Question

Determine whether the following statements are true and give an explanation or counterexample. a. A pyramid is a solid of revolution. b. The volume of a hemisphere can be computed using the disk method. c. Let \(R_{1}\) be the region bounded by \(y=\cos x\) and the \(x\) -axis on \([-\pi / 2, \pi / 2] .\) Let \(R_{2}\) be the region bounded by \(y=\sin x\) and the \(x\) -axis on \([0, \pi] .\) The volumes of the solids generated when \(R_{1}\) and \(R_{2}\) are revolved about the \(x\) -axis are equal.

Short answer

Question: Determine whether the following statements are true or false and provide explanations or counterexamples to support your answers. (a) A pyramid is a solid of revolution. (b) The volume of a hemisphere can be computed using the disk method. (c) The volumes of two solids, generated when regions R1 and R2 are revolved about the x-axis, are equal, with R1 bounded by y = cos(x) from x = -π/2 to x = π/2 and R2 bounded by y = sin(x) from x = 0 to x = π.

Step-by-step solution

Statement (a) - A pyramid is a solid of revolution

A solid of revolution is a 3-dimensional figure that can be obtained by rotating a 2-dimensional region around an axis. To examine this statement, let's visualize a pyramid. A pyramid has a base and triangular faces that meet at a common vertex. If we attempt to rotate a 2-dimensional region around an axis, we will not obtain a pyramid. So, the statement is false. A pyramid is not a solid of revolution.

Statement (b) - The volume of a hemisphere can be computed using the disk method

A hemisphere is a half of a sphere, and its volume can be found using the disk method. To use the disk method, we need to integrate the cross-sectional area of the solid along the axis of rotation. A hemisphere can be obtained by rotating a semicircle around its diameter, so we can find its volume using the disk method. Let \(y = \sqrt{R^2 - x^2}\) be the equation of the semicircle with radius \(R\). The disk method formula is given by: \(V = \pi \int_0^R y(x)^2 dx = \pi \int_0^R (\sqrt{R^2 - x^2})^2 dx = \pi \int_0^R (R^2 - x^2) dx\). Solving the integral, we get: \(V = \pi [R^2x - \frac{x^3}{3}]_0^R = \pi [(R^3 - \frac{R^3}{3}) - (0^3-\frac{0^3}{3})] = \pi [R^3 - \frac{R^3}{3}] = \pi R^3 (\frac{2}{3})\). This result represents the volume of a hemisphere, confirming that the statement is true. The volume of a hemisphere can be calculated using the disk method.

Statement (c) - Volumes of solids generated by revolving regions \(R_1\) and \(R_2\) around the x-axis are equal

The volume of the solid generated by revolving region \(R_1\) around the x-axis is given by: \(V_1 = \pi \int_{-\pi/2}^{\pi/2} (\cos x)^2 dx\). The volume of the solid generated by revolving region \(R_2\) around the x-axis is given by: \(V_2 = \pi \int_0^{\pi} (\sin x)^2 dx\). We will now compare the volumes \(V_1\) and \(V_2\). The integrals involving \(\cos^2x\) and \(\sin^2x\) are known as the average values of \(sin^2 x\) and \(cos^2 x\) on their respective intervals. Over the intervals \([-\pi/2, \pi/2]\) and \([0, \pi]\), the average values of both \(\cos^2x\) and \(\sin^2x\) are equal to 1/2. Thus, \(V_1 = \pi \int_{-\pi/2}^{\pi/2} (\frac{1}{2}) dx = \frac{\pi}{2} [\pi] = \frac{\pi^2}{2}\), and \(V_2 = \pi \int_0^{\pi} (\frac{1}{2}) dx = \frac{\pi}{2} [\pi] = \frac{\pi^2}{2}\). Since \(V_1 = V_2\), the statement is true. The volumes of the solids generated when \(R_1\) and \(R_2\) are revolved about the x-axis are equal.

Key concepts

Disk Method

The disk method is a technique to find the volume of a solid of revolution. It's a straightforward method that uses the concept of slicing. Imagine you have a 2D region that you revolve around an axis to create a volume. By using the disk method, you slice this volume into thin disks or circular plates.
Each disk has a small thickness and a radius defined by the function you are integrating. By summing up the volumes of these small disks via integration, you find the entire volume of the solid.
For example, consider finding the volume of a hemisphere using the disk method. Visualize cutting the hemisphere into a stack of disks, where each disk's volume is a tiny part of the hemisphere.The equation for a semicircle with radius \( R \) is \( y = \sqrt{R^2 - x^2} \). Using the disk method, the volume \( V \) is calculated as:\[ V = \pi \int_0^R (R^2 - x^2) \, dx \]By solving this integral, you obtain the hemisphere's volume as \( \frac{2}{3} \pi R^3 \). In this way, the disk method effectively calculates the volume of solids of revolution by stacking infinitesimal disks.

Volume of a Hemisphere

Understanding how to calculate the volume of a hemisphere is crucial for various applications. A hemisphere is simply half of a full sphere. This volume can be determined both geometrically and through calculus methods such as the disk method.
Geometrically, the volume of a full sphere of radius \( R \) is \( \frac{4}{3}\pi R^3 \). Therefore, the volume of a hemisphere would be half of that:\[ V = \frac{1}{2} \times \frac{4}{3} \pi R^3 = \frac{2}{3} \pi R^3 \]To calculate this using the disk method, revolve a semicircle around its diameter. Each cross-section perpendicular to the axis of revolution forms a disk. By integrating the area of these disks, the same result is achieved, further validating the accuracy of the disk method.Thus, both geometrical and calculus-based analyses confirm that the volume of a hemisphere can be easily computed and verified through proper understanding and application of methods.

Comparison of Volumes

Comparing the volumes of different solids can enhance our understanding of their properties and relationships. In the exercise, the volumes of solids generated from two regions, \( R_1 \) and \( R_2 \), were compared. These regions were bounded by \( y = \cos x \) and \( y = \sin x \), respectively.
To find the volumes, both regions are rotated about the \( x \)-axis. The volume for each is computed using the integral of the square of the function over its respective interval. For \( R_1 \), the volume is:\[ V_1 = \pi \int_{-\pi/2}^{\pi/2} (\cos x)^2 \, dx \]For \( R_2 \), it is:\[ V_2 = \pi \int_0^{\pi} (\sin x)^2 \, dx \]The expected result may not appear obvious at a glance, but through careful evaluation of the integrals, we find these volumes equal. This surprising fact shows how integral calculus allows us to uncover insights that are not immediately evident through geometric intuition alone. Therefore, comparison of volumes through definite integrals is a powerful technique in mathematics.