Problem 31 - Thomas Calculus

Question

In Exercises \(31-36,\) find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so. The triangle with vertices \((1,1),(1,2),\) and \((2,2)\) about $$\begin{array}{ll}{\text { a. the } x \text { -axis }} & {\text { b. the } y \text { -axis }} \\ {\text { c. the line } x=10 / 3} & {\text { d. the line } y=1}\end{array}$$

Step-by-Step Solution

Answer

Volumes: a. \(\frac{7\pi}{3}\), b. \(\pi\), c. \(\frac{7\pi}{3}\), d. \(\frac{\pi}{3}\).

Step 1: Identify the region and method

The region is a triangle with vertices at \((1,1)\), \((1,2)\), and \((2,2)\). We need to determine if the disk or washer method applies in each part of the question.

Step 2: Revolve about x-axis

For part (a), we revolve the triangle about the x-axis. Given that the triangle is bounded by y = x for \(1 \leq x \leq 2\) and y = 2, we integrate using the washer method from \(x=1\) to \(x=2\).The outer radius \(R(x) = 2\) and inner radius \(r(x) = x\), so the washer volume formula is \[V = \pi \int_{1}^{2} [(2)^2 - (x)^2] \, dx = \pi \int_{1}^{2} (4 - x^2) \, dx\].Evaluating the integral, we have \[ V = \pi \left[ 4x - \frac{x^3}{3} \right]_{1}^{2} = \pi \left( 8 - \frac{8}{3} - (4 - \frac{1}{3}) \right) = \frac{7\pi}{3} \].

Step 3: Revolve about y-axis

For part (b), we revolve the triangle about the y-axis. The equations are x = 1, y = 2, and y = x, so we use the disk method.The disk radius is \(R(y) = 1\), and integrating from \(y=1\) to \(y=2\), \[V = \pi \int_{1}^{2} (1)^2 \, dy = \pi [y]_{1}^{2} = \pi (2 - 1) = \pi\].

Step 4: Revolve about x = 10/3

For part (c), reflect across \(x = 10/3\) and use the washer method. The outer radius is \(R(x) = 10/3 - 1 = 7/3\) and the inner radius is \(r(x) = 10/3 - x\) for \(1 \leq x \leq 2\).Then the volume is \[V = \pi \int_{1}^{2} \left[ \left( \frac{7}{3} \right)^2 - \left( \frac{10}{3} - x \right)^2 \right] \, dx\].Simplifying, \[ V = \pi \int_{1}^{2} \left( \frac{49}{9} - \frac{(10 - 3x)^2}{9} \right) \, dx\].Evaluating, \[ V = \pi \left( x \left( \frac{49}{9} \right) - \frac{1}{9} [ (10 - 3x)^3/3 ] \right)_{1}^{2} = \frac{7\pi}{3} \].

Step 5: Revolve about y = 1

For part (d), reflect across \(y=1\) using the washer method with outer radius \(R(x) = x - 1\) and inner radius zero, integrate over \(x = 1\) to \(x = 2\).Volume is\[V = \pi \int_{1}^{2} \, [(x-1)^2] \, dx\]Insert expression, evaluate:\[ V = \pi \left[ \frac{x^3}{3} - x^2 \right]_{1}^{2} = \pi \left( \frac{8}{3} - 4 - (\frac{1}{3} - 1) \right) = \frac{\pi}{3} \].

Key Concepts

Disk Method

The Disk Method is a useful technique to find the volume of a solid of revolution. This comes into play when we revolve a region around an axis, creating a 3D solid.

With the Disk Method, we imagine slicing the solid into very thin circular disks perpendicular to the axis of rotation. The volume of each disk is given by its thickness times its cross-sectional area.

• The formula to calculate the volume by the Disk Method when revolving around the x-axis is: \[ V = \pi \int_{a}^{b} [R(y)]^2 \ dy \]
  Here, \( R(y) \) is the radius of the disk function in terms of \( y \).
• Revolve around the y-axis, and it's symmetrical: invoke \( R(x) \).

In step by step solution for step 3 of the exercise, where the region is revolved about the y-axis, we employed the Disk Method. We calculated the radius of the disk as constant, \( R(y) = 1 \), and integrated over \( y \) from 1 to 2, computing the volume as \( \pi \). This straightforward technique helps to visualize and mathematically determine the volume of solids you might encounter.

Washer Method

The Washer Method is a variation of the Disk Method, introduced when there's a hole in the middle of the solid. Essentially, it’s like a disk but with an inner region carved out. Hence, the cross-section resembles a washer.

To calculate the volume using the Washer Method:

• The formula changes to account for both an outer and inner radius: \[ V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \ dx \]
• \( R(x) \) is the outer radius, while \( r(x) \) is the inner radius.

In the step by step solution, for several parts of the exercise, like revolving the region about the x-axis and the line \( x = 10/3 \), the Washer Method accounted for the difference in radii. For instance, revolving the triangle about x-axis involved different outer (2) and inner (x) radii, resulting in integral \[ \pi \int_{1}^{2} (4 - x^2) \ dx \]. Evaluating this gave volume, illustrating the Power of the Washer Method in handling regions with removed central portions.

Integration Techniques

Integration is the mathematical tool at the heart of both the Disk and Washer Methods. It calculates the area under a curve, crucial in determining volumes of revolved solids.

With volume problems using these methods, we often need to evaluate definite integrals:

• For instance, calculating integral bounds is essential, often between intercepts or limits of the region being revolved.
• Substituting radius functions appropriately into the integral is crucial: \[ V = \pi \int_{a}^{b} (expression \, to \, be \, integrated) \, dx \]

Each of these must be done carefully, as shown in solutions. Consider revolving around axes, integrals account for differing functions representing various radius distances. Deeper understanding of integration bolsters your ability to solve complex volume calculations, appreciating its role in deriving a neat algebraic expression summing up infinite slices across a solid's volume.