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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.R.68
Chapter 6, Problem 6.R.68

Surface area of a cone Find the surface area of a cone (excluding the base) with radius 4 and height 8 using integration and a surface area integral.

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Recall that the lateral surface area of a cone can be found by revolving a line segment (the slant height) around the axis. To use integration, first express the cone's slant height as a function. Set up a coordinate system where the cone's height extends along the x-axis from 0 to 8, and the radius varies linearly from 0 to 4 along this height.
Find the equation of the line representing the radius as a function of height x. Since the radius increases linearly from 0 at x=0 to 4 at x=8, the radius function is \(r(x) = \frac{4}{8}x = \frac{x}{2}\).
The surface area generated by revolving a curve \(y = f(x)\) about the x-axis from \(x=a\) to \(x=b\) is given by the formula \(S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\). Here, \(y = r(x) = \frac{x}{2}\).
Compute the derivative \(\frac{dy}{dx}\) of the radius function: \(\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x}{2} \right) = \frac{1}{2}\). Substitute \(y\) and \(\frac{dy}{dx}\) into the surface area integral to get \(S = \int_0^8 2\pi \left( \frac{x}{2} \right) \sqrt{1 + \left( \frac{1}{2} \right)^2} \, dx\).
Simplify the integrand and set up the integral for evaluation. The integral becomes \(S = \int_0^8 \pi x \sqrt{1 + \frac{1}{4}} \, dx = \int_0^8 \pi x \sqrt{\frac{5}{4}} \, dx\). This integral can then be evaluated to find the lateral surface area.

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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 Solid of Revolution

The surface area of a solid formed by revolving a curve around an axis can be found using an integral formula. For a curve y = f(x) revolved about the x-axis, the surface area is the integral of 2π times the radius times the arc length differential. This approach generalizes to various shapes, including cones.
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Example 1: Minimizing Surface Area

Equation of the Cone's Slant Line

To set up the integral, express the cone's slant height as a function relating radius and height. For a cone with radius 4 and height 8, the radius changes linearly with height, forming a straight line. This linear function is essential to define the curve to revolve and compute the surface area.
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Equations of Tangent Lines

Arc Length Differential (ds)

The arc length differential ds represents an infinitesimal segment of the curve and is given by ds = sqrt(1 + (dy/dx)^2) dx or similarly in terms of y. It accounts for the slant of the surface when integrating to find surface area, ensuring the curved surface is measured accurately.
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Related Practice
Textbook Question

Use the general slicing method to find the volume of the following solids.

The solid whose base is the region bounded by the semicircle y=√1−x^2 and the x-axis, and whose cross sections through the solid perpendicular to the x-axis are squares

Textbook Question

Surface area and volume Let f(x) = 1/3 x³ and let R be the region bounded by the graph of f and the x-axis on the interval [0, 2].


c. Find the volume of the solid generated when R is revolved about the x-axis.

Textbook Question

43–55. Volumes of solids Choose the general slicing method, the disk/washer method, or the shell method to answer the following questions.


The region bounded by the curves y = sec x and y=2, for 0 ≤ x ≤ π/3, is revolved about the x-axis. What is the volume of the solid that is generated? 

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Textbook Question

Find the area of the following regions, expressing your results in terms of the positive integer n≥2.


Let Aₙ be the area of the region bounded by f(x)=x^1/n and g(x)=x^n on the interval [0,1], where n is a positive integer. Evaluate lim n→∞ Aₙ and interpret the result. br

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Textbook Question

Suppose a cut is made through a solid object perpendicular to the x-axis at a particular point x. Explain the meaning of A(x).

Textbook Question

27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure) are formed by the graphs of y = 2√x,y = 3−x,and x=3.


Find the volume of the solid obtained by revolving region R₂ about the y-axis.

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