Problem 42 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\)-axis. \(y=\sec x, \quad y=\tan x, \quad x=0, \quad x=1\)

Step-by-Step Solution

Answer

The volume is \( \pi \) cubic units.

Step 1: Identify the region to be rotated

The curves given are \( y = \sec x \) and \( y = \tan x \), bounded by \( x = 0 \) and \( x = 1 \). The area between these curves needs to be identified and calculated for revolution about the \( x \)-axis.

Step 2: Express the volume element

The volume of a solid of revolution can be calculated using the disk or washer method. Since we are revolving around the \( x \)-axis, we'll use the washer method: \[ V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \, dx \]where \( R(x) = \sec x \) and \( r(x) = \tan x \) are the outer and inner radii respectively.

Step 3: Set up the integral

Substitute the bounds \( a = 0 \) and \( b = 1 \), and expressions for \( R(x) \) and \( r(x) \) into the formula:\[ V = \pi \int_{0}^{1} [\sec^2 x - \tan^2 x] \, dx \].

Step 4: Simplify the integrand

Using the trigonometric identity \( \sec^2 x - \tan^2 x = 1 \), simplify the integrand:\[ V = \pi \int_{0}^{1} 1 \, dx \].

Step 5: Integrate and calculate the volume

The integral of 1 with respect to \( x \) over \( [0, 1] \) is simply \( x \):\[ V = \pi [x]_{0}^{1} = \pi [1 - 0] = \pi \].

Step 6: Interpret the result

The volume of the solid formed by revolving the region bounded by the curves and lines about the \( x \)-axis is \( \pi \) cubic units.

Key Concepts

Washer Method

The washer method is a reliable technique used when finding the volume of a solid of revolution. This method is particularly useful when you are dealing with an area between two curves. Imagine slicing the solid into thin, flat washers or rings as you revolve the region around an axis. Here's how it works:

• Identify the outer and inner radii: In the washer method, you need to determine the outer radius, \( R(x) \), and the inner radius, \( r(x) \). This involves finding the difference between the two curves that bound your region.
• Setup the integral: You then express the volume \( V \) as \( \pi \int_{a}^{b} [R(x)^2 - r(x)^2] \, dx \), where \([a, b]\) is the interval over which you're rotating the region.
• Calculate: Simplifying and integrating this expression gives you the total volume of the solid.

For the given exercise, you rotate around the \( x \)-axis and identify \( R(x) = \sec x \) and \( r(x) = \tan x \). This setup allows you to evaluate the integral easily, thanks to a clever trigonometric identity.

Volume Integral

The volume integral is a crucial aspect when calculating volumes of solids of revolution. Essentially, it represents summing up infinitesimally small volume elements over a specific interval. Think of it as adding up small pieces of the solid, slice by slice.

• Setting the bounds: These integrals range from \( a \) to \( b \), where these limits often represent where the curves intersect or where the region of interest ends.
• Function of the radius: You'll notice in our solution, \( V = \pi \int_{0}^{1} [\sec^2 x - \tan^2 x] \, dx \). The radii squared are handled inside the integral, providing crucial limits to the revolving process.
• Compute the total: The integration takes into account the entire region being revolved, ensuring the volume is calculated accurately and completely.

In our case, the bounds, \(0\) to \(1\), along the \( x \)-axis, were set based on lines that bound the specified region together with the curves.

Trigonometric Identities

Trigonometric identities often come up when working with functions involving sinusoidal and cosine asymptotic curves like \( \sec x \) and \( \tan x \). Understanding these identities simplifies otherwise complex problems.

• Basic identity: One of the foundational identities is \( \sec^2 x - \tan^2 x = 1 \). This is derived from the Pythagorean identity and makes calculations more straightforward.
• Simplification: In the exercise, recognize this identity when simplifying \( \sec^2 x - \tan^2 x \). This reduces the integral to a much simpler form: \( 1 \).
• Common scenarios: Such identities frequently appear in volume problems, propulsion of curves, and solving differential equations.

Here, employing the identity simplifies the integration process to \( V = \pi \int_{0}^{1} 1 \, dx \), making it an approachable and neat solution.