Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider an infinitely long wire coinciding with the positive \(x\)-axis and having mass density \(\delta(x)=\left(1+x^{2}\right)^{-1}\), \(0 \leq x<\infty\). (a) Calculate the total mass of the wire. (b) Show that this wire does not have a center of mass.

Step-by-Step Solution

Answer

Total mass is \( \frac{\pi}{2} \); no center of mass exists as the integral diverges.

Step 1: Set Up the Integral for Total Mass

The total mass of the wire can be calculated by integrating the mass density function \( \delta(x) = \frac{1}{1+x^2} \) over the interval \( [0, \infty) \). Therefore, the total mass \( M \) is given by the integral:\[ M = \int_0^{\infty} \frac{1}{1+x^2} \, dx. \]

Step 2: Solve the Integral for Total Mass

The integral \( \int \frac{1}{1+x^2} \, dx \) is a standard integral that evaluates to the arctangent function, i.e., \( \tan^{-1}(x) + C \). Thus, we can write:\[M = \left[ \tan^{-1}(x) \right]_0^{\infty} = \tan^{-1}(\infty) - \tan^{-1}(0).\]Since \( \tan^{-1}(\infty) = \frac{\pi}{2} \) and \( \tan^{-1}(0) = 0 \), the total mass is:\[ M = \frac{\pi}{2}. \]

Step 3: Formulate the Center of Mass Integral

The center of mass \( \bar{x} \) for the wire can be found using the formula:\[ \bar{x} = \frac{1}{M} \int_0^{\infty} x \delta(x) \, dx = \frac{2}{\pi} \int_0^{\infty} \frac{x}{1+x^2} \, dx. \]

Step 4: Solve the Integral for Center of Mass

To evaluate \( \int_0^{\infty} \frac{x}{1+x^2} \, dx \), let's use the substitution \( u = 1 + x^2 \) such that \( du = 2x \, dx \). This gives:\[ \int_0^{\infty} \frac{x}{1+x^2} \, dx = \frac{1}{2} \int_1^{\infty} \frac{1}{u} \, du. \]The integral \( \int \frac{1}{u} \, du \) results in \( \ln |u| + C \). Therefore:\[ \frac{1}{2} \left[ \ln(u) \right]_1^{\infty} = \frac{1}{2} \left( \ln(\infty) - \ln(1) \right) = \frac{1}{2} \cdot \infty. \]Since \( \ln(\infty) \) approaches infinity, the integral diverges.

Step 5: Conclusion on Center of Mass

Since the integral for calculating \( \bar{x} \) diverges, it indicates that the center of mass for the wire does not exist. Even though the wire has a finite mass, the center of mass is undefined because the contribution of mass times distance cannot converge.

Key Concepts

Integral Calculus

Integral calculus deals with the process of integration, which is essentially the reverse of differentiation. It is used to find areas, volumes, central points, and many other quantities. In the context of this exercise, integral calculus helps us determine quantities like the total mass of an object when given its mass density distribution across a certain domain.

When dealing with mass density, the integration of the density function over a given interval provides the total mass. For the infinitely long wire along the positive x-axis, the density function is \( \delta(x) = \frac{1}{1+x^2} \).

To find the total mass, we integrate from 0 to infinity:

• Set up the integral: \( M = \int_0^{\infty} \frac{1}{1+x^2} \, dx \).
• Calculate the integral: This particular integral equals \( \tan^{-1}(x) \).
• Evaluate the limits: Subtract \( \tan^{-1}(0) \) from \( \tan^{-1}(\infty) \).

The result shows the total mass is finite, specifically \( \frac{\pi}{2} \). This application illustrates how integral calculus is vital in calculating physical quantities over continuous spaces.

Mass Density

Mass density is a measure of the amount of mass per unit length, volume, or area. In this problem, mass density function is crucial for understanding how mass is distributed along the wire.

The given mass density is expressed as:\( \delta(x) = \frac{1}{1+x^2} \).

This means that the mass per unit length decreases as x increases.

• The structure of this function tells us that mass density diminishes with the square of the distance along the wire.



• It is essential to understand that even with an infinitely long wire, a declining density like this can result in a finite total mass.

This is a prime example of how different distributions can drastically affect the total mass even when the size is infinite. Understanding mass density and how it operates is key to solving real-world physics problems.

Center of Mass

The center of mass represents a point where the total mass could be thought to be concentrated. However, for this wire with mass density decreasing along the x-axis, calculating the center of mass becomes complex.

To find the center of mass \( \bar{x} \), you would typically use:

\( \bar{x} = \frac{1}{M} \int_0^{\infty} x \delta(x) \, dx \).

This integration measure assesses how mass is distributed relative to different positions:\

• Setup involves \( \frac{2}{\pi} \int_0^{\infty} \frac{x}{1+x^2} \, dx \).
• A mathematical substitution \( u = 1 + x^2 \) transforms the integral.
• Outcome shows that \( \bar{x} \) does not converge to a real number, indicating that no center exists.

For the described system, the effect of increased positions without increased masses renders the center of mass non-existent. This is an interesting conclusion, showing that tangible masses can exist without a definitive center of mass.

Divergent Integrals

Divergent integrals are those that do not settle at a finite value, often approaching infinity. In this example, solving the integral associated with the center of mass for the infinitely long wire resulted in a divergent value.

When you evaluate the integral for finding \( \bar{x} \), the substitution \( u = 1 + x^2 \) reduces it to:
\( \frac{1}{2} \int_1^{\infty} \frac{1}{u} \, du \),which simplifies to:

\( \frac{1}{2} \left[ \ln(u) \right]_1^{\infty} \).

• \( \ln(\infty) \) implies that the integral does not converge.



• Thus, \( \bar{x} \) can’t be determined because the mass contribution from increasingly farther points continues adding up indefinitely.

This presents an intriguing scenario where, despite finite mass, the center of mass remains undefined. Working with divergent integrals often brings up significant insights into analyzing infinite or boundless systems.