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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 6.4.19
Chapter 6, Problem 6.4.19

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 = 1−x²,x = 0, and y = 0, in the first quadrant; about the y-axis

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Identify the region R bounded by the curves: \(y = 1 - x^2\), \(x = 0\), and \(y = 0\) in the first quadrant. This means \(x \geq 0\) and \(0 \leq y \leq 1 - x^2\).
Since the solid is generated by revolving the region about the y-axis, use the shell method. The shell method formula for volume is: \(V = \int_a^b 2\pi (\text{radius})(\text{height}) \, dx\) where the shells are vertical slices parallel to the axis of revolution.
Determine the radius and height of a typical shell. Here, the radius is the distance from the y-axis to the shell, which is \(x\). The height is the value of \(y\) on the curve, which is \(1 - x^2\).
Set up the integral limits for \(x\). Since the region is in the first quadrant and bounded by \(x=0\) and \(y=0\), find the \(x\)-values where \(y=0\): \(1 - x^2 = 0\) implies \(x = 1\). So the limits are from \(0\) to \(1\).
Write the integral for the volume using the shell method: \(V = \int_0^1 2\pi x (1 - x^2) \, dx\). This integral represents the volume of the solid formed by revolving the region about the y-axis.

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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. This method is especially useful when revolving around vertical or horizontal axes where slicing perpendicular to the axis is complex.
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Finding Volume Using Disks

Region Bounded by Curves

Understanding the region bounded by y = 1 - x², x = 0, and y = 0 in the first quadrant is essential. This region lies between the parabola, the y-axis, and the x-axis, defining the limits of integration. Identifying these boundaries helps set up the integral correctly for volume calculation.
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Finding Area When Bounds Are Not Given

Revolution About the y-axis

Revolving the region about the y-axis means the shells are vertical, with radius equal to the x-value and height given by the function y = 1 - x². Recognizing the axis of revolution determines the radius and height expressions in the shell method integral.
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Disk Method Using y-Axis
Related Practice
Textbook Question

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

Textbook Question

Set up a sum of two integrals that equals the area of the shaded region bounded by the graphs of the functions f and g on [a, c] (see figure). Assume the curves intersect at x=b.

Textbook Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

y = (x²+2)^3/2 / 3 on [0, 1]

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Textbook Question

A 1.5-mm layer of paint is applied to one side of the following surfaces. Find the approximate volume of paint needed. Assume x and y are measured in meters. 


The spherical zone generated when the curve y=√8x−x^2 on the interval 1≤x≤7 is revolved about the x-axis 

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Textbook Question

For the following regions R, determine which is greater—the volume of the solid generated when R is revolved about the x-axis or about the y-axis.


R is bounded by y=x^2 and y=√8x.

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Textbook Question

Let f(x) = {x   if 0≤x≤2

      2x−2  if 2<x≤5

      −2x+18 if 5<x≤6. 


Find the volume of the solid formed when the region bounded by the graph of f, the x-axis, and the line x=6 is revolved about the x-axis.