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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 6.3.36

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=2−x^2; about the x-axis 

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First, identify the region R bounded by the curves \(y = |x|\) and \(y = 2 - x^2\). Since \(y = |x|\) is piecewise, consider \(y = x\) for \(x \geq 0\) and \(y = -x\) for \(x < 0\). The region is symmetric about the y-axis, so you can analyze for \(x \geq 0\) and then double the volume if needed.
Find the points of intersection between the curves by solving \(|x| = 2 - x^2\). For \(x \geq 0\), solve \(x = 2 - x^2\), which rearranges to \(x^2 + x - 2 = 0\). Find the roots to determine the limits of integration.
Set up the volume integral using the method of washers (disks with holes) since the region is revolved about the x-axis. The volume element is \(\pi (R_{outer}^2 - R_{inner}^2) \, dx\), where \(R_{outer}\) and \(R_{inner}\) are the distances from the x-axis to the outer and inner curves respectively.
Determine which curve is farther from the x-axis in the interval between the points of intersection. Since the revolution is about the x-axis, the radius is the y-value of the curve. For each \(x\), \(R_{outer} = \max(|y|)\) and \(R_{inner} = \min(|y|)\) between the two curves.
Write the integral for the volume as \(V = \pi \int_{a}^{b} \left[(2 - x^2)^2 - (|x|)^2\right] \, dx\), where \(a\) and \(b\) are the intersection points found earlier. Evaluate this integral to find the volume.

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Understanding the Region Bounded by Curves

To find the volume of a solid of revolution, first identify the region bounded by the given curves. This involves finding the points of intersection and understanding which curve lies above or below within the interval. For y = |x| and y = 2 - x², determine where these curves intersect to set the limits of integration.
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Finding Area When Bounds Are Not Given

Method of Disks and Washers

When revolving a region around the x-axis, the volume can be found using the disk or washer method. The disk method applies when the region touches the axis, while the washer method is used if there is a hole. The volume is computed by integrating the area of circular cross-sections perpendicular to the axis of revolution.
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Disk Method Using y-Axis

Handling Absolute Value Functions in Integration

The function y = |x| is piecewise defined, which affects the setup of the integral. Split the integral at x = 0 to handle y = x for x ≥ 0 and y = -x for x < 0 separately. This ensures correct evaluation of the volume by accurately representing the shape of the region.
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Percorso guidato
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Average Value of a Function