Question

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

Short answer

Answer: The volume of the solid is \(2\pi\) cubic units.

Step-by-step solution

Define the integrand

The integrand represents the volume of a single disk. It is calculated as the product of the area of the disk and its infinitesimal thickness, \(dx\). As the disk is being revolved around the \(x\)-axis, the radius of the disk will be \(f(x)\), and the area of the disk can be written as \(\pi*(f(x))^2\). The volume, then, will be \(\pi*(f(x))^2 dx\).

Set up the definite integral

To find the volume of the entire solid, we need to integrate the volume of each disk over the given interval (1, 2]. So, we will set up a definite integral from 1 to 2: \(\int_{1}^{2} \pi*(f(x))^2 dx\).

Substitute the function

Now, substitute the given function \(f(x) = (x - 1)^{-1/4}\) into the definite integral: \(\int_{1}^{2} \pi*((x - 1)^{-\frac{1}{4}})^2 dx\).

Simplify the integrand

Now, simplify the integrand inside the integral: \(\int_{1}^{2} \pi*(x - 1)^{-\frac{1}{2}} dx\).

Evaluate the definite integral

Now, we need to evaluate the integral: \(\pi*\int_{1}^{2} (x - 1)^{-\frac{1}{2}} dx\). This integral can be solved using a simple substitution (\(u = x - 1\)) or just considering the antiderivative of the function. The antiderivative of \((x - 1)^{-\frac{1}{2}}\) is \(2(x - 1)^{\frac{1}{2}}\). So, the integral becomes: \(\pi*[2(x - 1)^{\frac{1}{2}}]_{1}^{2}\)

Calculate the volume using the antiderivative

Now, evaluate the result using the antiderivative: \(\pi*[2(2 - 1)^{1/2} - 2(1 - 1)^{1/2}]\) \(\pi*[2(1)^{1/2} - 2(0)^{1/2}]\) \(\pi*[2 - 0]\) Thus, the volume of the solid is \(2\pi\) cubic units.

Key concepts

Definite Integral

The concept of a definite integral is pivotal in calculus, especially when it comes to finding areas, volumes, and other quantities that accumulate over an interval. When you see an expression like \(\int_{a}^{b} f(x) dx\), it represents the definite integral of the function \(f(x)\) with respect to \(x\), from \(x=a\) to \(x=b\). This mathematical procedure computes the accumulated value of \(f(x)\) as \(x\) varies within the limits between \(a\) and \(b\).

In our exercise, the definite integral calculates the volume of a solid by adding up infinitesimally thin disks' volumes, from the start of the interval, \(x=1\), to the end of the interval, \(x=2\). This process transforms a spatial problem into a manageable algebraic expression.

Understanding the definite integral is crucial because it represents the aggregate of infinitely many small quantities, which is the cornerstone of integral calculus and its applications to real-world problems.

Solids of Revolution

Solids of revolution are three-dimensional objects obtained by rotating a two-dimensional shape around an axis. The shape is commonly referred to as the 'generator' because it generates the entire solid through its rotation. Imagine the 2D area bounded by a function, such as \(f(x)\), and the \(x\)-axis; when this area is spun around that axis, it creates a 3D shape, much like how a potter's clay forms a pot on a wheel.

The exercise we’re exploring involves a solid formed by revolving the area under the curve \(f(x) = (x-1)^{-1 / 4}\) around the \(x\)-axis. Here, the curve itself and the \(x\)-axis define the boundaries of a region in the plane, which, when revolved, creates the solid of interest. By grasping this concept, we can visualize and calculate the volumes of complex shapes, making it a powerful technique in engineering, physics, and mathematics.

Disk Method

When we talk about the disk method, we're referring to a way of finding the volume of a solid of revolution. The method involves slicing the solid into thin, flat disks perpendicular to the axis of rotation. Each disk's volume is essentially that of a cylinder, \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height or thickness of the disk.

In the case of our problem, the 'height' of each disk is an infinitesimal \(dx\), which represents a tiny segment along the \(x\)-axis. The radius is given by the value of the function \(f(x)\), hence the volume of a thin disk becomes \(\pi (f(x))^2 dx\). This disk method simplifies a three-dimensional volume problem into a one-dimensional integration.

The total volume of the solid is found by integrating these disks' volumes along the given interval, capturing all the tiny contributions to the solid's volume as \(x\) sweeps from 1 to 2.

Antiderivative

The antiderivative, also known as the indefinite integral, is essentially the reverse operation of differentiation. It involves finding a function \(F(x)\) whose derivative is the given function \(f(x)\). Symbolically, if \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\). When calculating definite integrals, the antiderivative plays a key role because it allows us to evaluate the integral using the Fundamental Theorem of Calculus.

For the solid of revolution in the exercise, we use the antiderivative to calculate the exact volume. After setting up our definite integral, we find the antiderivative of the integrated function. We then evaluate this antiderivative at the upper and lower bounds of our interval, subtract the latter from the former, and multiply by \(\pi\) to get our final volume. This process is exemplified as we take the antiderivative of \( (x - 1)^{-1/2} \) and substitute the boundary values to find the solid's volume.