Problem 29 - Thomas Calculus

Question

For some regions, both the washer and shell methods work well for the solid generated by revolving the region about the coordinate axes, but this is not always the case. When a region is revolved about the \(y\) -axis, for example, and washers are used, we must integrate with respect to \(y .\) It may not be possible, however, to express the integrand in terms of \(y .\) In such a case, the shell method allows us to integrate with respect to \(x\) instead. Exercises 29 and 30 provide some insight. $$ \begin{array}{l}{\text { Compute the volume of the solid generated by revolving the region }} \\ {\text { bounded by } y=x \text { and } y=x^{2} \text { about each coordinate axis }} \\ {\text { using }}\end{array} $$ $$\text{a. the shell method}. \quad\text{ b. the washer method}.$$

Step-by-Step Solution

Answer

Using shells, the volume for y-axis is \(\frac{\pi}{2}\) and for x-axis \(\frac{5\pi}{6}\). With washers, only x-axis volume \(\frac{2\pi}{15}\) is straightforward.

Step 1: Understand the Problem

We need to find the volume of the solid created when the region bounded by the curves \(y = x\) and \(y = x^2\) is revolved about both the x-axis and y-axis using the shell and washer methods.

Step 2: Shell Method for y-axis

The shell method involves revolving vertical strips about the \(y\)-axis. The radius is \(x\) and the height is \(x - x^2\). The limits of integration are from \(x = 0\) to \(x = 1\) (the points where \(x = x^2\) and \(x = x\) intersect).The volume is given by:\[V = 2\pi \int_{0}^{1} x(x - x^2)\,dx\]Calculate:\[ = 2\pi \left[\frac{x^3}{2} - \frac{x^4}{4}\right]_0^1 = 2\pi \left(\frac{1}{2} - \frac{1}{4}\right) = \frac{\pi}{2}\]

Step 3: Shell Method for x-axis

Since we revolve around the x-axis, use the shell method by considering horizontal shells. The radius is \(y\) and the height is \(\sqrt{y} - y\). The integral runs from \(y = 0\) to \(y = 1\).Calculate:\[V = 2\pi \int_{0}^{1} y(\sqrt{y} - y)\,dy\]Calculation gives:\[= 2\pi \left[\frac{2y^{3/2}}{3} - \frac{y^3}{4}\right]_0^1 = 2\pi \left(\frac{2}{3} - \frac{1}{4}\right) = \frac{5\pi}{6}\]

Step 4: Washer Method for x-axis

Revolve the region about the x-axis using the washer method. The outer radius is \(x\), while the inner radius is \(x^2\). The limits of integration are from \(x = 0\) to \(x = 1\).Calculate the volume:\[V = \pi \int_{0}^{1} [(x)^2 - (x^2)^2] \,dx = \pi \int_{0}^{1} (x^2 - x^4) \,dx\]This results in:\[= \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi \left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}\]

Step 5: Washer Method for y-axis

Use the washer method for revolving around the y-axis. It becomes complex because expressing x in terms of y results in two functions \(x = \sqrt{y}\) and \(x = y\), leading to potential issues with setup.Thus, it's not feasible to use the washer method directly without complex nested functions.

Key Concepts

Shell Method

The shell method is a fantastic tool for finding the volume of solids formed by rotating a region around an axis. It becomes particularly useful when dealing with problems of rotation about the y-axis. Instead of considering sections, like slices in the washer method, we think of cylinders or shells.

For example, when revolving a region about the y-axis, consider a thin vertical strip at a distance 'x' from the y-axis. When this strip is revolved, it forms a cylindrical shell with a radius 'x.' The height of the shell is given by the difference between the two functions describing the region. In the specific case of the curves given by \( y = x \) and \( y = x^2 \), the difference is \( x - x^2 \).

The formula for computing the volume of such a shell is exactly where this technique shines:

• The formula is \( V = 2\pi \ imes \text{radius} \ imes \text{height} \ imes \text{thickness} \).
• When integrated, the volume becomes \( V = 2\pi \int_{a}^{b} x(x - x^2) \,dx \).

The computation using this integral allows for straightforward evaluation from \( x = 0 \) to \( x = 1 \), resulting in a final volume of \( \frac{\pi}{2} \). This method simplifies the problem, especially when one of the axes intersects, as aligning the strips parallel to the axis of rotation helps avoid complex setup.

Washer Method

The washer method is another integration technique for finding the volume of a solid obtained from revolving a region around an axis. However, unlike the shell method, the washer method uses slices or washers.

This method is often employed when revolving around either the x-axis or the y-axis. When revolving around the x-axis, the washer method captures successive cross-sections perpendicular to the axis. Each washer has an outer and inner radius, forming the disc and hole, respectively.

Take the curves \( y = x \) and \( y = x^2 \), revolving around the x-axis. The outer radius is \( y = x \) and the inner radius is \( y = x^2 \). The difference between these two radii gives the area of the washer, which is \( \pi[(x)^2 - (x^2)^2] \). The integral setup for the volume from \( x = 0 \) to \( x = 1 \) is:

• \( V = \pi \int_{0}^{1} (x^2 - x^4) \,dx \)

The integration results in a volume of \( \frac{2\pi}{15} \). However, the washer method's flexibility can sometimes be constrained if expressing the function solely in terms of the variable of integration is challenging, such as when attempting to revolve around the y-axis because it leads to complicated expressions.

Integration Techniques

Integration is a cornerstone technique when solving for the volume of solids of revolution. Both the shell and washer methods rely heavily on precise integration setup to successfully calculate volumes.

Understanding when to use each integration technique is crucial. In the shell method, the integral is set up such that the variable of integration should run perpendicular to the axis of rotation. Conversely, in the washer method, the variable of integration aligns parallel to the axis of rotation. This understanding dictates whether you should integrate with respect to \( x \) or \( y \), depending on how each axis relates to the problem.

Here's a brief guide when considering integration techniques:

• For shell method: The integral setup often involves multiplying by \( 2\pi \), as the circumference plays a role.
• For washer method: It's key to subtract areas within \( \pi \int [(\text{outer radius})^2 - (\text{inner radius})^2] \,dx \).

Techniques of integration also matter. You may need to apply basic polynomials, power rules, or trigonometric identities. Be comfortable with manipulating expressions, simplifying functions, and recognizing when terms cancel out or simplify considerably. Switching between methods and adjusting integrals make the process dynamic and fosters a deeper understanding of geometry and calculus working together.