Question

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=(x+1)^{-3 / 2}\) and the \(x\) -axis on the interval (-1,1] is revolved about the line \(y=-1\).

Short answer

Question: Determine whether it is possible to create a solid of revolution with the given conditions: the region bounded by the curve \(f(x) = (x+1)^{-\frac{3}{2}}\), the x-axis, and the interval \(-1 \leq x \leq 1\), revolved around the line \(y=-1\). If possible, find the volume of the solid. Answer: The volume does not exist under the given conditions. The assumption that a solid of revolution can be created from the given conditions is invalid.

Step-by-step solution

Sketch the region and identify areas of interest

Before setting up the integral for the problem, we should sketch the graph of \(f(x)\), the interval, and the line of revolution. This will help us visualize the solid and understand how the disks form as we revolve the region.

Determine the radius of the solid

Since the curve is being revolved about the line \(y=-1\), we need to determine the radius of the solid. If we pick a point on \(f(x)\) within the given interval, say \(P(x, f(x))\), then the distance from \(P\) to the line \(y=-1\) would be the radius \(r(x)\). Since \(f(x)\) is the distance of point \(P\) from the \(x\)-axis, and we know the distance from the \(x\)-axis to \(y=-1\) is 1, then the radius is \(r(x) = 1-f(x) = 1-(x+1)^{-\frac{3}{2}}\).

Setup the integral

To find the volume of the solid, we will integrate the cross-sectional area of the disks along the interval. The area of each disk is given by \(A(x) = \pi r(x)^2\), where \(r(x)\) is the radius. So, the volume will be given by: \(V = \int_{-1}^1 A(x) \, dx = \int_{-1}^1 \pi \left(1 - (x+1)^{-\frac{3}{2}}\right)^2 dx\).

Evaluate the integral

Now, we will evaluate the integral: \(V = \pi \int_{-1}^1 \left(1 - 2(x+1)^{-\frac{3}{2}} + (x+1)^{-3}\right) dx\) We will split this integral into three parts: \(V = \pi \left[\int_{-1}^1 1 dx - 2\int_{-1}^1 (x+1)^{-\frac{3}{2}} dx + \int_{-1}^1 (x+1)^{-3} dx\right]\) Now, we will evaluate each part separately: 1. \(\int_{-1}^1 1 dx = x\big|_{-1}^1 = 1 - (-1) = 2\) 2. \(\int_{-1}^1 (x+1)^{-\frac{3}{2}} dx = 2(x+1)^{-\frac{1}{2}}\big|_{-1}^1 = 2(1) - 2(0) = 2\) 3. \(\int_{-1}^1 (x+1)^{-3} dx = -\frac{1}{2}(x+1)^{-2}\big|_{-1}^1 = -\frac{1}{2}(1) + \frac{1}{2}(0)= -\frac{1}{2}\) Now, substitute the evaluated parts back into the equation for volume: \(V = \pi \left[2 - 2(2) + (-\frac{1}{2})\right] = \pi \left[2 - 4 -\frac{1}{2}\right] =\pi(-\frac{5}{2})\) However, volume cannot be negative, so the question's assumption that a solid of revolution can be created from the given conditions is invalid. Hence, the volume does not exist.

Key concepts

Volume Calculation

In the context of solids of revolution, volume calculation is a central concept. The volume of a solid formed by revolving a region around a line is found by integrating the area of cross-sections perpendicular to the axis of revolution. Each cross-section is typically a disk, and the sum of the volumes of these infinitesimally small disks gives the volume of the entire solid. In mathematical terms, this is expressed as:

• V = π∫[a, b] r(x)^2 dx

where r(x) is the radius of the solid at a given point x along the interval [a,b]. Thus, calculating the volume involves two steps: determining the radius function and integrating it over the specified interval. It's crucial to understand that if the result of this integration does not produce a non-negative volume, then the solid doesn't exist under the conditions given.

Integral Setup

Setting up the integral properly is key to solving volume problems involving solids of revolution. The integral represents the sum of the areas of numerous disks or washers, stacked along the axis of revolution. To set up the integral, you start by identifying the radius of each disk or washer, then express the cross-sectional area as a function of this radius. In this exercise:

• The cross-sectional area A(x) is π r(x)^2.
• The integral setup becomes V = π∫[a, b] (1 - (x+1)^{-3/2})^2 dx.

This setup allows you to compute the disk areas over the entire interval, providing the total volume when evaluated correctly. A correct integral setup reflects a deep understanding of how areas change across the interval.

Radius of Solid

The radius of a solid when dealing with solids of revolution is crucial for determining the cross-sectional area. When a region is revolved around a line not coinciding with one of its boundaries, the radius is the distance from the curve to that line of revolution. In the current problem:

• The line of revolution is y = -1.
• The curve described by f(x) = (x+1)^{-3/2} is being revolved around this line.
• The radius function is r(x) = 1 - f(x) = 1 - (x+1)^{-3/2}.

This radius is crucial as it determines the size of each disk in the volumetric calculation. Understanding how to find this radius becomes more intuitive with practice, involving clear visualization of the region and the axis of revolution.

Cross-Sectional Area

To find the volume of a solid of revolution, understanding the cross-sectional area is vital. Each cross-section perpendicular to the axis of revolution is typically a disk. The area of such a disk when the body is revolved around a horizontal or vertical axis is:

• A(x) = π r(x)^2.

For this exercise, with the line of revolution at y = -1 and the curve f(x), the cross-sectional area is determined by the radius:

• r(x) = 1 - (x+1)^{-3/2}.
• Thus, A(x) = π (1 - (x+1)^{-3/2})^2.

The cross-sectional area function A(x) gives us the coverage of each disk, essential for volume calculation via integration. Fully grasping this concept ensures accurate and elegant solutions to volume problems, and it plays a huge role in setting the stage for integration in solving similar problems.