Problem 6 - Thomas Calculus in SI Units

Question

Find the volumes of the solids. The solid lies between planes perpendicular to the \(x\) -axis at \(x=-\pi / 3\) and \(x=\pi / 3 .\) The cross-sections perpendicular to the \(x\) -axis are a. circular disks with diameters running from the curve \(y=\tan x\) to the curve \(y=\sec x\) b. squares whose bases run from the curve \(y=\tan x\) to the curve \(y=\sec x\)

Step-by-Step Solution

Answer

a. Evaluate the circular disk integral to find the volume. b. Evaluate the square cross-section integral, same as Step 3.

Step 1: Understanding the Cross-Section Shape for Part a

For part a, the cross-section is a circular disk. The diameter of the disk is the distance between the curves \(y = \tan x\) and \(y = \sec x\). To find the radius \(r\) of the disk, we calculate \(r = \frac{\sec x - \tan x}{2}\).

Step 2: Set Up the Integral for Volume of Circular Disk

The volume \(V\) of the solid is found by integrating the area of these disks from \(x = -\pi/3\) to \(x = \pi/3\). The area \(A\) of a circle is \(\pi r^2\). Thus, the volume is:\[V = \int_{-\pi/3}^{\pi/3} \pi \left(\frac{\sec x - \tan x}{2}\right)^2 \, dx.\] Simplifying gives:\[V = \frac{\pi}{4} \int_{-\pi/3}^{\pi/3} (\sec x - \tan x)^2 \, dx.\]

Step 3: Solving the Integral for Circular Disk

To solve the integral, expand \((\sec x - \tan x)^2\) and integrate each term separately:\[\int (\sec^2 x - 2 \sec x \tan x + \tan^2 x) \, dx.\]Using these integrals:- \(\int \sec^2 x \, dx = \tan x\)- \(\int \sec x \tan x \, dx = \sec x\)- \(\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x\)Evaluate these from \(-\pi/3\) to \(\pi/3\).

Step 4: Understanding the Cross-Section Shape for Part b

For part b, the cross-section is a square, and the base of the square is the distance between the curves \(y = \tan x\) and \(y = \sec x\). Thus, the side \(s\) of the square is \(\sec x - \tan x\).

Step 5: Set Up the Integral for Volume of Square Cross-Section

The volume \(V\) of the solid is found by integrating the area of these squares from \(x = -\pi/3\) to \(x = \pi/3\). The area \(A\) of a square is \(s^2\). Thus, the volume is:\[V = \int_{-\pi/3}^{\pi/3} (\sec x - \tan x)^2 \, dx.\] This simplifies to:\[V = \int_{-\pi/3}^{\pi/3} (\sec^2 x - 2 \sec x \tan x + \tan^2 x) \, dx.\]

Step 6: Solving the Integral for Square Cross-Section

This integral setup is exactly the same as in Step 3. Use the same antiderivatives for the terms, then evaluate from \(-\pi/3\) to \(\pi/3\) to find the volume of the solid.

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus that is used to find the exact area under a curve over a particular interval. In our exercise, we are using the definite integral to calculate the volumes of solids formed by cross-sections.
In this problem, the volumes are calculated by integrating across the range of values from \(-\pi/3\) to \(\pi/3\), where our cross-sections are positioned perpendicularly to the \(x\)-axis.
The definite integral helps in summing the infinitesimally small areas of cross-sections as you move along the \(x\)-axis, this effectively gives us the entire volume of the solid.

• To perform this process, we integrate the area of each cross-section - a circle or square - along the interval from \(-\pi/3\) to \(\pi/3\).
• The formula applied in the solution is structured as: \(V = \int_{a}^{b} A(x) \, dx\), where \(A(x)\) is the area of the cross-section at \(x\), and \(a\) and \(b\) are the bounds of the integration.

Trigonometric Functions

Trigonometric functions play a crucial role in the given problem, specifically the functions \(y = \tan x\) and \(y = \sec x\). These functions define the boundaries of the cross-sections.
The tangent function, \(\tan x\), represents the ratio of the opposite to the adjacent side in a right-angled triangle. The secant function, \(\sec x\), is the reciprocal of the cosine function.
These functions oscillate and exhibit different behaviors, impacting the shape and size of the cross-sections as \(x\) varies.

• Understanding \(\tan x\) and \(\sec x\) helps us accurately determine the diameter or the side length of the cross-sections.
• Due to their periodic nature, these functions vary within the given limits of \(-\pi/3\) and \(\pi/3\), thus influencing the overall volume calculations.

Cross-Sectional Area

The cross-sectional area is the area of a slice of the solid perpendicular to its length. For this exercise, cross-sections can either be circular disks or squares. The area of each is calculated differently and directly impacts the volume computation.
In the first case, cross-sections are circular disks, where the diameter is given by the difference \(\sec x - \tan x\). Therefore, the radius is \(\frac{\sec x - \tan x}{2}\), and the area is \(\pi r^2\).
In the second situation, cross-sections are squares, and the side length is simply \(\sec x - \tan x\), giving an area of \((\sec x - \tan x)^2\).

• These cross-sectional areas are crucial parts of the integrands that are used in calculating the total volume of the solids.
• The choice of circular or square cross-section changes the form of the integral and thus affects the calculation direction.

Areas of Circles and Squares

Calculating areas of geometric shapes is central to setting up the integrals in this problem.
For circles, the area is given by the formula \(A = \pi r^2\), where \(r\) is the radius. This formula incorporates the components of circular geometry and applies to cross-sections when they are circular disks.
For squares, the area is calculated as \(s^2\), where \(s\) is the side length of the square. This simple formula highlights the regular nature of square areas.

• In cross-sectional disks, determining the radius involves dividing the difference between two functions by two, which is then integrated over the specified limits.
• In squares, calculating the area involves squaring the difference \(\sec x - \tan x\), a simpler straightforward calculation that impacts the volume of the solid.

These principles form the groundwork for transforming the problem of finding the volume of solids into a calculus problem solvable through integration.