Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

The region bounded above by \(y=\frac{1}{x}\), below by the \(x\) -axis, and laterally by \(x=\frac{1}{2}\) and \(x=5\) is rotated about the \(x\) -axis. Find the volume of the funnel generated.

Step-by-Step Solution

Answer

Therefore, the volume of the funnel generated is \(\frac{9\pi}{5}\) cubic units.

Step 1: Understand the method

The volume of a solid of revolution can be found using the disk method or the washer method. The choice between them depends whether we have a hole in our figure or not. In this case, we don't have any hole, so we will use the disk method. The volume is the integration of \(\pi y^2\) over the interval [1/2, 5], i.e., \(\int_{1/2}^{5} \pi y^2 dx\). Since our equation is y = 1/x, we substitute y and get an integration equation in terms of x alone.

Step 2: Setup the Integration

So substituting the y in our integral form, we get \(V = \int_{1/2}^{5} \pi (\frac{1}{x})^2 dx\). This can be simplified as \(V = \int_{1/2}^{5} \frac{\pi}{x^2} dx\). Now we're ready to compute this definite integral.

Step 3: Compute the Integral

Now we have to compute the definite integral. The antiderivative of \(\frac{1}{x^2}\) is -1/x. At the upper limit, the value is \(-\frac{\pi}{5}\), and at the lower limit, the value is -2\(\pi\). So the result of the integral would be \(-\frac{\pi}{5} - (-2\pi) = \frac{9\pi}{5}\).

Key Concepts

Disk Method

The Disk Method is a powerful tool in calculus used to calculate the volume of a solid of revolution. Imagine taking a shape and rotating it around a particular axis to form a 3D object. The Disk Method helps find this volume by slicing the object into small, flat disks, much like stacking many coins together.

Each disk has a thickness (infinitesimally small) and a radius determined by the graph of the function you're rotating. In our exercise, the function is given by \( y = \frac{1}{x} \). Since we rotate around the \( x \)-axis, the radius of each disk is simply \( y \), and the thickness is an infinitesimal change in \( x \), or \( dx \).

To calculate the volume, we sum up (integrate) the volumes of these disks over a specified interval—from \( x = \frac{1}{2} \) to \( x = 5 \) in this case. The volume of each disk is \( \pi y^2 \) because the area of a circle is \( \pi r^2 \). Combining this concept with the integration process gives you the volume of the entire revolved solid.

Integral Calculus

Integral Calculus is a branch of mathematics that deals with the concept of integration. It involves calculating the total accumulation of quantities, such as area under a curve, or as in our exercise, the volume of a generated solid.

Integration is essentially the opposite of differentiation (finding a derivative), and it's tied to the accumulation of change. When dealing with functions, integration helps find the total value that arises as an input changes. In the case of the Disk Method, integration helps us accumulate the volume of tiny disks that make up a solid formed by revolving a function around an axis.

In our example, after substituting \( y = \frac{1}{x} \) into the volume formula, the problem reduces to finding the integral of \( \pi (\frac{1}{x})^2 \) from \( x = \frac{1}{2} \) to \( x = 5 \). This integral represents the total summed volume of the stacked disks.

Definite Integral

A Definite Integral is a specific type of integral used to calculate the net area under a curve, or as in the exercise, the volume within limits. It has defined upper and lower limits which specify the exact section of the curve you are interested in. These limits also frame the region over which we're revolving the function to form the solid.

In our exercise, the definite integral \( \int_{\frac{1}{2}}^{5} \pi (\frac{1}{x})^2 \, dx \) calculates the volume of the region rotated between \( x = \frac{1}{2} \) and \( x = 5 \) about the \( x \)-axis. This involves finding the antiderivative of \( \frac{1}{x^2} \), which is \( -\frac{1}{x} \), and evaluating it at the specified upper and lower limits.

The calculation \( -\frac{\pi}{5} - (-2\pi) \) reveals this definite integral's evaluation results in a net volume of \( \frac{9\pi}{5} \). This outcome signifies the total volume of the funnel-shaped solid formed by the revolution.