Problem 46 - Thomas Calculus

Question

Find the volume of the solid generated by revolving the triangular region bounded by the lines \(y=2 x, y=0,\) and \(x=1\) about a. the line \(x=1 . \quad\) b. the line \(x=2\)

Step-by-Step Solution

Answer

Part a: Volume is \(\frac{2\pi}{3}\). Part b: Volume is \(\frac{8\pi}{3}\).

Step 1: Describe the Problem

In this exercise, we are tasked with finding the volume of a solid generated by rotating a triangular region around two different lines. The triangular region is bounded by the lines \( y = 2x \), \( y = 0 \), and \( x = 1 \). We'll use the method of cylindrical shells for both parts of the problem.

Step 2: Set Up Integral for Part a (About x=1)

When revolving around \( x = 1 \), the radius of the shell is the distance from the line \( x = 1 \) to \( x \). That is, the radius is \( 1 - x \). The height of the shell is given by \( y = 2x \). Thus, the volume is \[V = \int_{0}^{1} 2\pi (1 - x) (2x)\, dx = 4\pi \int_{0}^{1} (1-x)x\, dx.\]

Step 3: Simplify the Integral for Part a

First, expand the expression inside the integral: \((1-x)x = x - x^2\). Therefore, the integral becomes: \[V = 4\pi \int_{0}^{1} (x - x^2)\, dx.\] We'll solve this integral in the next step.

Step 4: Compute the Integral for Part a

Compute the indefinite integral: \[\int (x - x^2)\, dx = \frac{x^2}{2} - \frac{x^3}{3}.\] Evaluate from 0 to 1: \[\left. \left( \frac{x^2}{2} - \frac{x^3}{3} \right) \right|_0^1 = \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{1}{6}.\] Thus, the volume for part (a) is \[V = 4\pi \times \frac{1}{6} = \frac{2\pi}{3}.\]

Step 5: Set Up Integral for Part b (About x=2)

When revolving around \( x = 2 \), the radius of the shell is \( 2 - x \). Using the same height, \( y = 2x \), the volume is \[V = \int_{0}^{1} 2\pi (2-x)(2x)\, dx = 4\pi \int_{0}^{1} (2x - x^2)\, dx.\]

Step 6: Simplify the Integral for Part b

Expand the expression inside the integral: \[(2x - x^2) = 2x - x^2.\] Therefore, the integral becomes \[V = 4\pi \int_{0}^{1} (2x - x^2)\, dx.\]

Step 7: Compute the Integral for Part b

Compute the indefinite integral: \[\int (2x - x^2)\, dx = x^2 - \frac{x^3}{3}.\] Evaluate from 0 to 1: \[\left. \left( x^2 - \frac{x^3}{3} \right) \right|_0^1 = \left( 1 - \frac{1}{3} \right) = \frac{2}{3}.\] Therefore, the volume for part (b) is \[V = 4\pi \times \frac{2}{3} = \frac{8\pi}{3}.\]

Key Concepts

Understanding Cylindrical Shells

Have you ever wondered how we can find the volume of a solid that results from rotating a shape? The method of cylindrical shells is a powerful technique to solve such problems.

The idea is simple: imagine slicing the solid into thin cylindrical shells. These shells can be "peeled off" the solid. Each shell has a tiny thickness, determined by the width of an infinitesimally small piece along the axis of rotation.

For the given exercise, the triangular region is revolved around a vertical line, which is a common scenario for cylindrical shells. When revolving around a line parallel to the y-axis, the radius of each shell is the horizontal distance from the line of rotation to any point on the region.

• The radius of a shell helps determine its circumference, which is necessary for finding its lateral surface area.
• The formula for the volume of one shell: Volume = Circumference x Height x Thickness.
• The circumference is given by \( 2\pi \times \text{radius} \), height is the y-value of the function, and the thickness corresponds to a small change in x, noted as \( dx \).

To find the entire volume, sum the volumes of all shells, expressed as an integral. This method often simplifies calculations, especially when the solid’s shape fits nicely with a cylindrical shell approach.

The Process of Integration

Integration is like a mathemagical way of putting together infinite little pieces to get a big picture. When we rotate shapes to find volumes, we use definite integrals to sum up an infinite number of tiny volumes.

In this exercise, integration helps us sum the volumes of cylindrical shells. Here’s how we go about it:

• First, setup the integral: Identify limits of integration, usually the x-values or y-values that bound the shape.
• Next, express the volume of each shell as a function that needs integration. This involves the radius and height found earlier.
• Finally, integrate the function using the limits. This step essentially adds up all tiny shell volumes to find the total volume.

The integration process requires practice, as it involves transforming and simplifying functions. For example, like in this solution where terms like \( 4\pi \int (1-x)x \) are integrated by simplifying first.

Integration is central to solving this type of problem, providing a structured approach to handle volumes with precision.

Diving into Integral Calculus

Integral calculus is the language we use for describing and analyzing changes and areas. When it comes to solids of revolution, one of its many applications, we employ integral calculus to merge the continuous into a calculable quantity. This branch of calculus delves into two main types: definite and indefinite integrals.

In this context:

• Definite integrals give us specific numerical values for volume calculations within set limits. They are particularly helpful in computing actual quantities like volume, as done in the example.
• Indefinite integrals deal with general antiderivatives. During our problem-solving, indefinite integrals represent the integration process without specified bounds, helping us understand the function derived from differentiation.

The intersection of geometry and calculus transforms our approach toward understanding and solving real-world volume problems, such as rotating shapes in mathematics.

Integral calculus enables us to approach intricate volumes through systematic techniques—especially useful for evaluating complex shapes that would otherwise be difficult to compute.