Problem 37 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the disk or washer method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$y=x^{3}, y=0, x=2$$

Step-by-Step Solution

Answer

Question: Determine the volume of the solid generated when the region R bounded by the curves \(y=x^3\), \(y=0\) (the x-axis), and \(x=2\) is revolved around the \(y\)-axis using the disk or washer method. Answer: The volume of the solid generated when R is revolved about the y-axis is \(32\pi\) cubic units.

Step 1: Sketch the region R

Before diving into the problem, start by sketching the region R to gain a better understanding of the shape being revolved. The region is bounded by \(y=x^3\), \(y=0\), and \(x=2\). The x-axis bounds R from below, so there is no "hole" in the solid when revolved.

Step 2: Identify the Disk Thickness, Outer Radius, and Inner Radius

Rotate R about the \(y\)-axis. Since we're using the disk/washer method, take thin disks perpendicular to the axis of rotation (the y-axis) with thickness \(\Delta y\). The outer radius of each disk is \(x = 2\). Since R is bounded by \(y=0\) on the bottom, there is no hole in the resulting solid, and thus the inner radius is 0.

Step 3: Set up the Volume Integral

Using the disk method, the volume of a single disk is given by \(\pi(\text{outer radius}^2 - (\text{inner radius})^2)\Delta y\). To find the volume of the entire solid, integrate the volume of each disk along the y-axis. In this case, \(x=2\) is the outer boundary, and \(y=0\) at the \(x\)-axis is the inner boundary. The integral takes the following form: $$V = \pi \int_{y=a}^{y=b} [(2)^2 - (0)^2]dy$$

Step 4: Determine the Limits of Integration

To find the limits of integration, we need to express \(y\) in terms of \(x\). We know that \(y=x^3\), \(x=0\), and \(x=2\). With k boundaries are already defined in terms of y coordinates for us: 1. When \(x=0\), \(y=0^3=0\) 2. When \(x=2\), \(y=2^3=8\) Thus, the limits of integration are \(y=0\) to \(y=8\). The integral becomes: $$V = \pi \int_{0}^{8} [(2)^2 - (0)^2]dy$$

Step 5: Evaluate the Integral and Compute the Volume

Now, we can evaluate the integral. The volume V is constant with respect to y, so it can be factored out of the integral. $$V = \pi [(2)^2 - (0)^2] \int_{0}^{8} dy$$ After the integration, we find: $$V = \pi [(2)^2 - (0)^2] [y|_{0}^{8}] = \pi [(2)^2 - (0)^2] [8 - 0]$$ Finally, compute the volume: $$V = \pi (4)(8) = 32\pi$$ The volume of the solid generated when R is revolved about the y-axis is \(32\pi\) cubic units.

Key Concepts

Calculus

Calculus is a branch of mathematics focused on the study of change and motion. Through its two major sub-fields, differential calculus and integral calculus, it gives us the tools to analyze and solve problems involving curves, areas, volumes, and rates of change.

In the context of the disk method volume calculation problem, calculus comes into play when we need to determine the volume of a solid of revolution. Using calculus, specifically integral calculus, we can sum an infinite number of infinitesimally thin disks to find the total volume of the solid. This process of summation is what integration is all about: adding up pieces to find a whole.

Solid of Revolution

A solid of revolution is a three-dimensional object obtained by rotating a two-dimensional shape around an axis. The resulting solid has a symmetry around the axis of rotation, similar to objects like vases or bottles.

The exercise provided involves creating a solid of revolution by revolving the region bounded by the curves around the y-axis. The shape we get is akin to a cylinder but with one end possibly being larger or differently shaped than the other. The disk method is particularly useful for calculating volumes of solids of revolution when the revolving axis is either the x-axis or the y-axis, as in this example.

Volume Integral

The volume integral is a way to calculate the volume of a three-dimensional object using integration. For solids of revolution, we often use the disk or washer methods, which involve slicing the solid into thin pieces, computing the volume of each slice, and then summing these volumes via integration.

In our exercise, the volume integral of the solid is simplified because the inner radius is zero, and the outer radius is a constant 2, leading to an integral that is straightforward to evaluate. The integral quantifies the accumulation of all the disk volumes from the bottom to the top of the solid, providing us with the total volume.

Limits of Integration

The limits of integration in an integral denote the boundaries over which the integration is performed. These are crucial for evaluating the definite integral, as they specify the starting and ending points.

For volume calculations, these limits correspond to the bottom and top of the solid of revolution along the axis of rotation—the y-axis in our example. Setting these limits accurately is essential for obtaining the correct volume. In the given problem, the region R's lowest and highest points on the y-axis are 0 and 8, respectively, providing us with the boundaries needed to solve the integral and determine the solid's volume.