Problem 28 - Calculus Single Variable

Question

In each of Exercises 25-30, use the method of cylindrical shells to calculate the volume \(V\) of the solid that is obtained by rotating the given planar region \(\mathcal{R}\) about the \(y\) -axis. \(\mathcal{R}\) is the region below the graph of \(y=\left(1+x^{2}\right)^{-2}\), \(0 \leq x \leq 1\) and above the \(x\) -axis.

Step-by-Step Solution

Answer

The volume of the solid is \( \pi \).

Step 1: Understand the problem

We are given the function \( y = (1 + x^2)^{-2} \) and need to find the volume of the solid formed by rotating the region \( \mathcal{R} \) under this curve, around the \( y \)-axis for \( 0 \leq x \leq 1 \). The method of cylindrical shells will be used.

Step 2: Formula for volume using cylindrical shells

The formula for the volume of a solid of revolution using cylindrical shells is \[ V = \int_{a}^{b} 2\pi x f(x) \, dx \]. Here, \( f(x) = (1 + x^2)^{-2} \), \( a = 0 \), and \( b = 1 \).

Step 3: Set up the integral

Substitute into the formula: \[ V = \int_{0}^{1} 2\pi x (1 + x^2)^{-2} \, dx \]. This integral represents the volume of the solid of revolution.

Step 4: Solve the integral

To solve \( \int_{0}^{1} 2\pi x (1 + x^2)^{-2} \, dx \), use the substitution \( u = 1 + x^2 \), then \( du = 2x \, dx \). Transform the limits of integration: when \( x = 0 \), \( u = 1 \), and when \( x = 1 \), \( u = 2 \). The integral then becomes \[ V = 2\pi \int_{1}^{2} u^{-2} \, du \].

Step 5: Compute the integral

The antiderivative of \( u^{-2} \) is \( -u^{-1} \). Apply the limits of integration: \[ V = 2\pi \left[ -\frac{1}{u} \right]_{1}^{2} = 2\pi \left( -\frac{1}{2} + \frac{1}{1} \right) = 2\pi \left( \frac{1}{2} \right) = \pi. \]

Step 6: Conclusion

The volume of the solid obtained by rotating the region \( \mathcal{R} \) below \( y = (1 + x^2)^{-2} \) from \( x = 0 \) to \( x = 1 \) about the \( y \)-axis is \( \pi \).

Key Concepts

Volume of Revolution

When we talk about the "Volume of Revolution," we're discussing the volume of a three-dimensional object created by rotating a two-dimensional region about an axis. Imagine taking a flat shape and spinning it around a line. The solid formed by this rotation has volume, and calculus helps us calculate it precisely.

In our exercise, the planar region beneath the curve \( y = (1 + x^2)^{-2} \) from \( x = 0 \) to \( x = 1 \), i.e., region \( \mathcal{R} \), is rotated about the \( y \)-axis. By applying the method of cylindrical shells, we can find this volume of revolution. This technique is quite useful when dealing with shapes like these, as it simplifies the calculation process by breaking down the solid into simpler cylindrical increments.

Some key highlights of using this method include:

• Provides a systematic approach for solving complex volume problems.
• Great for finding volumes of solids when rotating around vertical or horizontal axes.
• Based on the principle of summing up infinitely thin cylinders to approximate the shape of the solid.

Integral Calculus

Integral calculus is the backbone of calculating quantities like areas under curves, accumulated values, and in this context, volumes of solids of revolution. Integration takes a modest function of one variable, analyzes its behavior over an interval, and provides us with valuable information about the "sum" of this behavior. In terms of geometric shapes, it can tell us how much space, or volume, a particular 3D object occupies.

The essential idea when calculating the volume of a solid of revolution is to setup an integral that represents the summation of small segments of volume. For the cylindrical shells method, this formula is given by:

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

This integral equation represents the summing of many small cylindrical "shells" to calculate the total volume. Integral calculus enables us to perform these summations accurately, allowing us to understand and manipulate the results effectively.

Substitution Method

The substitution method is a handy procedure used in integral calculus to simplify the process of integration. When an integral is complex due to its structure, substitution can transform it into a more manageable form.

In the given problem, we need to evaluate \( \int_{0}^{1} 2\pi x (1 + x^2)^{-2} \, dx \). It's quite tedious to compute directly, so we opt for substitution to ease our calculation. By choosing \( u = 1 + x^2 \), and consequently \( du = 2x \, dx \), we simplify our original integral. The process involves changing the variables and adjusting the limits of integration to match the new variable, \( u \).

The whole substitution approach rotates our integral to:

\[ V = 2\pi \int_{1}^{2} u^{-2} \, du \]

This substitution transforms the complex integral into an easier one. Substitution is essential for breaking down complex integrals into solvable pieces, thus making the calculation of volumes, areas, and other integrals feasible and accurate.