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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 6.4.33
Chapter 6, Problem 6.4.33

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis. 


y = x³−x⁸+1,y=1; about the y-axis

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First, identify the region R bounded by the curves given: \(y = x^{3} - x^{8} + 1\) and \(y = 1\). Since the region is revolved about the y-axis, we will use the shell method with respect to \(x\).
Set up the shell radius and height. The radius of a typical shell is the distance from the y-axis, which is \(x\). The height of the shell is the vertical distance between the curves, which is \(y - 1 = (x^{3} - x^{8} + 1) - 1 = x^{3} - x^{8}\).
Determine the interval for \(x\) over which the region exists. Since the region is bounded by \(y = 1\) and \(y = x^{3} - x^{8} + 1\), find the values of \(x\) where these two curves intersect by solving \(x^{3} - x^{8} + 1 = 1\), which simplifies to \(x^{3} - x^{8} = 0\).
Express the volume integral using the shell method formula: \(V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx = 2\pi \int_{a}^{b} x (x^{3} - x^{8}) \, dx\) where \(a\) and \(b\) are the intersection points found in the previous step.
Simplify the integrand to \(x^{4} - x^{9}\) and set up the definite integral for volume: \(V = 2\pi \int_{a}^{b} (x^{4} - x^{9}) \, dx\). The final step would be to evaluate this integral to find the volume.

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Key Concepts

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

Shell Method for Volume

The shell method calculates the volume of a solid of revolution by integrating cylindrical shells. Each shell's volume is approximated by its circumference times height times thickness. When revolving around the y-axis, shells are vertical slices parallel to the axis, and the radius is the x-value of the shell.
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Finding Volume Using Disks

Setting up the Integral with Given Curves

To use the shell method, identify the region bounded by the curves y = x³ − x⁸ + 1 and y = 1. The height of each shell is the vertical distance between these curves, and the limits of integration correspond to the x-values where the region exists. Understanding the relationship between x and y is crucial.
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Revolution about the y-axis

Revolving a region about the y-axis means the shells are formed by rotating vertical slices around this axis. The radius of each shell is the horizontal distance from the y-axis to the slice, which equals the x-coordinate. This affects the integral's setup, as the radius function depends on x.
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Related Practice
Textbook Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=x and y=4√x; about the x-axis

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Textbook Question

Find the area of the region described in the following exercises.


The region bounded by y=|x−3|and y=x/2

Textbook Question

What is the area of the curved surface of a right circular cone of radius 3 and height 4?

Textbook Question

53–62. Choose your method Let R be the region bounded by the following curves. Use the method of your choice to find the volume of the solid generated when R is revolved about the given axis.

y = x²,y=2−x, and x=0, in the first quadrant; about the y-axis

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Textbook Question

Find the area of the region described in the following exercises.


The region bounded by y=2 / 1 + x^2 and y=1

Textbook Question

Determine the area of the shaded region in the following figures.

(Hint: Find the intersection point by inspection.)