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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 6.4.41
Chapter 6, Problem 6.4.41

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.


x =2

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Identify the region R bounded by the curves: \(y = x^{2}\), \(x = 1\), and \(y = 0\). This region lies between \(x=0\) and \(x=1\) above the \(x\)-axis.
Since we are revolving the region about the vertical line \(x = 2\), use the shell method with vertical shells parallel to the axis of revolution. The height of each shell is given by the function \(y = x^{2}\).
Determine the radius of a typical shell. The radius is the horizontal distance from the shell at position \(x\) to the line \(x=2\), which is \(2 - x\).
Write the volume element of a shell as \(dV = 2\pi \times (\text{radius}) \times (\text{height}) \times (\text{thickness}) = 2\pi (2 - x)(x^{2}) \, dx\).
Set up the integral for the volume by integrating \(dV\) from \(x=0\) to \(x=1\): \(V = \int_{0}^{1} 2\pi (2 - x)(x^{2}) \, dx\). This integral represents the volume of the solid generated.

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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 found by multiplying its circumference, height, and thickness. This method is especially useful when the axis of rotation is parallel to the axis of the function being integrated.
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Setting up the Shell Radius and Height

When revolving around a vertical line like x=2, the radius of each shell is the horizontal distance from the shell to the line x=2. The height of the shell corresponds to the function value, here y = x², bounded by y=0. Correctly identifying radius and height expressions is crucial for the integral setup.
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Limits of Integration

The limits of integration correspond to the interval over which the region extends along the axis perpendicular to the shells. Since the region is bounded by x=0 (implied by y=0 and x²) and x=1, the integral limits for x run from 0 to 1. Proper limits ensure the volume calculation covers the entire region.
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Related Practice
Textbook Question

Assume f is a nonnegative function with a continuous first derivative on [a, b]. The curve y=f(x) on [a, b] is revolved about the x-axis. Explain how to find the area of the surface that is generated.

Textbook Question

52–54. Force on a window A diving pool that is 4 m deep and full of water has a viewing window on one of its vertical walls. Find the force on the following windows. 


The window is circular, with a radius of 0.5 m, tangent to the bottom of the pool.

Textbook Question

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


The region in the first quadrant bounded by y=x^2/3 and y=4

Textbook Question

Assume f and g are continuous, with f(x) ≥ g(x) ≥ 0 on [a, b]. The region bounded by the graphs of f and g and the lines x=a and x=b is revolved about the y-axis. Write the integral given by the shell method that equals the volume of the resulting solid.

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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 line.


x=2−secy,x=2,y=π/3, and y=0; about x=2

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Textbook Question

13–20. Mass of one-dimensional objects Find the mass of the following thin bars with the given density function.


ρ(x) = {1 if 0≤x≤2 {2 if 2<x≤3