Problem 36 - Thomas Calculus

Question

In Exercises \(31-36,\) find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so. The region bounded by \(y=2 x-x^{2}\) and \(y=x\) about a. the \(y\) -axis \(\quad\) b. the line \(x=1\)

Step-by-Step Solution

Answer

Volumes are \( \frac{8\pi}{3} \) for the y-axis and \( \frac{\pi}{3} \) for the line x=1.

Step 1: Identify the Region

First, determine the intersection points of the given functions to find the region. Set \( y = 2x - x^2 \) equal to \( y = x \) and solve the equation: \( 2x - x^2 = x \). Rearrange to obtain \( x^2 - x = 0 \), factor to get \( x(x-1) = 0 \). Thus, the intersection points are \( x = 0 \) and \( x = 1 \). The region is bounded by these points.

Step 2: Revolve Around the Y-axis

To find the volume using the disk method around the \( y \)-axis, we express \( x \) in terms of \( y \). Using \( y = 2x - x^2 \), solve for \( x = \frac{2 \pm \sqrt{4 - 4y}}{2} \). Simplifying gives \( x = 1 \pm \sqrt{1-y} \). To revolve the region between \( y = 0 \) and \( y = 1 \), utilize washers: the outer radius is \( 1 + \sqrt{1-y} \), and the inner radius is \( 1-y \). The volume \( V \) is \[ V = \pi \int_0^1 ((1 + \sqrt{1-y})^2 - (1-y)^2) \, dy \].

Step 3: Compute the Integral for Y-axis

Calculate \( V = \pi \int_0^1 ((1 + \sqrt{1-y})^2 - (1-y)^2) \, dy \). Expanding the squares gives \( (1 + \sqrt{1-y})^2 = 1 + 2\sqrt{1-y} + (1-y) \) and \( (1-y)^2 = 1 - 2y + y^2 \). Simplify and integrate term by term: \( V = \pi \int_0^1 (2\sqrt{1-y} + 2y - y^2) \, dy \). Evaluate the integral using basic calculus techniques.

Step 4: Solve the Integral

The integrated values from the formula are \( \int (2\sqrt{1-y}) \, dy = -\frac{4}{3}(1-y)^{3/2} \), \( \int 2y \, dy = y^2 \), and \( \int y^2 \, dy = \frac{y^3}{3} \). Substituting these, evaluate from 0 to 1 and sum them to get the volume \( V = \frac{8\pi}{3} \).

Step 5: Revolve Around the Line x=1

For parts revolving around \( x = 1 \), use the washers with inner radius \( 1-x \) and outer radius \( 1-(2x-x^2) \). The volume \( V \) becomes \[ V = \pi \int_0^1 ((1-x)^2 - (1-(2x-x^2))^2) \, dx \]. Simplify and solve as before.

Step 6: Compute the Integral for x=1

Simplifying, the expression becomes \( (1-x)^2 - (1-(2x-x^2))^2 \). Evaluate each square and find their integral from 0 to 1, simplification results in \( \pi \int_0^1 (2x-x^2) \, dx \). Integrate term-by-term to get the volume.

Step 7: Solve the Integral

Perform the integration: \( \int 2x \, dx = x^2 \) and \( \int x^2 \, dx = \frac{x^3}{3} \). Substituting in the bounds from 0 to 1 and simplify gives \( V = \frac{\pi}{3} \).

Key Concepts

Disk Method

The disk method is a technique used in calculus to find the volume of a solid of revolution. Imagine meticulously slicing a solid into thin, coin-shaped disks. This method is most effective when the region is revolved around an axis, creating a straightforward, cylindrical shape.
To apply the disk method, begin by determining a "slice" of the solid. You'll visualize each slice's area as a disk with a small thickness. Then, for each disk, calculate the area which is based on the function that outlines the region. Here, the radius corresponds to the distance from the axis of revolution to the edge of the region.
The volume is determined by integrating the area of these slices across the bounds of the shape. The key formula for the disk method when revolving around the x-axis is:

• \[ V = \pi \int_a^b [f(x)]^2 \, dx \] where \( f(x) \) is the function representing the outer edge, \(a\) and \(b\) are the bounds.

The disk method is precise and organized, making it invaluable for straightforward revolution problems.

Washer Method

The washer method is a practical adaptation of the disk method, used when a solid has a hollow portion. This happens when you revolve an area around an axis and there is an empty space inside, like approximating the shape of a doughnut.
This method uses an 'outer' and an 'inner' radius to accommodate the gap in the middle. By subtracting the inner disk's volume from the outer disk, you find the actual volume you're interested in.
For the washer method, set up the integral similar to the disk method but account for the hollow section with a subtraction:

• \[ V = \pi \int_a^b ([f(x)]^2 - [g(x)]^2) \, dx \] where \( f(x) \) is the outer radius and \( g(x) \) is the inner radius.

Visualizing the problem as washers helps to identify both the parts that form the solid's outer surface and those that don't contribute to the solid's volume. It's especially useful when revolving functions where one is inside another, creating ring-shaped volumes.

Integral Calculus

Integral calculus is a branch of mathematical analysis that focuses on accumulation and the area under curves. It's the fundamental tool used to calculate areas, volumes, and their higher-dimensional counterparts.
This discipline revolves around two major ideas: integration and differentiation. While differentiation finds the rate of change or slope, integration is about generally summing up values to find a total.
There's a special integral calculus application in finding the volume of solids of revolution. Using definite integrals, one can calculate the volume by accumulating an infinite number of infinitesimal volumes, whether:

• Calculating the entire solid as one piece by using a single repetitive shape, like disks.
• Account for internal voids, adding complexity by using the washer method.

Integral calculus expresses these problems using integrals that are set up based on the bounds of the region and what that region is becoming when rotated.

Revolving Solids Around an Axis

Revolving a solid around an axis is a powerful concept in geometry and calculus. When you rotate a two-dimensional region about a line (axis), it transforms into a three-dimensional solid.
Depending on the axis you choose, the shape and complexity of the solid can change dramatically. Consider revolving:

• Around the x-axis, where each vertical slice forms a disk and builds up to a cylindrical shape.
• Around the y-axis, especially affecting problems where recalculating for x as a function of y is necessary.

Interestingly, problems can guide you to use the washer method as well when an axis passes through part of the region rather than fully outside it.
This rotational transformation expands flat regions into volumetric figures, allowing comprehensive analysis of their geometry and quantifiable characteristics.