Problem 39 - Calculus and its applications

Question

Find the volume generated by rotating about the \(x\) -axis the area bounded by the graphs of each set of equations and the \(x\) -axis. $$ y=\sqrt{\ln x}, x=e, x=e^{3} $$

Step-by-Step Solution

Answer

The volume is \(2\pi e^3\).

Step 1: Understand the Given Functions and Boundaries

The given function is \( y = \sqrt{\ln x} \). The data includes that the region is bounded by the lines \( x = e \) and \( x = e^{3} \) along with the \( x \)-axis. These set the limits for our integration to find the volume.

Step 2: Set Up the Integral for Disk Method

For finding the volume of a solid of rotation about the \( x \)-axis, we use the disk method formula: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] Here, \( f(x) = \sqrt{\ln x} \), and the limits of integration are \( x = e \) to \( x = e^3 \). Thus, the integral becomes:\[ V = \pi \int_{e}^{e^3} (\sqrt{\ln x})^2 \, dx \]

Step 3: Simplify the Integral Expression

Notice that \((\sqrt{\ln x})^2 = \ln x\), so the integral simplifies to:\[ V = \pi \int_{e}^{e^3} \ln x \, dx \] We will now proceed to compute this integral.

Step 4: Integrate the Function

Using integration by parts, where \( u = \ln x \) and \( dv = dx \). Thus, \( du = \frac{1}{x} dx \) and \( v = x \). Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:\[ \int \ln x \, dx = x \ln x - x + C \]We apply this from \( e \) to \( e^3 \).

Step 5: Evaluate the Definite Integral

Plug the limits of integration into the expression:\[ \left[x \ln x - x \right]_{e}^{e^3} = \left[e^3 \ln e^3 - e^3 \right] - \left[e \ln e - e \right] \] This simplifies further as:\[ = \left[3e^3 - e^3\right] - \left[e - e\right] \]\[= 2e^3 \]Therefore, the volume is:\[ V = \pi (2e^3) \]

Step 6: Finalize the Volume Calculation

Calculate \( \pi (2e^3) \) to find the volume of the solid:\[ V = 2\pi e^3 \] This represents the volume generated by rotating the region about the \( x \)-axis.

Key Concepts

Disk Method

The Disk Method is a powerful technique for finding the volume of a solid generated by rotating a region around an axis, typically the x-axis or y-axis. The method involves slicing the solid perpendicular to the axis of rotation to create thin circular disks. Each disk has a thickness \( dx \) and a radius \( f(x) \), the value of the function describing the boundary of the solid.

To use this method, you first need to understand and define the region you’re rotating. You identify the boundaries and write out the definite integral of the area of the disks created. For a region bounded above by the function \( y = f(x) \) and rotated around the x-axis, the volume \( V \) is determined using:

• \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
• The limits \( a \) and \( b \) are the points where you slice the region, which corresponds to the given x-boundaries.

This method is quite visual, helping you to see the solid as a series of small, stacked disks.

Integration by Parts

Integration by parts is a key technique in calculus used to integrate products of functions. In the context of finding the volume of revolution, it is especially useful when dealing with logarithmic and polynomial functions. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)
• Here, you choose \( u \) and \( dv \) from your integrand.

When you need to integrate functions like \( \ln x \) concern yourself with selecting \( u = \ln x \) since its derivative \( \frac{1}{x} \) is straightforward. The assignment \( dv = dx \) simplifies the other part of the integral. As seen in the problem solution, using these variables effectively simplifies the integration process, helping you find \( V = \pi \) integral quickly.

Definite Integral

The definite integral gives you the total accumulation of quantities under a curve, between two specific limits. This is at the heart of calculating volumes of solids of revolution and other applications demanding an exact numeric result.

When using the definite integral for finding volumes, as with the Disk Method, it’s key to remember it provides the means to calculate and evaluate the entire area under the curve \( f(x) \) between the limits \( a \) and \( b \):

• \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]
• Here, \( F(x) \) is the antiderivative of \( f(x) \).

The evaluation from \( a \) to \( b \) ensures that the integral you compute gives the volume for the specific portion of the function over which the solid exists. It’s important to always double-check your limits, as they precisely define the segment of the function contributing to the total volume.