Problem 30 - Essential Calculus

Question

Use the integration capabilities of a graphing utility to approximate the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(x\) -axis. $$ y=\ln x, \quad y=0, \quad x=1, \quad x=3 $$

Step-by-Step Solution

Answer

The volume of the solid can be approximated by \(\pi \int_1^3 [\ln x]^2 dx\) where the integration is done numerically using a graphing calculator.

Step 1: Understand and Graph the Equations

First, sketch the graphs of \(y = \ln x\), \(y = 0\), \(x = 1\), and \(x = 3\). These will form the boundaries of the region we're interested in. The region is simply the area under the curve \(y = \ln x\) from \(x = 1\) to \(x = 3\), above the x-axis which is represented by \(y = 0\).

Step 2: Apply Disk Method Formula

The general formula for the disk method is \[ V = \pi \int_a^b [R(x)]^2 dx \] where \(R(x)\) is the radius of a typical disc at \(x\) and \(a\) and \(b\) are, the endpoints of the interval.

Step 3: Find the Radius

For the given problem, \(R(x)\) is simply \(y\), so it would be \(R(x) = \ln x\).

Step 4: Substitute Values and Integrate

Putting all of this together, we get \[ V = \pi \int_1^3 [\ln x]^2 dx \] This is a problem for numerical approximation using a graphing utility.

Step 5: Approximate the Integration using Graphing utility

Approximate the integral using a graphing calculator or other mathematical software capable of numerical integration to find the result.

Key Concepts

Disk Method

The disk method is a technique in calculus for finding the volume of a solid of revolution. When you have a region bounded by a curve and revolve it around an axis, the shape of the solid resembles a stack of disks or washers. Each disk is like a slice of the solid, and by adding them up along the interval, you can calculate the entire volume.

In mathematical terms, the volume is calculated using the formula:

• \[ V = \pi \int_a^b [R(x)]^2 dx \]

Where \(R(x)\) is the radius of each disk, and \(a\) and \(b\) are the limits of integration along the axis of revolution.

For the exercise problem, the region described revolves around the \(x\)-axis. Here, \(R(x)\) is determined by the equation \(y = \ln(x)\), so the radius of each disk from the axis to the curve is exactly \(\ln(x)\). This is then squared as per the formula before integration.

Volume of Revolution

When we talk about the volume of revolution, we mean the process of rotating a two-dimensional shape to form a three-dimensional object. Imagine spinning a flat disk around an axis and seeing it become a cylinder. This is essentially what we're dealing with when solving problems like these.

The specific region involved in this problem is bounded by \(y = \ln x\), \(y = 0\), \(x = 1\), and \(x = 3\). By rotating this around the \(x\)-axis, we form a solid whose volume we can determine using the disk method. The method calculates the sum of volumes of infinitesimally thin disks from \(x = 1\) to \(x = 3\).

Solids of revolution are common in practical problems, such as when designing containers or other cylindrical objects, where accurately knowing the volume is critical.

Numerical Approximation

Numerical approximation is crucial when solving integrals that do not have an easy algebraic solution. In this exercise, after setting up the integral \(\pi \int_1^3 [\ln(x)]^2 dx\), we rely on numerical methods to evaluate it. Graphing calculators or computational software can help in calculating the integral through techniques such as Riemann sums or Simpson's rule.

Here’s why numerical approximation matters:

• Some integrals are too complex to solve analytically.
• It allows for practical calculation of definite integrals, especially when exact symbolic antiderivatives aren't available.
• It offers flexible approaches adaptable to computer algorithms.

By using software tools, students and professionals save significant time and effort, providing solutions with sufficient precision for most practical applications. In this problem, using a graphing utility converts the symbolic integral into a practical answer reflecting the volume of the solid specified.