Problem 29 - Thomas Calculus

Question

The region in the first quadrant bounded by the curve \(x=y-y^{3}\) and the \(y\)-axis about a. the \(x\) -axis b. the line \(y=1\)

Step-by-Step Solution

Answer

a) Volume: \( \frac{3\pi}{5} \). b) Volume: \( \frac{4\pi}{7} \).

Step 1: Identify the Curve and Region

The given curve is \( x = y - y^3 \). We need to analyze the region formed by this curve in the first quadrant, meaning \( y \geq 0 \) and \( x \geq 0 \). Let's find and verify the points where this curve intersects the \( y \)-axis. Since the \( y \)-axis is where \( x = 0 \), setting \( x = 0 \) in the equation gives \( 0 = y - y^3 \), which simplifies to \( y(1 - y^2) = 0 \). This yields intersection points at \( y = 0 \) and \( y = 1 \). Therefore, the region is bounded by the line \( x=0 \), the curve \( x = y - y^3 \), and extends from \( y = 0 \) to \( y = 1 \).

Step 2: Set Up the Integral for Rotation About the x-axis

To find the volume of the solid formed by rotating the region around the \( x \)-axis, we apply the disk method or washer method. With the disk method, if the shape revolves around the \( x \)-axis, the radius of a disk is the distance from the \( x \)-axis to the curve \( y = \sqrt[3]{x} \), expressed as \( y(x) = \sqrt[3]{x} \). Thus, the volume \( V \) is given by the integral:\[V = \pi \int_{y=0}^{y=1} (y)^2 \left|\frac{dx}{dy}\right| \, dy = \pi \int_{0}^{1} y^2 (3y^2) \, dy = \pi \int_{0}^{1} 3y^4 \, dy\]Calculate this integral.

Step 3: Evaluate the Integral for Rotation About the x-axis

Expanding the integral, we have:\[V = \pi \int_{0}^{1} 3y^4 \, dy = 3\pi \left[ \frac{y^5}{5} \right]_{0}^{1} = 3\pi \left( \frac{1^5}{5} - \frac{0^5}{5} \right)\]This simplifies to:\[V = \frac{3\pi}{5}\]Thus, the volume of the solid formed by rotating the region about the \( x \)-axis is \( \frac{3\pi}{5} \).

Step 4: Set Up the Integral for Rotation About the Line y=1

For rotation about the line \( y = 1 \), we use the washer method. Consider each horizontal slice of the area. The outer radius is from \( y = 1 \) to \( y = 0 \) and the inner radius is \( y = 1 - y \) from the line \( y \) to \( y = 1 \). So, the volume \( V \) is given by:\[V = \pi \int_{0}^{1} ((1)^2 - (1-y)^2) \left|\frac{dx}{dy}\right| \, dy = \pi \int_{0}^{1} (1 - (1 - 2y + y^2))3y^2 \, dy = \pi \int_{0}^{1} (2y - y^2)3y^4 \, dy\]

Step 5: Evaluate the Integral for Rotation About the Line y=1

Expanding the integral, we have:\[V = \pi \int_{0}^{1} (2y - y^2)3y^4 \, dy = 3\pi \int_{0}^{1} (2y^5 - y^6) \, dy\]Evaluate:\[= 3\pi \left[ \frac{2y^6}{6} - \frac{y^7}{7} \right]_{0}^{1} \]\[= 3\pi \left( \frac{2}{6} - \frac{1}{7} \right) = 3\pi \left( \frac{1}{3} - \frac{1}{7} \right)\]\[= 3\pi \left( \frac{4}{21} \right) \]\[= \frac{12\pi}{21} = \frac{4\pi}{7}\]Thus, the volume of the solid formed by rotating the region about \( y = 1 \) is \( \frac{4\pi}{7} \).

Key Concepts

Disk Method

The Disk Method is a way to find the volume of a solid of revolution, specifically when revolving a region around a horizontal axis, such as the x-axis. In this exercise, we use the disk method to determine the volume when the region bounded by the curve \( y = \sqrt[3]{x} \) is rotated around the x-axis. Imagine slicing the solid perpendicular to the axis of rotation. Each slice is a disk with a radius equal to the distance from the x-axis to the curve, which in this case is simply \( y \). The volume of each disk is the area of the circle (\( \pi r^2 \)) multiplied by the disk's thickness (a small change in y, written as \( dy \)). To find the total volume, we integrate across the specified interval (from \( y = 0 \) to \( y = 1 \)):

• Set up the integral of \( V = \pi \int_{y=0}^{y=1} y^2 \left| \frac{dx}{dy} \right| \, dy \).
• This equation takes the radius squared, accounts for the thickness, and sums this all over the interval.

Evaluating this integral simplifies to \( V = \frac{3\pi}{5} \), expressing the volume of the solid formed.

Washer Method

The Washer Method expands upon the disk method by allowing for a hollow region. This is perfect for cases where the solid is formed by rotating around a line that is not the boundary of the shape, like \( y = 1 \) in this example.Here, we'll have washers instead of disks because we have both an outer and inner radius. The outer radius reaches from \( y = 1 \) to the curve, while the inner radius is the space left near the line \( y = 1 \). Each washer's volume is found by

• calculating the area of the outer circle \( \pi \times (1)^2 \)
• subtracting the area of the inner circle \( \pi \times (1 - y)^2 \)
• including the thickness of the washer, \( dy \).

The integration process removes these slices and calculates their total volume as the washers rotate around the line \( y = 1 \). Integrating from \( y = 0 \) to \( y = 1 \), the complete expression becomes \[ V = \pi \int_{0}^{1} ((1)^2 - (1-y)^2) \cdot 3y^4 \, dy \].Solving yields \( V = \frac{4\pi}{7} \), giving the solid's volume.

First Quadrant

In mathematics, especially in coordinate geometry, the first quadrant refers to the section of the coordinate plane where both the x and y values are positive. This is key for problems where regions are described by their boundaries on the plane, as it impacts which limits of integration to use.For the curve \( x = y - y^3 \) in the first quadrant:

• The x-axis and y-axis form two boundaries.
• We set up our bounds and integrals looking at \( y \) values stretching from 0 to 1.

Understanding the first quadrant context ensures that any solid revolving around the x or a parallel axis does not delve into any negative x or y territories, simplifying the calculation of both disk and washer methods in such scenarios.