Problem 34 - 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 in the first quadrant bounded by \(x=y-y^{3}, x=1\) and \(y=1\) about $$\begin{array}{ll}{\text { a. the } x \text { -axis }} & {\text { b. the } y \text { -axis }} \\ {\text { c. the line } x=1} & {\text { d. the line } y=1}\end{array}$$

Step-by-Step Solution

Answer

Volumes are: a) \(\pi/2\); b) \(2\pi/21\); c) \(5\pi/42\); d) 0.

Step 1: Define the region and visualise the solid

The given region is bounded by the curves \(x = y - y^3\), \(x = 1\), and \(y = 1\) in the first quadrant. To find the volume of the solid generated by revolving this region about different axes, visualize how this region stretches and forms shapes when revolved. The region is a bounded area inside first quadrant where these curves meet.

Step 2: Setup for Part a - Revolving about the x-axis

When revolving around the x-axis, consider vertical slices (parallel to the y-axis). These slices will form washers where the outer radius is determined by \(y = 1\) and the inner radius by \(y = \sqrt[3]{1-x}\). Because \(x = y - y^3\) simplifies to \(y = \sqrt[3]{1-x}\), the limits from 0 to 1 are chosen from the x-axis for calculations.The washer volume formula is:\[ V = \pi \int_{0}^{1} [R(x)^2 - r(x)^2] \, dx \]where \(R(x) = 1\) and \(r(x) = (1-x)^{1/3}\).

Step 3: Solve for Part a

Substitute \(R(x)\) and \(r(x)\) into the volume integral:\[ V = \pi \int_{0}^{1} [1^2 - (1-x)^{2/3}] \, dx \]Evaluating this, simplify the integration and calculate to find the volume.The integral turns into:\[ V = \pi \left([x]^{1}_{0} - \int_{0}^{1} (1-x)^{2/3} \, dx \right) \]This calculates the volume to be\( V = \frac{1}{2} \pi \).

Step 4: Setup for Part b - Revolving about the y-axis

Here, use horizontal slices (parallel to the x-axis) to form washers. These washers will have an outer radius determined by \(x=1\) and an inner radius defined by \(x = y - y^3\). Limits of integration are from \(y=0\) to \(y=1\).The formula becomes:\[ V = \pi \int_{0}^{1} [(1)^2 - (y - y^3)^2] \, dy \]where \(1\) is the outer radius and \(y - y^3\) is the inner radius.

Step 5: Solve for Part b

Calculate the volume using the integral setup:\[ V = \pi \int_{0}^{1} [1 - (y - y^3)^2] \, dy \]Simplify the integral and evaluate:\[ V = \pi (1 - \int_{0}^{1} (y^2 - 2y^4 + y^6) \, dy) \]Perform the integration:\( V = \pi \left([y]^{1}_{0} - \frac{1}{3} [y^3]^{1}_{0} + \frac{1}{5} [y^5]^{1}_{0} - \frac{1}{7} [y^7]^{1}_{0}\right) \).Calculate this to find volume equals \( \frac{2}{21} \pi \).

Step 6: Setup for Part c - Revolving about the line x=1

To rotate around the line \(x=1\), segmentate using horizontal washers similar to part b. The outer radius becomes \(y - y^3\) from line \(x=1\), and there's no inner circle since we just rotate to form a cylindrical solid out to \(x=1\).Integral Setup:\[ V = \pi \int_{0}^{1} ((1-y^3 - y) - 0)^2 \, dy \] which simplifies to a clear radius through original \(x\) stays the same with the formulation driving a focus across \(h\).

Step 7: Solve for Part c

Perform the integral calculations:\[ V = \pi \int_{0}^{1} [y^3-y]^2 \, dy \]Break apart the \((1-y^3-y)^2\) terms:\[ V = \pi \int_{0}^{1} [y^6-2y^4+y^2] \, dy \]Address the definite integral:\( V = \frac{5}{42} \pi \).

Step 8: Setup for Part d - Revolving about the line y=1

For rotation about \(y=1\), the component measuring the revolution needs calculation on difference topside between \((1-y^3)-y\) and full curve tracking. Use horizontal slices and washers between line and curve:Outer Radius: \((1)\), Inner: \(\left|0 - (\sqrt[3]{x})\right|\).Calculate:\[ V = \pi \int_{0}^{1} [(1)-(0 - y)] dy = 0\] (as it creates none form behind solid along line axis)

Step 9: Solve for Part d

Complete the value calculations for intensity, as full circle lacks cross glue to yield continued pie:\( V=0 \).This cylinder at this angle produced quiet rounded zero.

Key Concepts

Volume of Solids of Revolution

When we talk about the volume of solids of revolution in calculus, we refer to shapes created by rotating a plane region around a specific axis. Imagine taking a flat region, like a piece of paper, and spinning it around a line. The resulting shape is what we call a solid of revolution.

In problems of this nature, the axis around which we revolve the shape is crucial. Depending on whether it's the x-axis, y-axis, or another line, the volume calculation can differ significantly. This concept is widely used because it enables us to find the volume of intricate shapes which are otherwise hard to measure using conventional methods.

Think of common objects like a vase or a doughnut. By defining the generating curve and the axis of rotation, we can calculate the precise volume of these objects. This is an example of how calculus helps translate geometric intuition into mathematical formulae through integration.

Definite Integrals

Definite integrals are key tools in calculus for finding areas under curves or volumes of shapes. They "sum up" an infinite number of infinitesimally small pieces to find the total. When we say "definite," we mean the integral has specific start and end points on the x or y-axis. These bounds define the region we're interested in.

For instance, if you want to calculate the volume of a shape formed by rotating a curve from x = 0 to x = 1, the definite integral helps in this process. It takes each small slice of the curve as it is revolved by the axis and adds them together to find the full volume.

The process involves setting up an integral with the appropriate bounds, integrating the function representing the shape, and evaluating it at the specified limits. This is the key mathematical operation for determining areas, volumes, and even in more advanced topics like finding accumulated quantities over time.

Washer Method

The washer method is a popular technique in calculus for calculating the volume of solids of revolution. It's especially useful when the solid has a hole or is not uniform all the way through, much like a doughnut or a washer.

The idea is to envision the shape as being made of several "washers"—thin disk-like slices stacked along the axis of rotation. Each washer has two radii: an outer-radius R and an inner-radius r. As you stack these washers from the start to end of the definite integral, you fill up the entire volume.

The volume of each individual washer is given by the formula:

• \(V = \pi \int_a^b [R(x)^2 - r(x)^2] \, dx\)

This formula accounts for subtracting the smaller inner circle from the larger outer circle for each washer. By integrating this expression from point a to b, we get the total volume of the solid.

Using this method enables precise calculations involving complex shapes that are not uniform throughout, making it a critical tool for many practical engineering and physics applications.

Integration Techniques

Integration can sometimes be challenging, but with the right techniques, it becomes manageable. In the context of finding volumes, various integration approaches may be applied depending on the functions involved.

Firstly, substitution is a common technique. If the integral seems difficult at first glance, substituting variables could simplify it, making integration straightforward. For example, substituting a term like \((1-x)^{1/3}\) with a simple variable can greatly ease the process.

Secondly, integration by parts is another strategy used primarily when the product of functions appears in the integral. While not always necessary in washer method calculations, it's handy elsewhere in calculus.

Also, recognizing standard integral forms is crucial. Function shapes such as polynomials or trigonometric forms often have straightforward rules that can be applied directly once identified.

Overall, choosing the right technique depends on the function and the context provided by the problem, and sometimes a combination of methods will be the best approach to tackle complex integrals efficiently.