Problem 11 - Single Variable Calculus

Question

Use the method of cylindrical shells to find the volume of the solid generated by revolving the region bounded by the graphs of the equations and/or inequalities about the indicated axis. Sketch the region and a representative rectangle. \(y=\frac{1}{x}, \quad y=0, \quad x=1, \quad x=2 ; \quad\) the \(y\) -axis

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by the graphs of \(y=\frac{1}{x}\), \(y=0\), \(x=1\), and \(x=2\) around the y-axis is \(2\pi\) cubic units. This result is obtained using the method of cylindrical shells, which involves integrating the product of the circumference, the height, and the width of the cylindrical shells formed by revolving the region.

Step 1: Sketch the region and representative rectangle

To begin, let's sketch the given region to better understand the problem. The region is bounded by the curve \(y=\frac{1}{x}\), the lines \(y=0\), \(x=1\), and \(x=2\). Mark the points of intersection on the graph and draw a representative rectangle perpendicular to the axis of rotation (in this case, parallel to the x-axis).

Step 2: Set up the volume integral with the method of cylindrical shells

To find the volume using the method of cylindrical shells, we'll integrate the circumference of each cylindrical shell, multiplied by its height, and the width of the shell. The circumference of a cylindrical shell is given by \(2\pi x\) (where x is the distance of the shell from the axis of rotation), its height is the function value \(y\), and the shell's width is \(\Delta x\). So, the volume of a cylindrical shell is given by \(2\pi x y \Delta x\). Now, we'll integrate this expression over the range of x-values defining the region (x=1 to x=2): \[V = \int_{1}^{2} 2\pi x \left(\frac{1}{x}\right) dx\]

Step 3: Simplify and solve the volume integral

Now let's simplify the integral and solve it: \[V = \int_{1}^{2} 2\pi \left(\frac{1}{x}\right) x dx = \int_{1}^{2} 2\pi dx\] The integral of \(2\pi\) with respect to x is \(2\pi x\), so we have: \[V = 2\pi x \Big|_{1}^{2} = 2\pi (2 - 1) = 2\pi\] So, the volume of the solid formed by revolving the region bounded by the given curves around the y-axis is \(2\pi\) cubic units.

Key Concepts

Volume of Solids of Revolution

Understanding the volume of solids of revolution involves visualizing how a shape is created by rotating a planar region around an axis. In single variable calculus, this concept is essential for solving real-world problems involving rotational symmetry. For instance, to find the volume of a vase or a bowl, you might slice it into thin discs or shells and then sum up the volume of these small parts.

Using the method of cylindrical shells, as in the provided exercise, we consider thin, hollow cylinders ('shells') with variable radii, heights, and thickness. By revolving a rectangle around an axis (in this case, the y-axis), we form one such shell. We then calculate the volume of each shell and integrate this over the interval to find the total volume of the solid. It is a practical approach that simplifies complex volume calculations.

Definite Integral

The definite integral is a fundamental concept in calculus that represents the accumulation of quantities and can be thought of as the area under a curve between two points on the x-axis. When we integrate a function from point 'a' to point 'b', we are essentially adding up an infinite number of infinitely small quantities between these points.

In our exercise, the integral \[\int_{1}^{2} 2\pi x \left(\frac{1}{x}\right) dx\] represents the sum of the volumes of all cylindrical shells between x=1 and x=2. This process yields the total volume of the solid formed by revolution. The 'definite' part means we are looking at a specific interval, rather than the entire area under the curve.

Single Variable Calculus

Single variable calculus is an area of mathematics focused on functions of one variable and includes differentiation and integration. It's used to analyze continuous processes such as growth, motion, and changing shapes. In our exercise, calculus helps to calculate the volume of complex shapes created by rotating a function around an axis.

Understanding concepts like limits, derivatives, and integrals in the context of a single variable is crucial for solving such problems. Single variable calculus lays the foundation for many applications, including physics, engineering, economics, and beyond, by enabling the calculation of rates of change and accumulation.

Volume Integral

A volume integral extends the concept of integration over a three-dimensional space and is often used to calculate the volume of solids with known cross-sectional areas. In the cylindrical shells method, the volume integral takes the form \[V = \int_{a}^{b} 2\pi \cdot \text{radius} \cdot \text{height} \cdot \text{width}\], with 'radius' being the distance from the shell to the axis of rotation, 'height' being the value of the function, and 'width' being the thickness of the shell (\(\Delta x\) in differentials).

By performing the volume integral over the specified range, and simplifying as necessary, we find the exact volume of the solid of revolution. The beauty of the method of cylindrical shells illustrated in our exercise is that it enables us to break down the complex geometry into calculable slices, showcasing the power of calculus in spatial analysis.