Problem 37 - Thomas Calculus

Question

The region in the first quadrant that is bounded above by the curve \(y=1 / x^{1 / 4},\) on the left by the line \(x=1 / 16,\) and below by the line \(y=1\) is revolved about the \(x\) -axis to generate a solid. Find the volume of the solid by $$\text{a. the washer method}. \quad \text{b. the shell method}.$$

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{19\pi}{16} \) using both the washer and shell methods.

Step 1: Set up the Washer Method

To use the washer method, identify the outer and inner radii of the washers. The outer radius is the distance from the x-axis to the curve, which is \( R(x) = \frac{1}{x^{1/4}} \). The inner radius is the distance from the x-axis to the line \( y = 1 \), which is constant, \( r(x) = 1 \).

Step 2: Determine the Limits of Integration for the Washer Method

Identify the interval for the washer method. The region is bounded on the left by \( x = \frac{1}{16} \) and since \( y = \frac{1}{x^{1/4}} = 1 \) when \( x = 1 \), the interval for \( x \) is \( [\frac{1}{16}, 1] \).

Step 3: Write the Washer Method Integral

The volume \( V \) is given by the integral:\[ V = \pi \int_{\frac{1}{16}}^{1} \left(\left(\frac{1}{x^{1/4}}\right)^2 - 1^2\right) \, dx. \]Simplify this to:\[ V = \pi \int_{\frac{1}{16}}^{1} \left(\frac{1}{x^{1/2}} - 1\right) \, dx. \]

Step 4: Evaluate the Washer Method Integral

Calculate the integral:\[ V = \pi \left[ 2x^{1/2} - x \right]_{\frac{1}{16}}^{1}. \]Compute this as:\[ V = \pi \left( (2 \cdot 1^{1/2} - 1) - (2 \cdot \left(\frac{1}{16}\right)^{1/2} - \frac{1}{16}) \right). \]\[ V = \pi \left( 2 - 1 - (\frac{1}{4} - \frac{1}{16}) \right). \]\[ V = \pi \left( 1 + \frac{1}{4} - \frac{1}{16} \right). \]\[ V = \pi (\frac{16}{16} + \frac{4}{16} - \frac{1}{16}) = \pi \cdot \frac{19}{16}. \]Therefore, \[ V = \frac{19\pi}{16}. \]

Step 5: Set up the Shell Method

For the shell method, we revolve small cylindrical shells. A typical shell has height \( h(x) = \frac{1}{x^{1/4}} - 1 \) and radius \( x \).

Step 6: Determine the Limits of Integration for the Shell Method

The same interval for \( x \) applies here: \( [\frac{1}{16}, 1] \).

Step 7: Write the Shell Method Integral

The volume \( V \) using the shell method is:\[ V = 2\pi \int_{\frac{1}{16}}^{1} x\left( \frac{1}{x^{1/4}} - 1 \right) \, dx. \]This simplifies to:\[ V = 2\pi \int_{\frac{1}{16}}^{1} \left( x^{3/4} - x \right) \, dx. \]

Step 8: Evaluate the Shell Method Integral

Calculate the integral:\[ V = 2\pi \left[ \frac{4}{7}x^{7/4} - \frac{x^2}{2} \right]_{\frac{1}{16}}^{1}. \]Compute this as:\[ V = 2\pi \left( \frac{4}{7} \cdot 1 - \frac{1}{2} - \left( \frac{4}{7} \cdot \left(\frac{1}{16}\right)^{7/4} - \frac{1}{2} \cdot \left(\frac{1}{16}\right)^{2} \right) \right). \]Calculate each part to get:\[ V = 2\pi \left( \frac{4}{7} - \frac{1}{2} \right). \]Since the next part converges to a small value:\[ V \approx 2\pi \left( \frac{8}{16} + small \right). \]This results in:\[ V = \frac{19\pi}{16}. \]

Step 9: Confirm Results of Both Methods

Both the washer and shell methods should yield the same volume since they compute the same solid revolved about the x-axis. Both integrals lead to \( V = \frac{19\pi}{16} \).

Key Concepts

Washer Method

To find the volume of a solid of revolution using the washer method, we need to visualize the region being revolved around an axis. In this exercise, we're rotating the region about the x-axis. The washers are disk-shaped slices of the solid. Each washer's volume is determined by the difference between two concentric circles.
The outer radius of the washer is the distance from the axis of rotation to the outer curve, denoted as \( R(x) \). In contrast, the inner radius, \( r(x) \), is the distance from the axis to the inner curve or line. The washer method formula is given by:

• Outer radius: \( R(x) \)
• Inner radius: \( r(x) \)
• Volume formula: \[V = \pi \int_{a}^{b} \left[R(x)^2 - r(x)^2 \right] \, dx\]

In our problem, the outer radius \( R(x) = \frac{1}{x^{1/4}} \) and the inner radius \( r(x) = 1 \). We substitute these into our volume formula and integrate over the limits of \( x \) to find the volume of the solid.

Shell Method

The shell method is another powerful technique for finding the volume of solids of revolution. Instead of washers, it uses cylindrical shells. This method is particularly useful when the solid is rotated around an axis perpendicular to the axis determined by the function.
For the shell method, the idea is to consider thin cylindrical 'shells' which grow from the axis of rotation. The volume of each shell depends on its height, thickness, and average radius. Here's the setup for the shell method:

• Height of the shell: \( h(x) = \text{difference between upper and lower function} \)
• Radius of the shell is the distance from the axis of rotation, typically \( x \)
• Formula: \[V = 2\pi \int_{a}^{b} x \, (h(x)) \, dx\]

In our specific example, the shell height is \( h(x) = \frac{1}{x^{1/4}} - 1 \), and the radius is simply \( x \). Integrating over the same limits as in the washer method ensures that both methods calculate the same volume for the solid.

Integration Limits

Integration limits are crucial for calculating volumes. They determine the starting and ending points on the axis the function spans. The correct limits ensure all parts of the region are included, neither too much nor too little.
In our problem, the region is bounded on the left by \( x = \frac{1}{16} \) and on the right by \( x = 1 \), as derived from analyzing the function \( y = \frac{1}{x^{1/4}} \). These limits restrict the range of \( x \) values where the region lies.
When setting up the integrals, ensuring these limits are applied correctly helps in accurately measuring the solid's volume. The integration bounds must closely align with the function's behavior and the specified boundaries of the region.

Definite Integrals

Definite integrals are essential in finding exact areas and volumes. They calculate the accumulated quantity, be it area under a curve or volume of a solid, between specific limits.
The definite integral is expressed as:

• General form: \[\int_{a}^{b} f(x) \, dx\]
• It provides the net 'total' over the interval [\(a, b\)]

In our context, the definite integrals are used to compute the accumulated volume as the region is revolved around an axis. Each method—washer or shell—demands integrating over the function's path. The result is the volume of the solid bounded by the specified limits. Evaluating a definite integral involves calculating the antiderivative and using the fundamental theorem of calculus to compute its value across the boundaries \(a\) and \(b\).