Question

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=\left(x^{2}-1\right)^{-1 / 4}\) and the \(x\) -axis on the interval (1,2] is revolved about the \(y\) -axis.

Short answer

The volume of the solid is \(\pi\sqrt{3}\) cubic units.

Step-by-step solution

Understand the problem

We are given a function \(f(x) = (x^2-1)^{-\frac{1}{4}}\) and we need to find the volume of the solid formed by revolving the region bounded by this function and the x-axis on the interval (1, 2] around the y-axis.

Setting up the integral

To determine the volume of the solid, we'll use the disk method. For the disk method, the volume of each disk is given by the formula: \(V = \pi \int_a^b [f(x)]^2 dx\). We'll need to find the integral of the square of the given function from 1 to 2 and multiply the result by \(\pi\).

Squaring the function

Consider the given function, \(f(x) = (x^2-1)^{-\frac{1}{4}}\). We want to square it: \[f(x)^2 = [(x^2-1)^{-\frac{1}{4}}]^2 = (x^2-1)^{-\frac{1}{2}}.\]

Integrate the squared function

Now, we should integrate the squared function on the interval (1, 2]: \[V = \pi \int_1^2 (x^2-1)^{-\frac{1}{2}} dx.\]

Substitution

To solve the integral, we will use substitution. Let \(u = x^2 - 1\), therefore, \(du = 2x dx\). Then, we have: \[\frac{1}{2}\pi \int_{u(1)}^{u(2)} u^{-\frac{1}{2}} du.\]

Update the limits

For the new integral, we need to update the limits. When \(x = 1\), \(u = (1^2 - 1) = 0\). When \(x = 2\), \(u = (2^2 - 1) = 3\). Now, we have: \[\frac{1}{2}\pi \int_0^3 u^{-\frac{1}{2}} du.\]

Integrate

Now we can solve the integral: \[\frac{1}{2}\pi \int_0^3 u^{-\frac{1}{2}} du = \frac{1}{2}\pi\left[2u^{\frac{1}{2}}\right]_0^3 = \pi\left[\sqrt{u}\right]_0^3 = \pi(\sqrt{3} - 0) = \pi\sqrt{3}.\]

Final Answer

The volume of the solid formed by revolving the region bounded by \(f(x) = (x^2-1)^{-\frac{1}{4}}\) and the x-axis on the interval (1, 2] around the y-axis is: \[V = \pi\sqrt{3}\] cubic units.

Key concepts

Disk Method

The disk method is an essential technique in calculus for finding the volume of a solid of revolution. This method is particularly useful when a shape is generated by rotating a region bounded by a curve around an axis. To visualize it, imagine slicing the solid into many thin disks perpendicular to the axis of rotation. Each disk's volume can be calculated using the formula for the volume of a cylinder, \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height (thickness of the disk).

For our problem, where the region under the curve \(f(x) = (x^2-1)^{-\frac{1}{4}}\) is revolved around the y-axis, each disk's radius corresponds to the value of \(f(x)\) at a given point, and its thickness corresponds to an infinitesimal change in \(x\), \(dx\). We then sum up the volumes of all the infinitely thin disks using a definite integral from the lower to upper bounds of \(x\).

Definite Integral

The definite integral is a fundamental concept in calculus, providing the net area under a curve between two points. When we talk about the area, we are usually referring to the region between the graph of a function and the x-axis, within specific limits, \(a\) and \(b\).

The symbolic representation of the definite integral of a function \(f(x)\) between limits \(a\) and \(b\) is \(\int_a^b f(x) dx\). The operation of taking the integral, known as integration, gives us the accumulated sum of the areas of infinitesimally thin rectangles with height \(f(x)\) and width \(dx\). In the context of finding volumes of solids of revolution, we are essentially summing up the cross-sectional areas of the solid at every point along the axis of rotation, which provides the total volume of the solid.

Substitution Method

The substitution method in calculus is a strategy to simplify an integral, making it easier to solve. It's akin to changing the variables to find a new, often simpler, integral that we are capable of evaluating. In essence, we choose a new variable \(u\) to stand in place of a more complex expression in the original integral. Then we express \(du\) in terms of \(dx\) (or vice-versa), which allows us to switch variables and perform the integration with respect to \(u\) instead of \(x\).

After substituting and integrating, the final step is to revert back to the original variable if necessary. This method is incredibly valuable when the function we're integrating has a composite structure, like in our problem, where we put \(u = x^2 - 1\) to simplify the integral of \(f(x)^2\).

Calculus

Understanding Calculus

Calculus is a branch of mathematics that deals with rates of change (differential calculus) and accumulation of quantities (integral calculus). It's the mathematical study of continuous change, just as geometry is the study of shapes and algebra the study of generalizations of arithmetic operations.

At the heart of calculus is the concept of the limit, which helps us handle quantities that approach one another infinitesimally closely. For instance, when finding the volume of a solid of revolution, we think of it as an accumulation of infinitely many infinitesimally thin disks or shells, which is made possible through integral calculus. Regardless of whether we are analyzing the motion of planets or the growth of populations, calculus provides the language to describe and understand the dynamics of change.