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In Exercises 39-42, find the volume of the solid analytically. 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 \(y=\sec x\).

Short Answer

Expert verified
The answers are the values of the integrals calculated in steps 2 and 4.

Step by step solution

01

Set up the Integrand for Part (a)

We are dealing with discs, so the formula for the volume of the solid is given by \(\int_{a}^{b} A(x) dx\), where \(A(x)\) is the area of the cross section at \(x\). The value \(A(x)\) is given by the formula for the area of a circle, \(A(x) = \pi r^2\), where \(r\) is the radius of the disc. In our case, the radius \(r\) is the difference of the equations \(y=\sec x\) and \(y=\tan x\). So, \(A(x) = \pi (\sec x - \tan x)^2\).
02

Evaluate the Integral for Part (a)

We evaluate the definite integral \(\int_{-\pi / 3}^{\pi / 3} \pi (\sec x - \tan x)^2 dx\) to find the volume of the solid.
03

Set up the Integrand for Part (b)

Now we are dealing with squares. The formula for the volume of the solid remains the same, but this time, \(A(x)\) is \((\sec x - \tan x)^2\), which is the area of a square with side length equal to the difference between the curves \(y=\sec x\) and \(y=\tan x\).
04

Evaluate the Integral for Part (b)

We evaluate the definite integral \(\int_{-\pi / 3}^{\pi / 3} (\sec x - \tan x)^2 dx\) to find the volume of the solid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integration
Definite integration is a powerful mathematical tool used to calculate the accumulation of quantities, such as areas under curves and the volumes of solids. In this context, we are using definite integration to determine the volume of a solid that is bounded by specific functions along the x-axis.

The idea here is quite simple: to find the volume of a solid, we take cross-sections perpendicular to an axis (in this case, the x-axis) and sum up the areas of these cross-sections over a given interval. This sum involves integrating the area function of the cross-section.

For example, if the cross-sections are circular disks, the formula for the area of each disk is \(A(x) = \pi r^2\), where \(r\) is the radius. We then integrate this area function over the specified interval of \(x\). The definite integral, written as \(\int_{a}^{b} A(x) \, dx\), represents this accumulation and provides the total volume of the solid.

Definite integration is crucial for solving problems related to volumes of solids as it helps in systematically adding up infinite slices, giving us precise results.
Cross Sections
Cross sections are the key to understanding the volume of irregular solids. Imagine slicing a loaf of bread – each slice is a cross-section of the loaf. Similarly, solid objects can be thought of as a series of congruent or varying cross-sections stacked together.

In our exercise, we're focusing on two types of cross sections:
  • Circular Disks: These are formed between the curves \(y = \tan x\) and \(y = \sec x\). The diameter of these disks is the vertical distance between these curves, and the radius is half this distance. The area of each disk can be calculated with the formula \(\pi (\sec x - \tan x)^2\).
  • Squares: Instead of being round, these cross-sections are squared. The side length of each square is the same as the distance between the curves \(y = \tan x\) and \(y = \sec x\). Hence, the area of each square is \( (\sec x - \tan x)^2\).

As you integrate the areas of these cross-sections from \(x = -\pi/3\) to \(x = \pi/3\), you sum up each slice to get a total volume for the solid. Understanding cross-sections is pivotal for visualizing and solving volume problems.
Trigonometric Functions
Trigonometric functions often serve as boundaries or defining equations for solids in integration problems. They add complexity and a periodic nature to the problems. In this exercise, the functions involved are \(y = \tan x\) and \(y = \sec x\).

These specific functions have unique properties:
  • Tangent Function \(y = \tan x\): This function represents a curve that rises indefinitely and is undefined at odd multiples of \(x = \pi/2\). It's important because it serves as one boundary for the solid in the exercise.
  • Secant Function \(y = \sec x\): This function is the reciprocal of the cosine and is also undefined at \(x = \pi/2\) intervals. The secant curve often lies above the tangent curve, forming the second boundary for the solid.

Understanding how these trigonometric functions behave is critical when setting up your integrals. The trigonometric functions can lead to specific points where the functions might not graph normally, so it becomes essential to identify these points to set up bounds correctly without encountering undefined regions.

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Most popular questions from this chapter

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