Problem 25 - Thomas Calculus in SI Units

Question

Find the volume of the solid generated by revolving the region about the given line. The region in the first quadrant bounded above by the line \(y=\sqrt{2},\) below by the curve \(y=\sec x \tan x,\) and on the left by the \(y\) -axis, about the line \(y=\sqrt{2}\).

Step-by-Step Solution

Answer

The volume is calculated by the integral from 0 to \(\frac{\pi}{4}\) of \( \pi (2\sec x \tan x - (\sec x \tan x)^2) \).

Step 1: Understand the Problem

We need to find the volume of a solid created by rotating a region in the first quadrant around the horizontal line \(y = \sqrt{2}\). The region is bounded by the curve \(y = \sec x \tan x\), the line \(y = \sqrt{2}\), and the \(y\)-axis.

Step 2: Set Up the Integral

To find the volume of the rotated solid, we'll use the method of washers. The volume \(V\) of the solid is given by the integral: \[ V = \pi \int_0^a (R(x)^2 - r(x)^2) \, dx \] where \(R(x)\) is the distance from the axis of rotation to the outer curve \(y = \sqrt{2}\), and \(r(x)\) is the distance from the axis of rotation to the inner curve \(y = \sec x \tan x\).

Step 3: Identify the Limits of Integration

The region is bounded by the \(y\)-axis and \(y = \sqrt{2}\). Since we revolve around \(y = \sqrt{2}\), \(x\) ranges from 0 to the point where \(\sec x \tan x = \sqrt{2}\). Solving \(\sec x \tan x = \sqrt{2}\), we notice \(x = \frac{\pi}{4}\). So the limits are 0 to \(\frac{\pi}{4}\).

Step 4: Determine Radii Functions

The outer radius \(R(x)\) is the distance from \(y = \sqrt{2}\) to \(y = 0\), which is \(\sqrt{2}\). The inner radius \(r(x)\) is the distance from \(y = \sqrt{2}\) to \(y = \sec x \tan x\), given as \(\sqrt{2} - \sec x \tan x\).

Step 5: Perform Integration

Substitute \(R(x) = \sqrt{2}\) and \(r(x) = \sqrt{2} - \sec x \tan x\) into the integral: \[ V = \pi \int_{0}^{\frac{\pi}{4}} \left( (\sqrt{2})^2 - (\sqrt{2} - \sec x \tan x)^2 \right) dx \]This simplifies to: \[ V = \pi \int_{0}^{\frac{\pi}{4}} \left( 2 - (2 - 2\sec x \tan x + (\sec x \tan x)^2) \right) dx \]\[ = \pi \int_{0}^{\frac{\pi}{4}} \left( 2\sec x \tan x - (\sec x \tan x)^2 \right) dx \]

Step 6: Calculate the Result

Solve the integral:\[ V = \pi \int_{0}^{\frac{\pi}{4}} \left( 2\sec x \tan x - (\sec x \tan x)^2 \right) dx \]Integrate term by term: compute the antiderivative of \(2\sec x \tan x\) and \((\sec x \tan x)^2\), then evaluate from 0 to \(\frac{\pi}{4}\). This gives the volume of the solid.

Key Concepts

Washer Method

The washer method is a technique used in calculus to find the volume of a solid of revolution. This method is particularly helpful when the solid has a hollow center, like a doughnut or a cylindrical shell. In this approach, you revolve a defined shape around an axis, forming a series of washers, which are essentially circles with holes in the middle.
To calculate the volume using the washer method, you set up a definite integral that accounts for both the outer and inner radii of the washers. For example, if you revolve a region around a horizontal line, your integral will take the form:

• \( V = \pi \int_a^b (R(x)^2 - r(x)^2) \, dx \)

Where:

• \( R(x) \) is the outer radius, the distance from the axis to the outer curve.
• \( r(x) \) is the inner radius, the distance from the axis to the inner curve.

This method effectively slices the solid perpendicular to the axis of rotation, helping to derive the volume by adding up the areas of all washers within the specified bounds.

Calculus Integration

Integration is a fundamental concept in calculus that's used to find a variety of values, such as area, volume, and length. When dealing with solids of revolution, integration is key to calculating their volumes. Calculus integration comes in different forms, but in the context of volumes and the washer method, we specifically use definite integrals.
By setting up an integral, you capture the continuous sum of infinitesimally small elements (like washers), to find a total value. In this case, you're summing the areas of these after being squared and summed, which (when multiplied by pi) gives the volume of the solid.
To perform the integration, you generally follow steps like identifying both the function to integrate and the limits of integration. Next, you solve the integral either through standard techniques or with the help of complex functions, such as trigonometric identities, where necessary.

Definite Integral

A definite integral is a precise mathematical computation that allows you to measure the accumulated quantity, such as area under a curve or volume of a solid between two boundaries. It's defined as the limit of a sum of function values, each multiplied by the width of an interval.
The definite integral in the context of revolving solids is used to compute volumes. Specifically, you integrate \( \pi \int_a^b (R(x)^2 - r(x)^2) \, dx \), representing the volume of washers stacked along the length of the solid. Here:

• \( a \) and \( b \) are the lower and upper bounds of integration, often determined by points of intersection or boundaries of the shape.
• The integral simplifies by first expanding terms, then solving for the total volume from start to end points (like \( 0 \) to \( \frac{\pi}{4} \) in our case).

Evaluating the definite integral gives the exact volume that is enclosed by the region, revolutioning around the line.

Trigonometric Functions

Trigonometric functions often play a crucial role in calculus, especially when dealing with regions bounded by curves like sine, cosine, tangent, and their reciprocals (secant, cosecant, and cotangent). In the exercise, the curve \( y = \sec x \tan x \) presents a challenge that must be addressed using trigonometry.
These functions define how angles relate to side lengths, and, in the method of washers, they emerge frequently when setting up the radius functions. With our function \( y = \sec x \tan x \), trigonometric identities and derivatives are used to simplify and solve the subsequent calculations.
Moreover, computing the critical points where \( \sec x \tan x = \sqrt{2} \) requires understanding how these functions behave and knowing their basic properties on specific intervals. Solving these will give us necessary limits of integration and ensure accuracy in our final calculations of volume.