Skip to main content
Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.6.135a
Chapter 7, Problem 7.6.135a

Find the volumes of the solids in Exercises 135 and 136.
135. The solid lies between planes perpendicular to the x-axis at x=-1 and x=1. The cross-sections perpendicular to the x-axis are
a. circles whose diameters stretch from the curve y=-1/√(1+x²) to the curve y=1/√(1+x²).

Verified step by step guidance
1
Identify the interval over which the solid extends along the x-axis, which is from \(x = -1\) to \(x = 1\).
Determine the length of the diameter of each circular cross-section at a given \(x\). The diameter stretches from \(y = -\frac{1}{\sqrt{1+x^2}}\) to \(y = \frac{1}{\sqrt{1+x^2}}\), so the diameter length is the difference between these two values.
Calculate the diameter length as \(D(x) = \frac{1}{\sqrt{1+x^2}} - \left(-\frac{1}{\sqrt{1+x^2}}\right) = \frac{2}{\sqrt{1+x^2}}\).
Find the radius of the circular cross-section as half the diameter: \(r(x) = \frac{D(x)}{2} = \frac{1}{\sqrt{1+x^2}}\).
Write the formula for the volume of the solid using the integral of the cross-sectional area: \(V = \int_{-1}^{1} \pi [r(x)]^2 \, dx = \int_{-1}^{1} \pi \left(\frac{1}{\sqrt{1+x^2}}\right)^2 \, dx\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
9m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Volume of a Solid with Known Cross-Sections

To find the volume of a solid with cross-sections perpendicular to an axis, integrate the area of each cross-section along that axis. The volume V is given by V = ∫ A(x) dx, where A(x) is the area of the cross-section at position x. This method applies when the shape of the cross-section is known as a function of x.
Recommended video:
05:38
Introduction to Cross Sections

Area of a Circle from Diameter

The area of a circle is A = πr², where r is the radius. If the diameter d is known, the radius is r = d/2, so the area becomes A = π(d/2)² = (π/4)d². In this problem, the diameter is the vertical distance between two curves, which varies with x.
Recommended video:
05:06
Finding Area When Bounds Are Not Given

Using Curves to Determine Cross-Section Dimensions

The diameter of each circular cross-section is the distance between two curves y = f(x) and y = g(x). This distance is |f(x) - g(x)|. Here, the curves y = 1/√(1+x²) and y = -1/√(1+x²) define the endpoints of the diameter, so the diameter is the vertical distance between these two functions at each x.
Recommended video:
05:38
Introduction to Cross Sections
Related Practice
Textbook Question

Evaluate the integrals in Exercises 67–74 in terms of

b. natural logarithms.

73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)

1
views
Textbook Question

In Exercises 41–44:

a. Find f⁻¹(x).


41. f(x) = 2x + 3, a = −1

Textbook Question

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.

7. a. sec^(-1)(-√2)

Textbook Question

In Exercises 41–44:

a. Find f⁻¹(x).


42. f(x) = (x + 2) / (1 − x), a = 1/2

Textbook Question

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:


b. Solve the equation y=f(x) for x as a function of y, and name the resulting inverse function g.


70. y= x³/(x²+1), -1 ≤ x ≤ 1, x_0=1/2

1
views
Textbook Question

21. a. Show that ln(x) grows slower as x→∞ than x^(1/n) for any positive integer n, even x^(1/1,000,000).