Problem 44 - Calculus Single Variable

Question

In each of Exercises 43-48, use the method of cylindrical shells to calculate the volume \(V\) of the solid that is obtained by rotating the given planar region \(\mathcal{R}\) about the \(x\) -axis. \(\mathcal{R}\) is the region that is to the right of \(y=5 x-4,\) to the left of \(y=10-x^{2},\) and above \(y=1\)

Step-by-Step Solution

Answer

The volume of the solid is evaluated by computing \( V = 2\pi \) times the integral of the shell's radius and height from \( x = -7 \) to \( x = 2 \).

Step 1: Identify the Region of Integration

The region \( \mathcal{R} \) is bounded by the curves \( y = 5x - 4 \) and \( y = 10 - x^2 \), and lies above \( y = 1 \). First, find their points of intersection to identify the bounds. Set \( 5x - 4 = 10 - x^2 \). This simplifies to \( x^2 + 5x - 14 = 0 \). Solve this quadratic equation to find the \(x\) values which are the limits of integration.

Step 2: Solve the Quadratic Equation

The quadratic equation \( x^2 + 5x - 14 = 0 \) is solved using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 5 \), and \( c = -14 \). This yields \( x = \frac{-5 \pm \sqrt{25 + 56}}{2} = \frac{-5 \pm \sqrt{81}}{2} = \frac{-5 \pm 9}{2} \). The solutions are \( x = 2 \) and \( x = -7 \). The physical region between these x-values is used for integration.

Step 3: Setup the Integral for Shell Method

Use the method of cylindrical shells with the formula \( V = 2\pi \int_{a}^{b} (shell\ radius) \times (shell\ height)\ dx \). The shell radius is \( y \) and the shell height is \( (5x - 4) - 1 \). Thus, the volume integral is \( V = 2\pi \int_{2}^{-7} y((5x - 4) - 1)\ dy \). However, integration is in terms of \( x \), so instead solve in terms of \( x \) with \( y = x \). After setting up the cylindrical shell, switch bounds of integration to the appropriate limits of \( y \).

Step 4: Calculate the Shell Height and Radius

First note that the shell height is simply the difference between \( 10 - x^2 \) (the top of the region) and \( 5x - 4 \) (the bottom of the region): \( (10 - x^2) - (5x - 4) = 14 - x^2 - 5x \). Use this expression as the height in the integral: \( 2\pi \int_{-7}^{2} x (14 - x^2 - 5x)\ dx \). The shell radius, being the distance from the axis of rotation, is \(x\).

Step 5: Evaluate the Integral

Substitute the expressions for height and radius into the integral: \( V = 2\pi \int_{-7}^{2} x (14 - x^2 - 5x)\ dx \). Expand the integral: \( V = 2\pi \int_{-7}^{2} (14x - x^3 - 5x^2)\ dx \). Integrate term-by-term to get \( V = 2\pi \left[ 7x^2 - \frac{x^4}{4} - \frac{5x^3}{3} \right]_{-7}^{2} \). Substitute \( x = 2 \) and \( x = -7 \) into this expression and calculate the results.

Step 6: Substitute the Limits and Solve

Compute the definite integral: \( V = 2\pi \left[ (7(2)^2 - \frac{(2)^4}{4} - \frac{5(2)^3}{3}) - (7(-7)^2 - \frac{(-7)^4}{4} - \frac{5(-7)^3}{3}) \right] \). Simplify each term and find the resultant difference. This yields the final volume \( V \).

Key Concepts

Volume Calculation

To calculate the volume of a solid of revolution, we use a method known as the cylindrical shell method. This technique is particularly useful when the solid is created by rotating a region around an axis. The basic idea is to imagine the solid as made up of many cylindrical shells, each with a small thickness.
The volume of each shell can be calculated, and then the total volume is found by adding up all these individual shell volumes. The formula for the volume of a cylindrical shell is given by:

• Volume of a shell = Perimeter × Height × Thickness

In mathematical terms, when rotating a region about the x-axis, the volume is determined using the integral:

• \[ V = 2\pi \int_{a}^{b} (shell\ radius) \times (shell\ height)\ dx \]

For this problem, the shell radius is the distance of the shell from the axis of rotation, and the shell height is the difference between the two bounding functions. The limits of integration, 'a' and 'b', are the x-values where these functions intersect.

Definite Integrals

Definite integrals are a crucial concept in calculating volumes, especially when using techniques like the cylindrical shell method. A definite integral represents the signed area under a curve within a specific interval. In the context of volume calculation through integration, it sums up all the infinitely small elements, like the cylindrical shells, over the given interval to calculate the total volume.
When setting up the integral for the shell method, we must properly define the integral's limits, as well as the functions that represent the radius and height of the shells.
The formula for finding the total volume takes into account the structure of the definite integral:

• \[ V = 2\pi \int_{a}^{b} x(14 - x^2 - 5x)\ dx \]

This formula integrates the product of the shell radius (x) and the shell height \((14 - x^2 - 5x)\) from the lower bound \(x = -7\) to the upper bound \(x = 2\), illustrating how definite integrals are utilized in real-world applications.

Quadratic Equations

Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \). Solving these equations is essential for identifying the points at which curves intersect, which in turn determine the limits of integration for many problems involving curves, including the calculation of volumes. In this exercise, the quadratic equation formed is:
\[ x^2 + 5x - 14 = 0 \]
We solve this using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 5 \), and \( c = -14 \). This computation yields the solutions \( x = 2 \) and \( x = -7 \).
These solutions represent the x-values at which the bounding functions intersect, crucial for evaluating the volume integral correctly.
By understanding quadratic equations and their solutions, students can better grasp how these equations help define the regions they are working with in various mathematical problems.