Problem 38 - Calculus Early Transcendentals

Question

Use either the washer or shell method to find the volume of the solid that is generated when the region in the first quadrant bounded by \(y=x^{2}, y=4,\) and \(x=0\) is revolved about the following lines. $$x=-1$$

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\frac{54 \pi}{5}\) cubic units.

Step 1: 1. Sketch the region and identify the radii

First, let's sketch the region bounded by the given curves. The parabola \(y = x^2\) is an upward-opening parabola. The line \(y = 4\) is a horizontal line, and the line \(x = 0\) is the y-axis. The region will be revolved around \(x = -1\). When revolving this region around x = -1, the outer radius is the distance from the line x = -1 to the curve y = 4, and the inner radius is the distance from the line x = -1 to the curve y = \(x^2\). The outer radius is given by \((x+1)\), and the inner radius is given by \(((x^2)+1)\).

Step 2: 2. Set up the integral

Using the washer method, we will now set up the integral to find the solid's volume. We must integrate along the x-axis, from the leftmost point (x = 0) to the rightmost point (x = 2) of the region. The volume can be found using the formula: \(V = \pi \int_{a}^{b} [R(x)^{2} - r(x)^{2}] dx\) Here, \(R(x) = (x+1)\) is the outer radius and \(r(x) = ((x^2)+1)\) is the inner radius. So, \(V = \pi \int_{0}^{2} [((x+1)^{2} - ((x^2)+1)^{2})] dx\)

Step 3: 3. Integrate and solve for the volume

Now, we will find the antiderivative and evaluate the integral to find the volume: \(V = \pi \int_{0}^{2} [(x+1)^{2} - ((x^2)+1)^{2}] dx\) \(V = \pi \left [ \frac{1}{3}(x+1)^3 - \frac{1}{5}((x^2)+1)^{5/2} + \frac{2}{3}x^3\right ]_0^2\) After evaluating the definite integral, we get: \(V = \pi \left[\frac{1}{3}(2+1)^3 - \frac{1}{5}((2^2)+1)^{5/2} + \frac{2}{3}(2)^3 \right]\) \(V = \frac{54 \pi}{5}\) So, the volume of the solid is \(\frac{54 \pi}{5}\) cubic units.

Key Concepts

Washer Method

The Washer Method is a technique used to find the volume of a solid of revolution. This approach is particularly useful when a region is revolved around a horizontal or vertical axis and the region is bounded by two curves resulting in a hollow center. Imagine a bunch of washers stacked side by side, with varying inner and outer diameters. Each washer represents a cross-section of the solid.

For the Washer Method, you'll need to:

• Determine the outer radius, which is the distance from the axis of rotation to the outer edge of the region.
• Determine the inner radius, which is from the axis of rotation to the inner edge.
• Subtract the square of the inner radius from the square of the outer radius to compute the area of each washer.
• Integrate these washer areas over the range of rotation to find the total volume.

This method proves efficient when your region is between two curves, and you're rotating about an axis outside of the area covered by the region, such as in our exercise.

Shell Method

The Shell Method, another technique for finding the volume of a solid of revolution, involves imagining the solid as a series of cylindrical "shells". Instead of washing machine disks, visualize vertical or horizontal cylindrical tubes stacked inside one another. This method is most effective when rotating around an axis parallel to the line of function, usually making it simpler when dealing with horizontal lengths.

Applying the Shell Method involves:

• Considering a tiny slice perpendicular to the axis of rotation, determining its radius and height.
• Setting up a formula based on the height of the shell (given by the function), times the circumference of the shell (which is based on the distance from the axis of rotation).
• Integrating along the correct axis.

This approach becomes particularly handy when the Washer Method is too complex due to the way the function behaves. In certain problems like our example, using the Shell Method laterally would lead to an unnecessarily complicated scenario due to the need for horizontal layers instead of washers.

Definite Integral

A Definite Integral is a powerful tool used in calculus that provides the total "net area" under a curve within a given interval. It also serves a pivotal role in computing volumes, especially volumes of revolutions. Here, the integral aspect uses the endpoints or limits of the region to precisely calculate the build-up of all infinitesimally small pieces.

Understanding this concept involves:

• Knowing the bounds: You set the limits of integration which usually range from the smallest to the largest value of the variable.
• Setting up the expression inside the integral: This can be as simple as the function squared, or a combination of functions especially when working with rotations.
• Solving the integral: This involves finding the antiderivative, then evaluating it at the endpoints.

In our specific problem, the definite integral is key to summing up all the volumes of the "washers" from the range 0 to 2, giving us the total volume of the solid.

Calculus Problem Solving

Tackling calculus problems efficiently revolves around understanding the basics, converting the problem into solvable parts, and applying the right techniques. There’s always a journey from understanding the graphical area involved, concocting a plan to set up integrals, and finally carrying out the numerical integrations or algebraic simplifications.

When solving calculus problems involving volumes of revolutions:

• Start with a clear visualization: Sketch your region and axis of rotation to clarify the revolved shape.
• Select the appropriate method: Decide between washer and shell methods based on simplicity and complexity of the area.
• Set up proper integrals: These expressions need to accurately reflect the volume as calculated by either washers or shells.
• Thoroughly solve and evaluate: Ensure you've not only set up the problem correctly, but also follow through each step of integration and evaluation for accuracy.

By following these steps, you'll build confidence in deconstructing problems and ensure solutions are both accurate and insightful in nature.