Problem 133 - Calculus Volume 1

Question

For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and y = 0 around the x-axis. \(y=\ln (x), x=1,\) and \(x=e\)

Step-by-Step Solution

Answer

The volume is \( \frac{\pi (e^2 + 1)}{2} \).

Step 1: Identify the Region

We need to find the volume of the solid formed by rotating the region bounded by the curve \( y = \ln(x) \), the line \( y = 0 \), and the vertical lines \( x = 1 \) and \( x = e \) about the x-axis.

Step 2: Determine the Shell's Radius and Height

For the shell method, consider a vertical strip at a distance \( x \) from the y-axis. The height of this strip is \( y = \ln(x) \). The radius of the potential shell is \( y \) because we're rotating around the x-axis.

Step 3: Express the Volume of a Shell Element

The volume \( dV \) of a cylindrical shell is given by the formula \[ dV = 2\pi (\text{radius})(\text{height})(\text{thickness}) = 2\pi (\ln(x))(x)dx \] where \( dx \) is the thickness of the shell.

Step 4: Determine the Limits of Integration

The region to be integrated extends from \( x = 1 \) to \( x = e \). Consequently, our limits of integration for the integral are \( x = 1 \) to \( x = e \).

Step 5: Set up and Evaluate the Integral

The volume \( V \) is the integral of \( 2\pi x \ln(x) \) from \( x = 1 \) to \( x = e \):\[ V = \int_{1}^{e} 2\pi x \ln(x) \, dx \]To solve this integral, use integration by parts, where \( u = \ln(x) \) and \( dv = x \, dx \). ThenLet \( u = \ln(x) \) and \( dv = x \, dx \), which gives \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \). Using integration by parts:\[ \int u \, dv = uv - \int v \, du \]\[ \int \ln(x) \, x \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \]\[ = \frac{x^2}{2} \ln(x) - \frac{1}{2} \int x \, dx \]\[ = \frac{x^2}{2} \ln(x) - \frac{1}{2} \cdot \frac{x^2}{2} + C \]\[ = \frac{x^2}{2} \ln(x) - \frac{x^2}{4} + C \]Evaluate from 1 to \( e \):\[ \left.\frac{x^2}{2} \ln(x) - \frac{x^2}{4} \right|_{1}^{e} \]\[ = \left[ \frac{e^2}{2} \ln(e) - \frac{e^2}{4} \right] - \left[ \frac{1}{2} \ln(1) - \frac{1}{4} \right] \]\[ = \left[ \frac{e^2}{2} \cdot 1 - \frac{e^2}{4} \right] + \frac{1}{4} \]\[ = \left[ \frac{2e^2}{4} - \frac{e^2}{4} \right] + \frac{1}{4} \]\[ = \left[ \frac{e^2}{4} \right] + \frac{1}{4} \]Simplifying gives \( \frac{e^2 + 1}{4} \). So, the volume \( V = 2\pi \cdot \frac{e^2 + 1}{4} = \frac{\pi (e^2 + 1)}{2} \).

Key Concepts

Shell Method

The shell method is a powerful technique used in calculus to find the volume of solids of revolution. When dealing with problems that involve rotating a region about an axis, it can become complex to figure out the resulting volume. Unlike the disk or washer method, the shell method involves visualizing the solid as a collection of cylindrical shells.

• Each shell is formed by considering a small vertical or horizontal slice of the region that will be rotated.
• For our exercise, since we are rotating around the x-axis, the slices are vertical.
• The radius of each shell corresponds to how far the shell is from the axis of rotation.
• The height is determined by how high the function curve reaches at that particular slice.

To compute the volume using the shell method, you multiply the circumference of each shell by the height and thickness, resulting in the expression for the volume element, which is then integrated over the region. This method is particularly useful when the axis of rotation is parallel to the axis along the bounding curves. Once the formula for the differential volume (\[ dV = 2\pi (\text{radius})(\text{height})(\text{thickness}) \]) is set up, the challenge is to evaluate the resulting integral across the given limits.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation and is used when dealing with integrals that involve products of functions. In our volume of revolution problem, it was essential for solving the integral when the function included a natural logarithm.

• The formula for integration by parts is: \( \int u \, dv = uv - \int v \, du \).
• Choosing the right \( u \) and \( dv \) is crucial for simplifying the integral.
• A typical strategy is to let \( u \) be a function that becomes simpler when differentiated (like \( \ln(x) \)), and \( dv \) a function easily integrable (like \( x \, dx \)).

Once \( u \) and \( dv \) are selected, compute \( du \) and \( v \), then substitute into the formula. This approach reduces the complexity of the integrand, making it possible to find the integral solutions. In the context of volume of revolution problems, integration by parts is often a key step in evaluating the definite integral to find the volume.

Natural Logarithm

The natural logarithm, denoted by \( \ln(x) \), is a mathematical function that is the inverse of the exponential function \( e^x \). It is an essential function in calculus and appears frequently in integration problems, particularly involving growth and decay.

• The base of the natural logarithm is the irrational number \( e \), approximately equal to 2.71828.
• The function \( \ln(x) \) increases slowly, which is important when examining its behavior and impacts in calculus problems.
• Key properties include \( \ln(1) = 0 \) and \( \ln(e) = 1 \), which simplify computations and evaluations.

In the context of the shell method exercise, understanding \( \ln(x) \) is crucial for determining the height of the cylindrical shell. Since the height of each shell was determined as \( y = \ln(x) \), the natural logarithm's characteristics directly influenced the setup of the problem and the integration process. Recognizing the properties of \( \ln(x) \) allows us to handle the logarithmic expressions more readily throughout the calculations.