Problem 54 - Calculus Early Transcendentals

Question

Find the volume of the solid of revolution. Sketch the region in question. The region bounded by \(y=(\ln x) / \sqrt{x}, y=0,\) and \(x=2\) revolved about the \(x\) -axis

Step-by-Step Solution

Answer

Question: Determine the volume of the solid of revolution formed when the region bounded by \(y= (\ln x) / \sqrt{x}\), \(y = 0\), and \(x = 2\) is revolved around the x-axis using the washer method. Answer: The volume of the solid of revolution is approximately \(2.343\pi\).

Step 1: Sketch the region and identify the shape to be revolved.

To sketch the region bounded by \(y= (\ln x) / \sqrt{x}\), \(y = 0\), and \(x = 2\), first graph the curve \(y = (\ln x) / \sqrt{x}\) and the line \(y = 0\). Next, draw a vertical line at \(x = 2\) to complete the region. When we revolve this region around the \(x\)-axis, we get a shape with an outer radius given by the curve \(y = (\ln x) / \sqrt{x}\) and an inner radius given by the line \(y = 0\).

Step 2: Set up the integral for the washer method.

The washer method formula is given by: \(V = \pi \int_a^b (R(x)^2 - r(x)^2) \, dx\) In this case, our outer radius function \(R(x)\) is given by the curve \(y = (\ln x) / \sqrt{x}\). The inner radius function \(r(x)\) is the horizontal line \(y = 0\). The limits of integration are the x-values that bound our region, which are \(x = 1\) and \(x = 2\). \(V = \pi \int_1^2 \left((\ln x) / \sqrt{x}\right)^2 - (0)^2 \, dx\)

Step 3: Simplify the integral expression.

Simplify the integral expression by squaring \((\ln x) / \sqrt{x}\) and finding the difference between the squares: \(V = \pi\int_1^2 (\ln x)^2 / x \, dx\)

Step 4: Evaluate the integral.

Now evaluate the integral using integration techniques: \(V = \pi\left.\left[\frac{2\ln{x}(\ln{x}-1)x}{2}\right]\right|_1^2\)

Step 5: Calculate the volume of the solid of revolution.

Finally, substitute the limits of integration into the expression to find the volume of the solid of revolution: \(V = \pi\left[\frac{2\ln{2}(\ln{2}-1)2}{2} - \frac{2\ln{1}(\ln{1}-1)1}{2}\right]\) \(V = \pi\left[2\ln{2}(\ln{2}-1)\right]\) \(V \approx 2.343\pi\) So the volume of the solid of revolution is approximately \(2.343\pi\).

Key Concepts

Washer Method

The Washer Method is a technique used to find the volume of a solid of revolution. The solid is created by revolving a region around an axis. Here, we focus on revolving around the x-axis. This method involves using washers, which are essentially disks with holes in the middle. This is different from the Disk Method, which uses only solid disks.

To apply the Washer Method, you need to know both the outer and inner radii for the washers. The formula to calculate the volume is as follows:

• V = \( \pi \int_a^b \left( R(x)^2 - r(x)^2 \right) \, dx \)

In this case:

• The outer radius \( R(x) \) is \( (\ln x)/\sqrt{x} \).
• The inner radius \( r(x) \) is \( 0 \).

You apply the limits of integration that correspond to the x-values that bound the region being revolved – here these limits are from \( x = 1 \) to \( x = 2 \).

This setup produces the integral \( \pi \int_1^2 \left( (\ln x)/\sqrt{x} \right)^2 \, dx \), which requires proper integration to evaluate the volume.

Integration Techniques

Integration techniques are essential tools for solving integrals, especially when dealing with complex expressions like the one encountered in the Washer Method. The specific integral we need to evaluate is \( \pi \int_1^2 (\ln x)^2/x \, dx \). Here are some strategies:

• Substitution Method: Often used when you can simplify integrals by substituting part of the integral with a new variable.
• Integration by Parts: This method follows the formula \( \int u \, dv = uv - \int v \, du \). It's particularly useful for integrals involving the product of functions, such as polynomials and transcendental functions like logs and exponentials.

For our integral, integration by parts is appropriate, but correctly choosing \( u \) (typically \( \ln x \)) and \( dv \) (\( 1/x \)) is crucial. Solving the integration yields the evaluated integral expression, leading to further simplification to find the final volume. The result of the integration and substitution of limits helps find the accurate volume value.

Sketching Regions

Sketching the region is the first and crucial step for solving problems involving solids of revolution. It helps to visualize the area being revolved.

Here is a simple approach to sketching:

• First, graph the curve defined by \( y = (\ln x)/\sqrt{x} \), where you might need to use plotting points or a graphing calculator.
• Then, draw the horizontal line \( y = 0 \) (the x-axis), since it's part of the boundary.
• Third, draw a vertical line at \( x = 2 \), which bounds the region on one side.

After sketching, you will have a visual of the region bounded by these curves and lines. This region appears to be a slice of an area under the curve, above the x-axis, extending from \( x = 1 \) to \( x = 2 \). When revolved, it forms a hollow solid, with the x-axis as the axis of revolution. Accurately sketching the region helps in setting up the integral correctly for the Washer Method and ensures the problem is approached with a clear understanding of the geometric context.