Question

Sketch the region \(R\) bounded by the graphs of the given equations, and show a typical vertical slice. Then find the volume of the solid generated by revolving \(R\) about the \(x\) -axis. \(y=\frac{1}{x}, x=2, x=4, y=0\)

Short answer

The volume of the solid is \( \frac{\pi}{4} \).

Step-by-step solution

Identify the Region

The region is bounded by the equations \( y = \frac{1}{x} \), \( x = 2 \), \( x = 4 \), and \( y = 0 \). First, sketch the graph of \( y = \frac{1}{x} \), a hyperbola in the first quadrant. This region is between the vertical lines at \( x = 2 \) and \( x = 4 \), above the x-axis (\( y = 0 \)), and below the curve \( y = \frac{1}{x} \).

Describe the Vertical Slice

A typical vertical slice is a rectangle at a position \( x \) between \( x = 2 \) and \( x = 4 \). The height of the slice is \( y = \frac{1}{x} \) and the thickness is an infinitesimally small change in \( x \), denoted as \( dx \).

Set up the Integral for Volume

Using the disk method to revolve around the x-axis, each slice forms a disk with radius \( \frac{1}{x} \). The volume of a typical disk is \( \pi \left(\frac{1}{x}\right)^2 \). Integrating this from \( x = 2 \) to \( x = 4 \) gives the total volume: \[ V = \pi \int_{2}^{4} \left(\frac{1}{x}\right)^2 \, dx \]

Solve the Integral

To find the volume, evaluate the integral: \[ V = \pi \int_{2}^{4} \frac{1}{x^2} \, dx = \pi \left[ -\frac{1}{x} \right]_{2}^{4} \]. Compute the antiderivative: \[ V = \pi \left( -\frac{1}{4} + \frac{1}{2} \right) = \pi \left( \frac{1}{2} - \frac{1}{4} \right) = \pi \left( \frac{1}{4} \right) \].

Final Result

After solving, the volume of the solid is \[ V = \frac{\pi}{4} \].

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with integrals and their properties. It's primarily concerned with the concept of integration, which is essentially the reverse operation of differentiation. In simpler terms, if differentiation enables you to find the rate at which a function changes, integration allows you to determine the total accumulation of quantities. For example, while differentiation could tell you how fast water is flowing into a tank, integration would tell you the total amount of water in the tank over time.

Some of the basic elements involved in integral calculus include:

• Definite integrals: These are used to compute the exact area under a curve within specified limits. For example, the definite integral of a function from a to b would give the total accumulation of the function from x = a to x = b.
• Indefinite integrals: These represent a family of functions and include a constant of integration. They don't have specific boundaries, unlike definite integrals.
• Properties of integrals: Include linearity, the fundamental theorem of calculus, and substitution methods.

In our exercise, we use a definite integral spanning from x = 2 to x = 4, corresponding to the boundaries of our region of interest.

Disk Method

The disk method is a powerful technique used in integral calculus to find the volume of solids of revolution. This method is applicable when a plane region is revolved around a line (in this case, the x-axis). By thinking of the solid as being composed of numerous thin disks stacked along the axis of revolution, we can approximate the volume using these disks.

Here's a simple breakdown:

• Each disk has a small "thickness" represented by an infinitesimal change in the x-value, denoted as \( dx \).
• The radius of a disk is given by the function value at that particular x, which is typically the distance from the axis of revolution to the edge of the region being revolved.
• The volume of a single disk can be calculated using the formula \( \pi r^2 \cdot \text{thickness} \) where \( r \) is the radius of the disk.

In this exercise, the radius is \( \frac{1}{x} \) and hence the volume of a small disk is \( \pi \left( \frac{1}{x} \right)^2 \cdot dx \). Integrate this from \( x = 2 \) to \( x = 4 \) to find the total volume.

Volume of Revolution

The volume of revolution is the volume of a 3D object obtained by rotating a 2D area around an axis. This concept is instrumental in calculating volumes for objects like vases, bowls, or other symmetrical objects. To determine this volume, we often rely on methods such as the disk method or its counterpart, the shell method.

When using the disk method, as in our exercise, the focus is on treating the 2D area as multiple disks stacked along the axis. Each disk contributes a small volume, and their sum gives the total volume of the solid
key aspects to consider:

• Region identification: Clearly sketch and identify the area to be rotated.
• Axis of rotation: Determine whether you are rotating around the x-axis or y-axis.
• Integrating appropriately: Set up the integral based on the area and limits given, matching the axis of rotation.

For our exercise, rotating the region bounded by \( y = \frac{1}{x} \), \( x = 2 \), \( x = 4 \), and the x-axis around the x-axis creates the solid whose volume we are interested in.

Antiderivatives

Antiderivatives, or indefinite integrals, are functions that represent the reverse process of differentiation. If you think of derivatives as expressing a rate of change, then antiderivatives give you the function whose derivative yields that rate. For instance, if the derivative of a function \( F(x) \) is \( f(x) \), then \( F(x) \) is the antiderivative of \( f(x) \).

Key points about antiderivatives include:

• Indefinite nature: They don't have defined limits as they represent a family of functions adding a constant of integration \( + C \).
• Connection to definite integrals: The antiderivative aids in computing definite integrals through the fundamental theorem of calculus.
• Common functions: Knowing antiderivatives for basic functions (like power functions, exponentials, etc.) is crucial for solving integral problems.

In the context of finding the volume in this exercise, we calculated the antiderivative of \( \frac{1}{x^2} \) to solve the integral \( \pi \int_{2}^{4} \frac{1}{x^2} \, dx \). Successfully evaluating this antiderivative at the bounds gives us the precise volume of the solid.