Problem 54 - Calculus Early Transcendentals

Question

Find the volume of the following solids of revolution. Sketch the region in question. The region bounded by \(y^{2}=\ln x, y^{2}=\ln x^{3},\) and \(y=2\) revolved about the \(x\) -axis

Step-by-Step Solution

Answer

Question: Find the volume of the solid formed by revolving the region bounded by the equations \(y^2 = \ln x\), \(y^2 = \ln x^3\), and \(y=2\) about the x-axis. Answer: The volume of the solid is \(V = 3\pi (e^3 - 1)\).

Step 1: Find the intersection points of the given equations

To find the limits of integration, we need to determine where the functions intersect. Let's find the intersection points of \(y^2 = \ln x\) and \(y^2 = \ln x^3\). Set the two equations equal to each other: \(\ln x = \ln x^3\) Simplify and solve for x: \(\ln x = 3\ln x\) \(\ln x / x^3 = 1\) The intersection points occur at \(x=e^0=1\) and \(x=e^3\) These intersection points denote the limits of integration.

Step 2: Express the equations in terms of x

To make it easier to apply the washer method, we'll need to convert the given equations to terms of x. For \(y^2=\ln x\), we have \(y=\sqrt{\ln x}\) and \(y=-\sqrt{\ln x}\). For \(y^2=\ln x^3\), we have \(y=\sqrt{3\ln x}\) and \(y=-\sqrt{3\ln x}\).

Step 3: Set up the integral for the washer method

Revolve the region bounded by the given equations and \(y=2\) about the x-axis. For the washer method, the volume V is given by the following integral: \(V = \pi \int_{x=a}^{x=b} (R(x))^2 - (r(x))^2 dx\) Where R(x) is the outer radius and r(x) is the inner radius. Since we're revolving around the x-axis, the outer radius R(x) will be \(y = 2\) and the inner radius r(x) will be the given equation \(y = \sqrt{\ln x}\). Then, our limits of integration are \(a=1\) and \(b=e^3\). Setting up the integral for the washer method: \(V = \pi \int_{x=1}^{x=e^3} (2^2 - (\sqrt{\ln x})^2) dx\)

Step 4: Evaluate the integral

Now, we need to evaluate the integral: \(V = \pi \int_{x=1}^{x=e^3} (4 - \ln x) dx\) Using antiderivatives, we have: \(V = \pi [(4x - x\ln x - x)|_{x=1}^{x=e^3}]\) \(V = \pi [4e^3 - e^3\ln e^3 - e^3 - (4 - \ln 1 - 1)]\) \(V = \pi [4e^3 - 3e^3 - e^3 - 3]\) \(V = \pi [3e^3 - 3]\) The volume of the solid of revolution is \(V = 3\pi (e^3 - 1)\).

Key Concepts

Solids of Revolution

When we talk about solids of revolution, we're referring to three-dimensional shapes formed by rotating a two-dimensional region around an axis. This concept is crucial in calculus and helps us find volumes of various shapes.

To visualize it, imagine taking a flat, paper-cut profile and swiveling it around a stick. The shape you get is a solid of revolution. In our exercise, the region bounded by given equations is revolved about the x-axis, creating a distinct 3D shape.

Understanding these solids involves identifying the boundaries or curves and the axis of rotation. This axis is where the 2D shape spins around to become 3D. By understanding the shape's boundaries, you can set up methods like the washer method to find the volume.

Washer Method

The washer method is a popular technique in calculus to find the volume of a solid of revolution, especially when the region has a hole in it. Picture it as numerous thin washers (or doughnut slices) stacked together.

Here's how it works: Each washer's volume is determined by subtracting the inner circle's area from the outer circle's area, multiplied by the thickness. In formula form, it's:

• Outer radius, \( R(x) \), and inner radius, \( r(x) \)
• Volume, \( V = \pi \int_{a}^{b} [(R(x))^2 - (r(x))^2] \, dx \)

The integral tallies up the volume of all washers from the start to end of the region. For the practice exercise, the outer radius is \( y = 2 \) and the inner is \( y = \sqrt{\ln x} \). Setting up the problem this way allows easier calculation of the volume.

Understanding this method is crucial for dealing with complex shapes in calculus.

Integration

Integration is a fundamental concept in calculus used to find areas, volumes, and more. Think of it as the process of summing up tiny parts to find the whole, like adding up pixels to create a full picture.

In solids of revolution, integration helps us accumulate the volume of thin slices or washers. We set limits of integration (the start and end points) which denote where we start and stop adding up those pieces.

In our problem, we set the limits from \( x=1 \) to \( x=e^3 \). The integral sums up the volume of all slices between these points. This essential tool takes the approximation of stacking many small pieces and turns it into an exact calculation.

Antiderivatives

An antiderivative is the reverse of a derivative, representing the original function given its rate of change. It's like finding the background story from a summary.

In calculus, we use antiderivatives when integrating, converting complex expressions into simpler ones for evaluation. In our exercise, the antiderivative helps transform the integral into a manageable form. For instance, integrating \( 4 - \ln x \) involves finding functions whose derivatives would give those terms.

The antiderivative of \( 4 \) is \( 4x \), and for \( \ln x \), it's \( x \ln x - x \). This step is integral to solving the volume problem accurately, as it allows us to calculate the definite integral over the given interval.