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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.R.102

102–105. Volumes The region R is bounded by the curve y = ln(x) and the x-axis on the interval [1, e]. Find the volume of the solid generated when R is revolved in the following ways.
102. About the y-axis

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Identify the region R bounded by the curve \(y = \ln(x)\) and the x-axis on the interval \([1, e]\). This means the region lies between \(x=1\) and \(x=e\), above the x-axis and below the curve \(y=\ln(x)\).
Since the solid is generated by revolving the region around the y-axis, consider using the method of cylindrical shells. The formula for the volume using shells is \(V = \int_a^b 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dx\).
In this problem, the radius of a shell is the distance from the y-axis to a point \(x\), which is simply \(x\). The height of the shell is the value of the function \(y = \ln(x)\).
Set up the integral for the volume as \(V = \int_1^e 2\pi x \ln(x) \, dx\). This integral represents the sum of the volumes of all cylindrical shells from \(x=1\) to \(x=e\).
To find the volume, evaluate the integral \(\int_1^e x \ln(x) \, dx\) using integration by parts, then multiply the result by \(2\pi\).

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The shell method calculates volume by integrating cylindrical shells formed when a region is revolved around an axis. For rotation about the y-axis, vertical slices parallel to the axis create shells with radius equal to the x-value and height given by the function. The volume is found by integrating 2π(radius)(height) dx over the interval.
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