computing the inner integral:\[\int_{0}^{\frac{1}{x^3}} x \cdot dy = x \cdot \frac{1}{x^3} = \frac{1}{x^2}\]Thus,\[\overline{x} = \frac{1}{M} \cdot \delta \int_{1}^{a} \frac{1}{x^2} \, dx = \frac{1}{M} \cdot \delta \left[-\frac{1}{x}\right]_{1}^{a} = \frac{1}{M} \cdot \delta \left(1 - \frac{1}{a}\right)\]
06

Substitute \(M\) and Simplify \(\overline{x}\)

Substituting the expression of \(M\):\[\overline{x} = \frac{\delta \left(1 - \frac{1}{a}\right)}{\delta \left(\frac{1}{2} - \frac{1}{2a^2}\right)} = \frac{2a(1 - \frac{1}{a})}{1 - \frac{1}{a^2}}\]Simplifying further:\[\overline{x} = \frac{2a - 2}{1 - \frac{1}{a^2}} = 2\frac{a - 1}{1 - \frac{1}{a^2}}\]
07

Calculate \(\lim _{a \rightarrow \infty} \overline{x}\)

Evaluate the limit:\[\lim_{a \to \infty} \overline{x} = \lim_{a \to \infty} 2\frac{a - 1}{1 - \frac{1}{a^2}}\]As \(a\) approaches infinity, \(-\frac{1}{a}\) and \(-\frac{1}{a^2}\) approach zero, so it becomes:\[\lim_{a \to \infty} \overline{x} = 2\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental concept in calculus used to determine the total accumulation of quantities, such as areas under curves, in a continuous interval. When solving for the center of mass of a region, integration helps compute quantities like total mass and density distributions over a specified area. In this exercise, we integrated with respect to two variables, first in terms of the function of y and then x, to find differentiated mass components, allowing us to calculate overall properties of the region. This process involves taking the integral of the density times the volume element, which is dy times dx, over the specified region denoted by the limits of integration.
Mass Distribution
Mass distribution is a key element in understanding centers of mass. It refers to how mass is spread out over a region. In the given problem, the region's mass distribution is symmetrical about the x-axis. When the density is constant, the center of mass calculation relies on integrating the volume element across the region. The symmetrical nature of the plate contributes to simplifying the calculations. To find the mass, we consider the area between the curves, using the value of the constant density and the integration limits. This approach finds the total effect of all mass elements distributed evenly across the region.
Limits of Integration
Limits of integration define the boundaries over which functions are integrated. In this problem, they help specify the region R where the center of mass is calculated. We consider limits along both x and y. The x-limits are given from 1 to a, while the y limits vary from 0 to 1/x^3, capturing the height change due to the curve y = 1/x^3 above the x-axis. This approach helps break down the integration into manageable components. Identifying these limits is crucial since they determine the specific portion of the graph and the corresponding mass contributing to the total mass and center of mass calculations.
Calculus
Calculus provides the tools to solve for the center of mass using continuous change principles. Specifically, it allows us to compute integrals that capture the mass, size, and behavior of particular regions using variables that constantly change. By using calculus, we apply differentiation and integration to achieve a deeper understanding of how the mass is distributed across a complex shape. It enables us to determine quantities such as the total mass of a region and specific positional attributes like the center of mass, defined as an average coordinate where the entire mass of the body can be considered concentrated. It ensures that we handle the values across the defined region with accuracy, especially as a approaches infinity in this problem, leading to insightful results like determining limits.

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Most popular questions from this chapter

In Exercises \(23-26,\) use the shell method to find the volumes of the solids generated by revolving the regions bounded by the given curves about the given lines. \(y=x^{3}, \quad y=8, \quad x=0\) $$ \begin{array}{ll}{\text { a. The } y \text { -axis }} & {\text { b. The line } x=3} \\ {\text { c. The line } x=-2} & {\text { d. The } x \text { -axis }} \\\ {\text { e. The line } y=8} & {\text { f. The line } y=-1}\end{array} $$

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