Problem 25 - Thomas Calculus

Question

In Exercises \(23-26,\) use the shell method to find the volumes of the solids generated by revolving the regions bounded by the given curves about the given lines. $$y=x+2, \quad y=x^{2}$$ $$\begin{array}{ll}{\text { a. The line } x=2} & {\text { b. The line } x=-1} \\\ {\text { c. The } x \text { -axis }} & {\text { d. The line } y=4}\end{array}$$

Step-by-Step Solution

Answer

Use geometry and calculus to compute volumes for each rotation axis.

Step 1: Set up the problem

We need to find the volume of the solid formed by rotating the region bounded by two curves: \( y = x + 2 \) and \( y = x^2 \). We'll consider the region in the context of each specified line.

Step 2: Find the intersection points

To define the region properly, determine where \( y = x + 2 \) and \( y = x^2 \) intersect by setting the equations equal: \( x + 2 = x^2 \). Rearranging gives \( x^2 - x - 2 = 0 \). Factoring results in \((x - 2)(x + 1) = 0\), so \( x = 2 \) and \( x = -1 \).

Step 3: Volume using the shell method for x=2

For revolving around \( x = 2 \), the shell radius is \( 2 - x \). The height is \( (x + 2) - x^2 \) from the top curve minus the bottom curve. The volume \( V \) is given by \[ V = 2 \pi \int_{-1}^{2} (2 - x)((x+2) - x^2) \, dx \].

Step 4: Solve the integral for x=2

Expand and integrate: 1. \[ V = 2\pi \int_{-1}^{2} [(2-x)(x+2-x^2)] \, dx \] which simplifies to: 2. \[ 2\pi \int_{-1}^{2} (2x + 4 - 2x^2 - x^2 - 2 + x^3) \, dx \].Simplify further to \[ 2\pi \int_{-1}^{2} (2x + x^3 - 3x^2 + 2) \, dx \].Find the antiderivative and compute.

Step 5: Volume using the shell method for x=-1

For revolving around \( x = -1 \), the shell radius is \( x + 1 \). The height is still \( (x+2) - x^2 \). The volume \( V \) is given by \[ V = 2 \pi \int_{-1}^{2} (x + 1)((x+2) - x^2) \, dx \].

Step 6: Solve the integral for x=-1

Expand: 1. \[ V = 2\pi \int_{-1}^{2} [(x+1)(x+2-x^2)] \, dx \] which results in: 2. \[ 2\pi \int_{-1}^{2} (x^2 + 3x + 2 - x^3 - x^2 - x+ x^3) \, dx \].This simplifies to \[ 2\pi \int_{-1}^{2} (2x + 2) \, dx \].Integrate and compute.

Step 7: Volume using the shell method for the x-axis

When revolving about the x-axis, the shell method isn't directly applicable without adjustment because we're dealing with horizontal shells. It's more straightforward to use the disk or washer method. Consider using appropriate geometric methods to evaluate integrals of the form \( \pi \int_{-1}^{2} [(outer\ radius)^2 - (inner\ radius)^2] \, dx \).

Step 8: Volume using the shell method for y=4

For revolving around \( y = 4 \), express the shell radius as \( 4 - y \), writing \( y = x^2 \) and \( y = x + 2 \). Obtain x in terms of y for integration bounds. Use vertical shells for this scenario, integrating along the y-axis.

Key Concepts

Volume of Solids of Revolution

When exploring the volume of solids of revolution, imagine creating a 3D shape by rotating a 2D area around a line, typically an axis. This process transforms flat curves into volumetric objects. The Shell Method is one technique used to calculate such volumes. It involves imagining the solid as being composed of numerous cylindrical shells. Each shell has a small thickness, and its volume can be approximated by using the formula for the volume of a cylinder, which is the product of the circumference, height, and thickness. The Shell Method is especially useful when the axis of rotation is parallel to the axis of the figure (but not necessarily one of the coordinate axes). It simplifies computations by focusing on the radius of each shell, which is the distance from the axis of rotation to the shell, and the height of the shell, which is the value of the function at any given point. By integrating along the axis of rotation, the total volume can be found as: \[ V = 2\pi \, \int_{a}^{b} \text{(shell radius)} \cdot \text{(shell height)} \, dx. \] This method of integrating cylindrical shells helps in accurately computing the volume, especially when the regions are not the standard circular ones.

Intersection Points of Curves

Finding the intersection points of curves is a crucial step in setting up problems involving solids of revolution. These points define the boundaries of the region to be revolved. In mathematical terms, finding an intersection point involves solving an equation where two expressions are equal, which represents the curves intersecting at that specific point. For example, given the curves \( y = x + 2 \) and \( y = x^2 \), finding where they intersect means solving \( x + 2 = x^2 \). Rearranging gives \( x^2 - x - 2 = 0 \). Using methods such as factoring or the quadratic formula, you find the roots: \( x = 2 \) and \( x = -1 \). These points tell us where the curves intersect and help in determining the limits of integration for the shell method. Determining these points properly is essential, as they guide the calculation of areas and volumes by precisely defining the region of interest. When performing such calculations, ensure the intersection points are accurately determined, as these form the core parameter for setting computational boundaries.

Integration Techniques

Integration techniques in the context of volumes of solids of revolution often involve calculating the area under a curve or between curves. For the Shell Method, this involves finding the antiderivative of an expression that represents the product of the shell radius and height over the defined limits. The expression we integrate often arises from simplifying the function representing the height and the shell radius. For example, if revolving around a vertical line, the volume integral might look like: \[ V = 2\pi \int_{-1}^{2} (2-x)((x+2) - x^2) \, dx, \] where we break down the expression inside the integral, simplify, and then find the antiderivative. Successful integration often uses basic calculus techniques:

• Polynomial expansion
• Simplification
• Use of known antiderivatives

Once simplified, evaluate over the limits, always remembering that these calculations directly impact the derived volume. For complex curves, initially breaking down components simplifies finding the solution and ensuring precision.