Problem 27 - Thomas Calculus

Question

The region bounded by the curves \(y=\pm 4 / \sqrt{x}\) and the lines \(x=1\) and \(x=4\) is revolved about the \(y\) -axis to generate a solid. a. Find the volume of the solid. b. Find the center of mass of a thin plate covering the region if the plate's density at the point \((x, y)\) is \(\delta(x)=1 / x\) . c. Sketch the plate and show the center of mass in your sketch.

Step-by-Step Solution

Answer

a. Volume is \( \frac{112\pi}{3} \). b. Center of mass is at \( (2, 0) \). c. Sketch shows symmetry around the x-axis.

Step 1: Set up the volume integral

To find the volume formed by revolving the region around the y-axis, we use the method of cylindrical shells. The shell method formula for volume is: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \cdot dx \] Since our curves are described by \( y = \pm \frac{4}{\sqrt{x}} \), the region's height is \( \frac{8}{\sqrt{x}} \). Thus, the integral setup becomes: \[ V = 2\pi \int_{1}^{4} x \cdot \frac{8}{\sqrt{x}} \cdot dx \].

Step 2: Simplify and evaluate the integral for volume

The integral \( 2\pi \int_{1}^{4} x \cdot \frac{8}{\sqrt{x}} \cdot dx \) simplifies to: \[ V = 16\pi \int_{1}^{4} \sqrt{x} \cdot dx \]Convert \( \sqrt{x} \) to its exponential form, \( x^{1/2} \), so the integral becomes: \[ V = 16\pi \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{4} \]Evaluating this we find: \[ V = 16\pi \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{112\pi}{3} \].

Step 3: Determine the expression for center of mass

The center of mass of the plate requires finding both \( \bar{x} \) and \( \bar{y} \). For \( \bar{x} \), we use: \[ \bar{x} = \frac{1}{M} \int_{1}^{4} x \cdot \frac{8}{\sqrt{x}} \cdot \frac{1}{x} \cdot dx \] Simplified, this becomes: \[ \bar{x} = \frac{1}{M} \int_{1}^{4} \frac{8}{\sqrt{x}} \cdot dx \] For \( \bar{y} \), we use symmetry since the region above and below the y-axis is the same, and thus \( \bar{y} = 0 \).

Step 4: Calculate total mass and \( \bar{x} \)

The mass \( M \) is given by the integral \[ M = \int_{1}^{4} \frac{8}{\sqrt{x}} \cdot \frac{1}{x} \cdot dx \]Simplify to: \[ M = 8 \int_{1}^{4} x^{-3/2} dx \] Evaluating, \[ M = 8 \left[ -2x^{-1/2} \right]_{1}^{4} = 8( -1/2 + 2/2 ) = 4 \]. Now calculate \( \bar{x} \): \[ \bar{x} = \frac{1}{4} \int_{1}^{4} 8x^{-1/2} dx = 4 \times [2x^{1/2}]_{1}^{4} = 4 \times 2 = 8 \]. Thus, \( \bar{x} = 2 \) and \( \bar{y} = 0 \).

Step 5: Sketch the plate and mark center of mass

Sketch the region bounded by the curves \( y = \pm \frac{4}{\sqrt{x}} \) and the lines \( x=1 \) and \( x=4 \). The center of mass is at point \( (2, 0) \). The plate is symmetrical about the x-axis and spans from \( x = 1 \) to \( x = 4 \), curving upwards and downwards to form the outer shell around the y-axis.

Key Concepts

Volume of Solids

The volume of a solid can be determined using several calculus techniques, depending on how the solid is generated. In particular, when a region is revolved around an axis, we often use the method of cylindrical shells to compute the volume. This method is beneficial for solids of revolution, especially when revolving around the y-axis.

To calculate the volume using the cylindrical shells method, we consider thin vertical slices of our 2D region and revolve them around the desired axis. Each slice essentially forms a cylindrical shell. The volume of these shells is accumulated through integration over a specified range, which was from 1 to 4 in our exercise.

In mathematical terms, the volume can be expressed as:

• \( V = 2\pi \int_{a}^{b} x \, f(x) \, dx \)

The integral takes into account the height of the slice (determined by the function), the radius (distance from the y-axis), and the thickness (infinitesimally small, represented by \( dx \)). The formula provides a powerful yet simple way of calculating even complex volumes when dealing with symmetrical regions.

Center of Mass

The center of mass of an object represents the average position of all the mass in a body. For a symmetric object like a plate, the center of mass can often be found at a point that equally balances the mass in all directions.

In our problem, the center of mass of the thin plate is calculated by using the density function, often represented as \( \delta(x) \). For regions defined by continuous functions, this calculation often involves two quantities, \( \bar{x} \) and \( \bar{y} \), which denote the average x and y positions of the mass.

The expression for the x-coordinate of the center of mass is:

• \( \bar{x} = \frac{1}{M} \int_{a}^{b} x \, \delta(x) \, f(x) \, dx \)

The total mass \( M \) is the integral of the density function over the region, which gives us a way to distribute the object's mass evenly.

Given our exercise, the symmetry of the region about the y-axis results in a zero average \( y \)-coordinate for the center of mass. This equates to \( \bar{y} = 0 \), providing a simplified focus on just the x-coordinate for balance.

Cylindrical Shells Method

The cylindrical shells method is a technique used in calculus to find the volumes of solids of revolution. It is particularly useful when revolving a region about an axisthat is parallel rather than perpendicular to the axis of the function. This scenario occurs in our problem, as we revolve around the y-axis.

The idea is to decompose the volume into many thin concentric cylindrical shells. The volume of each shell is based on three factors:

• The radius, which is the horizontal distance from the y-axis to the shell, typically represented by \( x \)
• The height of the shell, determined by the function \( f(x) \)
• The thickness, which is an infinitesimally small\( dx \)

This method integrates all these shells over a given interval to accumulate the total volume.

The formula we use is:

• \( V = 2\pi \int_{a}^{b} x \, f(x) \, dx \)

Each term represents a key aspect: the \( 2\pi \) accounts for the circular path of the shell, and the integration sums it all up over the region specified.

Density Function

The density function in this context provides a way to evaluate the distribution of mass across a plate or solid. When density varies with position, as in our exercise where density \( \delta(x) = 1/x \), it affects both the mass calculation and the center of mass.

The density influences the integral used to find the total mass \( M \) over a given region. Our problem involves:

• The mass function \( M = \int_{a}^{b} f(x) \, \delta(x) \, dx \)

This accounts for the variable density across the plate rather than assuming uniform mass distribution.

Understanding and applying the density function is fundamental for accurately identifying the center of mass, particularly when analyzing objects with changing density. This allows for a more realistic modeling of real-world objects, reflecting how mass is concentrated around specific parts depending on the density function.