Problem 82 - Thomas Calculus

Question

Volume The region enclosed by the curve \(y=\operatorname{sech} x,\) the \(x\) -axis, and the lines \(x=\pm \ln \sqrt{3}\) is revolved about the \(x\) -axis to generate a solid. Find the volume of the solid.

Step-by-Step Solution

Answer

The volume of the solid is \( \pi \).

Step 1: Identify the given boundaries and setup

The region is bounded by the curve \( y = \operatorname{sech} x \), the \( x \)-axis, and the lines \( x = \pm \ln \sqrt{3} \). We need to find the volume of the solid formed by revolving this region about the \( x \)-axis.

Step 2: Define the volume integral using the disk method

Since we are revolving around the \( x \)-axis, the volume \( V \) can be found using the disk method. The formula for the volume is: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] where \( f(x) = \operatorname{sech} x \), \( a = -\ln \sqrt{3} \), and \( b = \ln \sqrt{3} \).

Step 3: Substitute the function and limits into the formula

Substitute the function \( f(x) = \operatorname{sech} x = \frac{1}{\cosh x} \) and the limits into the volume integral formula: \[ V = \pi \int_{-\ln \sqrt{3}}^{\ln \sqrt{3}} \left( \frac{1}{\cosh x} \right)^2 \, dx \]

Step 4: Simplify the integral expression

Simplify the integrand: \( \left( \frac{1}{\cosh x} \right)^2 = \operatorname{sech}^2 x \). So the integral becomes: \[ V = \pi \int_{-\ln \sqrt{3}}^{\ln \sqrt{3}} \operatorname{sech}^2 x \, dx \]

Step 5: Integrate \( \operatorname{sech}^2 x \)

The antiderivative of \( \operatorname{sech}^2 x \) is \( \tanh x \). Therefore, evaluate the integral: \[ V = \pi \left[ \tanh x \right]_{-\ln \sqrt{3}}^{\ln \sqrt{3}} \]

Step 6: Evaluate the antiderivative at the limits

Evaluate \( \tanh x \) at the upper and lower limits: \[ \tanh(\ln \sqrt{3}) - \tanh(-\ln \sqrt{3}) = \frac{e^{\ln \sqrt{3}} - e^{-\ln \sqrt{3}}}{e^{\ln \sqrt{3}} + e^{-\ln \sqrt{3}}} \]

Step 7: Simplify using properties of exponential functions

Since \( e^{\ln \sqrt{3}} = \sqrt{3} \), we can simplify: \[ \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{\sqrt{3} + \frac{1}{\sqrt{3}}} = \frac{2}{2/\sqrt{3}} = 1 \]

Step 8: Calculate the volume

Substitute the result from Step 7 back into the volume formula: \[ V = \pi (1) = \pi \] Thus, the volume of the solid is \( \pi \).

Key Concepts

Volume of solids

When we talk about the "volume of solids", we're looking at the measure of how much space a three-dimensional object occupies. Calculating the volume becomes particularly interesting when the solid has an unconventional shape, like one generated by revolving a two-dimensional area around an axis. Understanding this concept involves knowing the object’s boundaries and the axis of rotation. This solid’s volume is often computed using methods from calculus.

For objects that have flat surfaces, volumes are calculated using simple formulas, like length × width × height for cubes. However, for more complex shapes, especially those generated by rotation, calculus helps us derive an exact volume.

• The shapes discussed in these calculus problems, like the one involving hyperbolic functions and rotations around an axis, usually require integration to find volumes.
• Integration helps in accumulating small slices of the solid to get the total volume.
• In this task, when a region is revolved around the x-axis, small disk-like slices of this region need to be considered individually before summing their volumes to get the total.

This calculus approach is essential for understanding complex geometries encountered in various fields, from engineering to natural sciences.

Disk method

The disk method is a powerful technique in calculus for finding the volume of a solid of revolution. When a shape within a certain boundary is spun around an axis, it forms a solid. The disk method models this by slicing the solid into tiny disk-shaped pieces.

Imagine slicing the solid into thin disks along the axis. Each disk has a small, measurable width (think of it like the thickness of a slice). The disk’s radius corresponds to the function value at each point along the axis.

• The formula for the volume of each disk is \( V = \pi [f(x)]^2 \, dx\).
• To get the total volume of the solid, we integrate these small volumes over the given bounds. This is represented as \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx\).
• Integration steps accumulate the volume of all these disks to find the full volume of the solid.

In our specific example, the function \(y = \operatorname{sech} x\) is used, and it is revolved about the x-axis between limits \(x = -\ln \sqrt{3}\) and \(x = \ln \sqrt{3}\). This range of x-values defines the bounds for integration. By calculating step-by-step, we get the volume of these compound shapes. The beauty of the disk method lies in its systematic approach to handle complex rotations by simplifying them into more manageable calculations.

Hyperbolic functions

Hyperbolic functions, much like trigonometric functions, have a vital role in calculus, particularly with shapes and forms encountered in engineering and physics. They're functions related to the hyperbola in the same way trigonometric functions relate to the circle.

Some key hyperbolic functions are:

• Hyperbolic cosine: \( \cosh x = \frac{e^x + e^{-x}}{2} \)
• Hyperbolic sine: \( \sinh x = \frac{e^x - e^{-x}}{2} \)
• Hyperbolic secant: \( \text{sech} x = \frac{1}{\cosh x} \)

These functions possess properties similar to their trigonometric counterparts but differ in terms of periodicity and behavior at infinity.

In our example problem, the function \(y = \operatorname{sech} x\) plays a major role. The \(\text{sech}\) function, standing for hyperbolic secant, has a crucial property that it's always positive and grows towards zero as \(x\) moves away from the origin. Thus, when finding volumes or performing integration with these functions, recognizing their behavior aids in anticipating the results and ensuring steps like integration or algebraic manipulations are correctly applied.
Understanding hyperbolic functions provides insights into complex problems beyond simple geometry, extending into real-world applications such as wave dynamics or electrical circuits.