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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 6.4.43
Chapter 6, Problem 6.4.43

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.


y = -2

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First, identify the region R bounded by the curves: \(y = x^{2}\), \(x = 1\), and \(y = 0\). This region lies between \(x=0\) and \(x=1\) since \(y=0\) is the x-axis and \(y=x^{2}\) is a parabola opening upwards.
Since we are revolving the region about the line \(y = -2\), which is horizontal and below the x-axis, we will use the shell method with vertical shells. The shells will be formed by vertical slices of the region at position \(x\) with thickness \(dx\).
The height of each shell is the vertical distance between the curves \(y = x^{2}\) (top) and \(y = 0\) (bottom), so the height is \(h(x) = x^{2} - 0 = x^{2}\).
The radius of each shell is the distance from the slice at \(y\) to the axis of rotation \(y = -2\). Since the shell is at height \(y\) between \(0\) and \(x^{2}\), the radius is the vertical distance from \(y\) to \(-2\). However, because we are integrating with respect to \(x\), the radius is the distance from the shell at \(y = x^{2}\) to \(y = -2\), which is \(r(x) = x^{2} - (-2) = x^{2} + 2\).
The volume of the solid is given by the integral using the shell method formula: \(V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx\). Substitute the expressions for radius and height and the limits \(a=0\) and \(b=1\): \(V = 2\pi \int_{0}^{1} (x^{2} + 2)(x^{2}) \, dx\).

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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 using the shell method, the radius is the distance from the axis of rotation to the shell, and the height is the length of the shell segment. For rotation about y = -2, the radius is the vertical distance from y = -2 to a point y on the region, and the height corresponds to the horizontal length of the region at that y.
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Region Boundaries and Limits of Integration

The region R is bounded by y = x², x = 1, and y = 0, defining the shape to be revolved. Understanding these boundaries helps determine the limits of integration and the expressions for height and radius in terms of the variable of integration, ensuring accurate volume calculation.
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Related Practice
Textbook Question

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

Textbook Question

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


A hole of radius r≤R is drilled symmetrically along the axis of a bullet. The bullet is formed by revolving the parabola y = 6(1−x²/R²) about the y-axis, where 0≤x≤R.

Textbook Question

Lengths of symmetric curves Suppose a curve is described by y=f(x) on the interval [−b, b], where f′ is continuous on [−b, b]. Show that if f is odd or f is even, then the length of the curve y=f(x) from x=−b to x=b is twice the length of the curve from x=0 to x=b. Use a geometric argument and prove it using integration.

Textbook Question

Explain the steps required to find the length of a curve x = g(y) between y=c and y=d.

Textbook Question

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

x = 2e^√2y + 1/16e^−√2y, for 0 ≤ y ≤ ln²/√2

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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. 


{Use of Tech} y = √50 -2x², in the first quadrant; about the x-axis