Question

Find the volume of the solid generated by revolving the region bounded by the curve \(y=\ln x\), the \(x\) -axis, and the vertical line \(x=e^{2}\) about the \(x\) -axis. (Express the answer in exact form.)

Short answer

The volume of the solid is \( 2\pi e^2 \).

Step-by-step solution

Set up the integral

The region is bounded by the curve \( y = \ln x \), the \( x \)-axis, and the vertical line \( x = e^2 \). When revolved around the \( x \)-axis, the method of disks (or washers) can be used. The formula for the volume \( V \) of the solid of revolution is \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \). Here, \( f(x) = \ln x \), \( a = 1 \), and \( b = e^2 \). Thus, the integral becomes \( \pi \int_{1}^{e^2} (\ln x)^2 \, dx \).

Rewrite the integral

We have \( V = \pi \int_{1}^{e^2} (\ln x)^2 \, dx \). Using a change of variables or integration by parts is needed since \( (\ln x)^2 \) is not a basic integral form. To solve this, use integration by parts with \( u = (\ln x)^2 \) and \( dv = dx \).

Apply integration by parts

Let \( u = (\ln x)^2 \) and \( dv = dx \). Then, \( du = 2 \ln x \, \frac{1}{x} \, dx = \frac{2 \ln x}{x} \, dx \) and \( v = x \). Apply integration by parts \( \int u \, dv = uv - \int v \, du \). Therefore, \( \int (\ln x)^2 \, dx = x (\ln x)^2 - \int x \frac{2 \ln x}{x} \, dx = x (\ln x)^2 - 2 \int \ln x \, dx \).

Integrate \( \int \ln x \, dx \)

To integrate \( \int \ln x \, dx \), apply integration by parts again. Let \( u = \ln x \) and \( dv = dx \). Thus, \( du = \frac{1}{x} \, dx \) and \( v = x \). Hence, \( \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C \).

Combine parts and solve the definite integral

Substitute back into the volume integral, \[ \int (\ln x)^2 \, dx = x (\ln x)^2 - 2(x \ln x - x) \]. Evaluate this from \( x = 1 \) to \( x = e^2 \):\[ \left[ x (\ln x)^2 - 2(x \ln x - x) \right]_1^{e^2} = \pi \left[ e^2 (2)^2 - 2(e^2 \cdot 2 - e^2) - (0 - 0) \right] \].

Calculate the final result

Compute the expression: \[ \pi \left[ e^2 \times 4 - 2(2e^2 - e^2) \right] \].Simplify it to: \[ \pi (4e^2 - 2e^2) = \pi \times 2e^2 \]. The exact volume is \( 2\pi e^2 \).

Key concepts

Method of Disks

The Method of Disks is used to find the volume of a solid of revolution. In this scenario, we revolve a region around an axis and generate disks perpendicularly. Each disk forms a small cylindrical segment of the solid.
To use the method of disks, we need a function that represents the radius of these disks. When revolving around the x-axis, the radius of each disk is determined by the function's value, denoted as \( f(x) \).
The formula used in the method of disks to find the volume \( V \) of the solid is:

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

In our specific problem, the curve is \( y = \ln x \), which becomes the radius of each disk. The region is bounded by \( x = 1 \) and \( x = e^2 \), so we set up our integral accordingly:
\( V = \pi \int_{1}^{e^2} (\ln x)^2 \, dx \).
This integral allows us to compute the volume of the solid formed by revolving this curve around the x-axis.

Integration by Parts

Integration by Parts is a technique for integrating products of functions by transforming the integral of a product into a simpler form.
The rule states:

• \( \int u \, dv = uv - \int v \, du \)

In this problem, we encounter \( \int (\ln x)^2 \, dx \), which is not directly integrable. Thus, we select parts as follows:

• \( u = (\ln x)^2 \), \( dv = dx \)
• \( du = \frac{2 \ln x}{x} \, dx \), \( v = x \)

Applying integration by parts gives:
\( x(\ln x)^2 - 2 \int \ln x \, dx \).
To solve \( \int \ln x \, dx \), we apply integration by parts again:

• \( u = \ln x \), \( dv = dx \)
• \( du = \frac{1}{x} \, dx \), \( v = x \)

This results in:
\( x \ln x - x \), allowing us to complete the integral for \( (\ln x)^2 \).

Definite Integral

A Definite Integral computes the accumulation of quantities over an interval \([a, b]\) and results in a real number.
We use definite integrals to calculate the volume in this context by evaluating between two bounds:\( x = 1 \) and \( x = e^2 \).
After applying integration by parts, the modified integral is:

• \( V = \int x(\ln x)^2 - 2(x \ln x - x) \, dx \)

Compute this from \( x = 1 \) to \( x = e^2 \), which involves using the resulting expressions and substitution into the integral's limits.
By evaluating:

• \[ \left[ x(\ln x)^2 - 2(x \ln x - x) \right]_1^{e^2} \]

We find the final volume to be \( 2\pi e^2 \), representing the total space filled by the solid formed through this revolution.