Problem 88 - Calculus Volume 1

Question

For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the \(y\) -axis. \(y=\frac{1}{x+1}, x=0,\) and \(x=2\)

Step-by-Step Solution

Answer

The volume is \( 4\pi - 2\pi \ln(3) \).

Step 1: Understand the Boundaries

We need to determine the region bounded by the given curves. The functions involved are the curve \( y = \frac{1}{x+1} \) and the vertical lines \( x = 0 \) and \( x = 2 \). This sets the interval for \( x \) from 0 to 2.

Step 2: Sketch the Region

Sketch the curve \( y = \frac{1}{x+1} \) and the lines \( x = 0 \) and \( x = 2 \). The curve \( y = \frac{1}{x+1} \) is a decaying function starting from \( y = 1 \) when \( x = 0 \) and approaches \( y = \frac{1}{3} \) as \( x \) approaches 2. The region is thus bounded on the left by \( x = 0 \), on the right by \( x = 2 \), and below by the curve.

Step 3: Set up the Volume Integral

For rotation around the \( y \)-axis, we'll use the method of cylindrical shells. The formula for the volume using cylindrical shells is\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \]where \( f(x) = \frac{1}{x+1} \), \( a = 0 \), and \( b = 2 \).

Step 4: Calculate the Integral

Evaluate the integral \[V = 2\pi \int_{0}^{2} x \cdot \frac{1}{x+1} \, dx\]which simplifies using substitution: let \( u = x + 1 \), giving \( du = dx \) and \( x = u - 1 \). Then, transform the integral:\[ V = 2\pi \int_{1}^{3} (u-1) \cdot \frac{1}{u} \, du \]\[ V = 2\pi \int_{1}^{3} \left(1 - \frac{1}{u}\right) \, du \]\[ V = 2\pi \left( \left[u - \ln|u|\right]_{1}^{3} \right) \].

Step 5: Evaluate the Antiderivative

Evaluate the antiderivative:\[ V = 2\pi \left( \left[ 3 - \ln(3) \right] - \left[ 1 - \ln(1) \right] \right) \]\[ V = 2\pi (3 - \ln(3) - 1 + 0) \]\[ V = 2\pi (2 - \ln(3)) \]So, the final result is \[ 4\pi - 2\pi \ln(3) \].

Key Concepts

Cylindrical Shells Method

The Cylindrical Shells Method is an essential technique for finding the volume of solids formed by revolving a region around an axis. It's particularly useful when revolving around the y-axis, and the bounded region is defined in terms of x. This method helps us find the volume by visualizing the solid as a collection of cylindrical shells, or thin rings.

To use this method, one calculates each cylindrical shell with a certain radius, height, and thickness. Here's the formula used:

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

Here, \(f(x)\) represents the function defining the upper boundary of our region, and \(x\) represents the distance from the axis of rotation (y-axis). The integral is evaluated over the interval \([a, b]\), where \(x = a\) to \(x = b\) delineates the range for the bounded area. This approach greatly simplifies the process, allowing us to handle complex shapes, especially those that are not easily managed using the disk or washer methods.

Curve Sketching

Curve sketching is a critical step in visualizing and solving problems involving volumes of revolution. By sketching the curves, we gain a better understanding of the region that we will be revolving. For the given exercise, we have the curve \( y = \frac{1}{x+1} \).

To sketch this curve, consider the following points:

• The curve starts from \( y = 1 \) at \( x = 0 \).
• As \( x \to 2 \), the function approaches \( y = \frac{1}{3} \).
• The x-axis and the vertical lines \( x = 0 \) and \( x = 2 \) define the boundaries.

Understanding these characteristics helps identify the shape and limits of the region being considered, ensuring that when we revolve this region around the y-axis, we accurately capture the intended volume.

Integration by Substitution

Integration by Substitution is a technique often used to simplify integrals, especially in cases involving transformations of the form \( u = g(x) \). It is useful when direct integration is complicated or when integrating a function that requires changing variables to a more manageable form.

In our exercise, we begin with the integral involved in the cylindrical shells method: \[ V = 2\pi \int_{0}^{2} x \cdot \frac{1}{x+1} \, dx \]We substitute \( u = x + 1 \), with \( du = dx \) and \( x = u - 1 \). Transforming the integral, we have:

• \( 2\pi \int_{1}^{3} (u-1) \cdot \frac{1}{u} \, du \)

This substitution reduces the complexity, changing the limits from \([0, 2]\) to \([1, 3]\), and makes the integrand much simpler to work with, leading to our next step in finding the volume.

Antiderivatives

Antiderivatives, or indefinite integrals, are functions that reverse the process of differentiation. They play a vital role in solving definite integrals during the computation of volumes. When we integrate the reduced integral using substitution, we aim to find the antiderivative.

From the previous step, our simplified integral becomes:

• \( 2\pi \int_{1}^{3} \left(1 - \frac{1}{u}\right) \, du \)

The antiderivative of \( 1 \) with respect to \( u \) is \( u \), and for \( \frac{1}{u} \) it's \( \ln |u| \). Thus, the computation leads to \( u - \ln |u| \). Evaluating this from 1 to 3, we find:

• \( V = 2\pi \left( [3 - \ln(3)] - [1 - \ln(1)] \right) \)

Finally, \( V = 2\pi (2 - \ln(3)) \), which is the final expression we seek. Antiderivatives make it feasible to compute areas and volumes under or between curves, critical in mathematics and its applications.