Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Problem 7.3.100

Chapter 7, Problem 7.3.100

Surface area of a catenoid When the catenary y = a cosh x/a is revolved about the x-axis, it sweeps out a surface of revolution called a catenoid. Find the area of the surface generated when y = cosh x on [–ln 2, ln 2] is rotated about the x-axis.

Verified step by step guidance

Identify the curve and the interval: The curve is given by \(y = \cosh x\) on the interval \([-\ln 2, \ln 2]\).

Recall the formula for the surface area of a surface of revolution about the x-axis: \(S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)

Compute the derivative of \(y = \cosh x\): \(\frac{dy}{dx} = \sinh x\)

Substitute \(y\) and \(\frac{dy}{dx}\) into the surface area formula: \(S = \int_{-\ln 2}^{\ln 2} 2\pi \cosh x \sqrt{1 + (\sinh x)^2} \, dx\)

Simplify the expression inside the square root using the identity \(1 + \sinh^2 x = \cosh^2 x\), so the integrand becomes \(2\pi \cosh x \cdot \cosh x = 2\pi \cosh^2 x\). Then set up the integral \(S = \int_{-\ln 2}^{\ln 2} 2\pi \cosh^2 x \, dx\) to be evaluated.

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Key Concepts

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

Surface Area of a Surface of Revolution

The surface area generated by revolving a curve y = f(x) about the x-axis is found using the integral formula S = ∫ 2π y √(1 + (dy/dx)^2) dx over the given interval. This formula accounts for the circumference of circular slices and the curve's slope.

Hyperbolic Cosine Function and Its Derivative

The catenary is defined by y = a cosh(x/a), where cosh x = (e^x + e^{-x})/2. Its derivative, sinh x, is essential for computing the slope dy/dx, which appears in the surface area integral. Understanding these functions helps evaluate the integral accurately.

Evaluating Definite Integrals with Logarithmic Limits

The problem involves integrating over the interval [–ln 2, ln 2]. Recognizing properties of logarithms and symmetry can simplify the integral. Proper evaluation of definite integrals with such limits is crucial for obtaining the exact surface area.

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lim x → ∞ (1 − coth x) / (1 − tanh x)

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Textbook Question

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Textbook Question

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