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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 6.4.13
Chapter 6, Problem 6.4.13

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.
Graph showing region R bounded by y = √x, y = 0, and x = 4, with a shell method illustration for volume calculation.
y = √x,y=0, and x=4; about the x-axis

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Identify the region R bounded by the curves: \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\). This region lies between the x-axis and the curve \(y = \sqrt{x}\) from \(x=0\) to \(x=4\).
Since the solid is generated by revolving the region R about the x-axis, and we are asked to use the shell method, we consider vertical shells parallel to the axis of revolution. However, the shell method is typically easier when revolving around the y-axis or a vertical line. Here, revolving around the x-axis, it is more natural to use shells in terms of \(y\).
Express \(x\) in terms of \(y\) from the curve \(y = \sqrt{x}\). Squaring both sides gives \(x = y^2\). The bounds for \(y\) are from \(0\) to \(2\) because when \(x=4\), \(y=\sqrt{4}=2\).
The shell radius is the distance from the shell to the axis of rotation (x-axis), which is simply \(y\). The shell height is the horizontal length of the shell, which is from \(x=0\) to \(x=y^2\), so the height is \(y^2\).
Set up the volume integral using the shell method formula: \(V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dy = 2\pi \int_0^2 y \cdot y^2 \, dy = 2\pi \int_0^2 y^3 \, dy\). This integral will give the volume of the solid.

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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 Integral with Respect to y

Since the region is revolved about the x-axis, and the shell method involves cylindrical shells perpendicular to the axis of rotation, the integral is set up with respect to y. The radius of each shell is the distance from y to the x-axis, and the height corresponds to the horizontal length of the region at that y-value.
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Understanding the Region Bounded by Curves

The region R is bounded by y = √x, y = 0, and x = 4. This defines a finite area under the curve y = √x from x = 0 to x = 4. Understanding these boundaries is essential to determine the limits of integration and the expressions for radius and height in the shell method.
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Related Practice
Textbook Question

64–68. Shell method Use the shell method to find the volume of the following solids.


The solid formed when a hole of radius 3 is drilled symmetrically along the axis of a right circular cone of radius 6 and height 9

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Textbook Question

Find the area of the surface generated when the given curve is revolved about the given axis.


y=√1−x^2, for −1/2≤x≤1/2; about the x-axis

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Textbook Question

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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Textbook Question

Determine the area of the shaded region bounded by the curve x^2=y^4(1−y^3) (see figure).

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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 axis.

y=ln x,y=ln x^2; and y=ln 8; about the y-axis

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Textbook Question

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

(Hint: Find the intersection point by inspection.)

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