Problem 7 - Thomas Calculus in SI Units

Question

Use the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines about the \(y\) -axis. $$y=x, \quad y=-x / 2, \quad x=2$$

Step-by-Step Solution

Answer

The volume is \( 8\pi \).

Step 1: Identify the Bounded Region

The given curves are two lines, \( y = x \) and \( y = -\frac{x}{2} \), and the vertical line \( x = 2 \). First, identify the points of intersection between \( y = x \) and \( y = -\frac{x}{2} \) by setting their equations equal: \( x = -\frac{x}{2} \). This gives \( x = 0 \), and since both equations pass through (0,0), the intersecting points are at \((0, 0)\). The region is further bounded by \( x = 2 \).

Step 2: Set Up the Shell Method Formula

The shell method revolves around the formula for finding volumes: \[ V = 2\pi \int_{a}^{b} (radius \times height) \, dx \]. Here, as revolving around the \( y \)-axis, the radius is \( x \) and the shell height is the difference between \( y = x \) and \( y = -\frac{x}{2} \): that is, \( x + \frac{x}{2} = \frac{3}{2}x \).

Step 3: Determine Limits of Integration

The limits for integration \( x \) are determined from \( x = 0 \) to \( x = 2 \), as these are the intersection and boundary points respectively within the problem context.

Step 4: Integrate the Shell Volume

Plug the information into the shell volume formula: \[ V = 2\pi \int_{0}^{2} x \left(\frac{3}{2}x\right) \, dx = 2\pi \int_{0}^{2} \frac{3}{2} x^2 \, dx \]. Simplify inside the integral: \[ V = 3\pi \int_{0}^{2} x^2 \, dx \].

Step 5: Evaluate the Integral

Calculate the integral: \[ V = 3\pi \left[\frac{x^3}{3}\right]_{0}^{2} = 3\pi \times \left(\frac{2^3}{3} - \frac{0^3}{3}\right) = 3\pi \times \left(\frac{8}{3}\right) = 8\pi \]. Thus, the volume of the solid is \( 8\pi \).

Key Concepts

Volume of Solids

The concept of volume is crucial in understanding how three-dimensional objects occupy space. When dealing with solids, especially those generated by rotation, calculating the volume can become a sophisticated task. In calculus, we use various techniques to find the volumes of these solids, particularly through integration methods. For figure shapes with simple geometric properties, formulas are typically straightforward. However, for most irregular or rotated shapes, integration becomes necessary.

The Shell Method is one such technique that is particularly useful for finding the volume of solids of revolution. It involves revolving a region around an axis to create a solid. In our context, we revolve a region about the y-axis. The method takes vertical slices (cylindrical shells) of this region, calculates the volume of each slice, and adds them up via integration.

• The formula for the shell method is: \[ V = 2\pi \int_{a}^{b} (radius \times height) \, dx \]
• This method is best used when slicing perpendicular to the axis of revolution is more complex.
• It relies on summing up volumes of cylindrical shells formed by vertical slices of the region.

Understanding the concept of volumes occupied in space and how integration can be used to sum infinitesimal components of volume is key for mastering the shell method.

Revolution about y-axis

Imagine rotating a simple flat shape around a vertical axis. This process is called revolution, and when the revolution is about the y-axis, it creates a three-dimensional shape or solid. This solid has a radial symmetry about the y-axis.

Consider a region bounded by several curves and lines—for example, the region between the lines you have, such as \( y = x \), \( y = -\frac{x}{2} \), and \( x = 2 \). When this region is rotated about the y-axis, it forms a hollow cylindrical shape, similar to a vase or a tunnel.

• This symmetry about the y-axis implies that each point on the boundary traces a circle.
• The objective is to measure how much space this solid occupies.
• It transforms a 2D problem into a 3D space, requiring integration along only one dimension.

This conceptual step is critical because it allows us to move from theoretical 2D calculus problems into understanding how these ideas manifest in the world around us.

Integration Techniques

Integration is a fundamental calculus tool used to accumulate or sum quantities, such as areas and volumes. When it comes to finding volumes of rotated solids, mastering integration techniques is essential. The core idea is to integrate small pieces of volume over a certain range to get the total volume.

In the Shell Method, the integral setup incorporates radius and height, which may change with each slice. The integral typically involves a polynomial or simple algebraic function, and the limits of integration define the bounds of summation.

• The limits are determined by the intersection points of the curves—here from \( x = 0 \) to \( x = 2 \).
• Integration simplifies the complexity of adding infinite slices, providing a precise volume.
• Techniques such as substitution, polynomial integration, and partial fractions might be used depending on the complexity.

By setting up and solving an integral correctly, you essentially break down the complex task of volume calculation into manageable mathematical operations, allowing for precise and meaningful results in understanding three-dimensional spaces.