Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(R\) be the region bounded by \(y=x^{2}\) and \(y=x .\) Find the volume of the solid that results when \(R\) is revolved about: (a) the \(x\) -axis; (b) the \(y\) -axis; (c) the line \(y=x\).

Step-by-Step Solution

Answer

(a) \(\frac{2\pi}{15}\), (b) \(\frac{\pi}{6}\), (c) \(\frac{\pi\sqrt{2}}{12}\).

Step 1: Identify Intersection Points

We begin by finding the intersection points of the curves \(y = x^2\) and \(y = x\). Set the equations equal to each other: \(x^2 = x\). This gives us \(x(x - 1) = 0\), which solves to \(x = 0\) and \(x = 1\). The bounds of integration are from \(x = 0\) to \(x = 1\).

Step 2: Volume Around the x-axis

Using the disk method, the volume of revolution about the \(x\)-axis is \( V = \pi \int_{0}^{1} [(x)^2 - (x^2)^2] \, dx\). Simplifying gives \( V = \pi \int_{0}^{1} [x^2 - x^4] \, dx\). Solving the integral gives \( V = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1} = \pi \left(\frac{1}{3} - \frac{1}{5}\right) = \pi \left(\frac{2}{15}\right)\), so the volume is \(\frac{2\pi}{15}\).

Step 3: Volume Around the y-axis

Using the shell method for rotation about the \(y\)-axis, the volume is \( V = 2\pi \int_{0}^{1} x[x - x^2] \, dx\). Simplify inside the integral: \( V = 2\pi \int_{0}^{1} [x^2 - x^3] \, dx\). Solve the integral: \( V = 2\pi \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} = 2\pi \left(\frac{1}{3} - \frac{1}{4}\right) = 2\pi \left(\frac{1}{12}\right)\), so the volume is \(\frac{\pi}{6}\).

Step 4: Volume Around the Line y=x

To find this volume, use the Pappus's centroid theorem, which states \( V = A \cdot 2\pi \cdot d\), where \(A\) is the area of \(R\) and \(d\) is the distance from the centroid. Area of \(R\) is \(\int_{0}^{1} (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{6}\). The centroid is at \((\frac{3}{4}, \frac{3}{4})\) based on symmetry. Distance from \(y = x\) is \(\sqrt{2} \cdot \frac{1}{4}\). Thus, the volume \( V = \frac{1}{6} \cdot 2\pi \cdot \sqrt{2}\cdot \frac{1}{4}\). This simplifies to \(\frac{\pi\sqrt{2}}{12}\).

Key Concepts

Intersection Points

To solve problems involving the volume of solids of revolution, one often begins by identifying the intersection points of the curves. It is crucial because these points define the limits of our integration. In our problem, we are given two curves:

• The parabola, described by the equation:\[ y = x^2 \]
• The line, represented by:\[ y = x \]

To find the points where these curves intersect, set their equations equal to each other: \[ x^2 = x \]Rearranging gives:\[ x(x - 1) = 0 \]Thus, the solutions are:

• \( x = 0 \)
• \( x = 1 \)

These solutions tell us that the curves intersect at \((0, 0)\) and \((1, 1)\), setting our interval from \( x = 0 \) to \( x = 1 \) for further computations.

Disk Method

The disk method is often used when you revolve a region around an axis, creating a series of disks. In this exercise, we'll revolve the region around the \( x \)-axis:

• We use integration to determine the volume:\[ V = \pi \int_{0}^{1} [(x)^2 - (x^2)^2] \, dx \]
• This integral represents a series of infinitesimally thin disks along the axis.

Upon simplifying the integral expression:

• \[ V = \pi \int_{0}^{1} [x^2 - x^4] \, dx \]
• Calculate the integral:\[ V = \pi \left( \frac{x^3}{3} - \frac{x^5}{5} \right)_{0}^{1} \]Solving this gives:\[ V = \frac{2\pi}{15} \]Thus, the volume of the solid when revolved around the \( x \)-axis is \( \frac{2\pi}{15} \).

Shell Method

The shell method is effective for finding the volume of a solid of revolution about a vertical line, like the \( y \)-axis. In this exercise, we use cylindrical shells:

• The formula used here is:\[ V = 2\pi \int_{0}^{1} x(x - x^2) \, dx \]
• The term \( 2\pi x \) represents the circumference of each shell, while the expression in the integral represents its height and thickness.

Simplifying gives:

• \[ V = 2\pi \int_{0}^{1} [x^2 - x^3] \, dx \]
• Upon evaluating:\[ V = 2\pi \left( \frac{x^3}{3} - \frac{x^4}{4} \right)_{0}^{1} \]This simplifies to:\[ V = \frac{\pi}{6} \]Thus, when the region is revolved about the \( y \)-axis, the volume of the solid is \( \frac{\pi}{6} \).

Pappus's Centroid Theorem

Pappus's centroid theorem offers an approach to find the volume of a solid of revolution. It is especially useful when the solid has a simple symmetrical shape. According to this theorem:

• The volume \( V \) generated by rotating a plane figure around an external axis not intersecting the figure is given by:\[ V = A \times 2\pi \times d \]
• Where \( A \) is the area of the figure and \( d \) is the distance from the centroid to the axis of rotation.

For our region \( R \), the area \( A \) is calculated as:\[ A = \int_{0}^{1} (x - x^2) \, dx = \frac{1}{6} \]The centroid is at the point:\( \left( \frac{3}{4}, \frac{3}{4} \right) \)due to symmetry.

• The distance from the line \( y = x \) is:\[ d = \sqrt{2} \times \frac{1}{4} \]

Thus, using the theorem:\[ V = \frac{1}{6} \times 2\pi \times \sqrt{2} \times \frac{1}{4} \]This simplifies to:\[ V = \frac{\pi\sqrt{2}}{12} \]So, the volume of the solid when revolved around the line \( y = x \) is \( \frac{\pi\sqrt{2}}{12} \).