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 \(y\) -axis on the interval (-1,1] is revolved about the line \(x=-1.\)

Short answer

Question: Find the volume of the solid generated by revolving the region bounded by \(f(x)=(x+1)^{-\frac{3}{2}}\) and the \(y\)-axis on the interval \((-1,1]\), about the line \(x=-1\). Answer: The volume of the solid does not exist.

Step-by-step solution

Graph the function and determine the region to revolve

Draw the graph of the function \(f(x) = (x+1)^{-\frac{3}{2}}\) and identify the region to be revolved, which is bounded by the function, the interval \((-1,1]\) and the \(y\)-axis. You will notice that the function has a vertical asymptote at \(x = -1\).

Set up the integral using the disk method

Since we are revolving the region around the line \(x=-1\), we will be summing up the volumes of infinitesimally thin disks perpendicular to the x-axis. The radius of each disk is the horizontal distance between the function and the line \(x=-1\). This distance is given by \((x+1)\), on the interval \((-1,1]\). The volume of each disk is \(π \cdot ((x+1))^2 \cdot dy\). To find the total volume, we'll integrate over the interval given: $$V = \pi\int_{-1}^1 (x+1)^2 \cdot dy$$

Express the integral in terms of y

To express the integral in terms of y, we'll need to replace x with an expression in y. Since we have \(y=(x+1)^{-\frac{3}{2}}\), we can rewrite it as: $$\frac{1}{\sqrt[3]{y^2}}=x+1$$ Now, solve for x in terms of y: $$x = \frac{1}{\sqrt[3]{y^2}} - 1$$ Now, replace x with this expression in the integral: $$V = \pi\int_{-1}^1 \left(\frac{1}{\sqrt[3]{y^2}} - 1\right)^2dy$$

Evaluate the integral

Now we can evaluate the integral: $$V = \pi\int_{-1}^1 \left(\frac{1}{\sqrt[3]{y^2}} - 1\right)^2 dy$$ To simplify and evaluate, we'll first expand the integrand: $$V = \pi\int_{-1}^1 \left(\frac{1}{y^{2/3}} - 2 + y^{2/3}\right) dy$$ Now integrate: $$V = \pi\left[\frac{3}{5}(y^{5/3}) - 2y + \frac{3}{5}(y^{5/3})\right]_{-1}^1$$ Finally, evaluate the integral at the limits: $$V = \pi\left[\frac{3}{5}(1^{5/3}) - 2\cdot 1 + \frac{3}{5}(1^{5/3}) - \frac{3}{5}((-1)^{5/3}) + 2\cdot (-1) + \frac{3}{5}((-1)^{5/3})\right]$$ This simplifies to: $$V = \pi\left[\frac{3}{5} - 2 + \frac{3}{5} +\frac{3}{5}-2-\frac{3}{5}\right]$$ $$V = \pi\left[-4 + \frac{6}{5}\right]$$ Therefore, the volume of the solid of revolution is: $$V = \pi\left[\frac{-14}{5}\right]$$ Since the volume is negative, it does not exist.

Key concepts

Disk Method

The disk method is a technique used in integral calculus to find the volume of a solid of revolution. Imagine a solid formed by rotating a region around a line. The disk method considers the volume as a stack of infinitesimally thin disks or washers, perpendicular to the axis of rotation.

To apply this method, one must first decide on the axis of rotation. In this exercise, the rotation is around the line \(x = -1\). Each disk's radius is determined by the distance from the axis of rotation to the outer edge of the region being revolved. For this exercise, this distance is the expression \((x+1)\).

The volume of each disk is given by \(\pi\cdot\text{(radius)}^2\cdot\text{(thickness)}\). When dealing with the x-axis, the thickness corresponds to \(dy\), and when integrating with respect to y, one needs to express x in terms of y. Summing the volumes of these disks across the interval results in a definite integral.

Integral Calculus

Integral calculus is a fundamental tool in mathematics used to compute the area under curves and, by extension, the volume and surface area of solids. In the context of this problem, integral calculus helps find the total volume of a solid of revolution generated by rotating a region about a specific line.

The definite integral is unmistakably important here, as it is used to compute the aggregate of infinite small disk volumes created by the rotation. In this problem, the integral is set up as \(\pi\int_{-1}^1 (x+1)^2\, dy\). Transforming from the spatial relation of x in terms of y allows for evaluating the integral effectively in terms of y.

Integral calculus also requires transforming expressions, as seen in this problem, making sure that the limits and integrands match the variable of integration.

Asymptote

Asymptotes refer to lines that a curve approaches infinitely close but never actually meets. They provide insight into the long-term behavior of a function.

In this exercise, the function \(f(x) = (x+1)^{-3/2}\) has a vertical asymptote at \(x = -1\). This is where the value of the function becomes unbounded, tending towards infinity as x approaches -1.

Understanding asymptotes is crucial when dealing with volumes of solids of revolution, as they may impact whether the volume converges to a finite value or diverges, indicating non-existence of a finite volume. In this instance, the vertical asymptote plays a pivotal role, contributing to why the integral eventually results in indicating that the solid does not have a valid, finite volume.

Definite Integral

The definite integral is central to calculating the volume in problems involving solids of revolution. It represents the accumulation of quantities, which makes it perfect for determining total areas and volumes. When a region in the plane is revolved around a line, the definite integral sums up the volumes of infinitesimal elements, like disks or washers.

For this exercise, the definite integral \(\pi\int_{-1}^1 ")" \left(\frac{1}{\sqrt[3]{y^2}} - 1\right)^2 dy\) is essential. The limits of integration (-1 and 1) define the span over which the volume is calculated. As you evaluate the integral, you actually account for how the function and the specified region behave across the entire interval, summing up all these small contributions to find the total volume.

It's important to pay keen attention to integration limits and ensure calculations address the correct range and variable. When care is not taken, as in cases affecting asymptotes or improper setup, the volume may not exist in a meaningful, finite way.