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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Briggs 3rd EditionCh. 6 - Applications of Integration문제 6.4.30
6장, 문제 6.4.30

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 = In x/x²,y = 0,x = 3, about the y-axis

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First, identify the region R bounded by the curves: \(y = \frac{\ln x}{x^2}\), \(y = 0\), and \(x = 3\). Since the region is revolved about the y-axis, we will use the shell method with respect to \(x\).
Recall the shell method formula for volume when revolving around the y-axis: \(V = 2\pi \int_a^b (\text{radius})(\text{height}) \, dx\). Here, the radius of a shell is the distance from the y-axis, which is \(x\), and the height is the function value \(y = \frac{\ln x}{x^2}\).
Set up the integral limits from \(x = 1\) to \(x = 3\) because \(y = \frac{\ln x}{x^2}\) is defined and positive between these points, and the region is bounded by \(y=0\) (the x-axis). Note that \(x=1\) is where \(y=0\) since \(\ln 1 = 0\).
Write the volume integral as: \(V = 2\pi \int_1^3 x \cdot \frac{\ln x}{x^2} \, dx\). Simplify the integrand to \(2\pi \int_1^3 \frac{\ln x}{x} \, dx\).
To find the volume, evaluate the integral \(\int_1^3 \frac{\ln x}{x} \, dx\) using integration techniques such as substitution or integration by parts, then multiply the result by \(2\pi\).

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주요 개념

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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. When revolving around the y-axis, shells are vertical slices parallel to the axis, and the radius is the x-value of the shell.
추천 영상:
04:48
Finding Volume Using Disks

Setting up the Integral with Given Curves

To use the shell method, identify the height and radius of each shell from the given curves. Here, the height is the function y = (ln x) / x², bounded below by y = 0, and the radius is the distance from the y-axis, which is x. The limits of integration are from x = 1 (where ln x / x² > 0) to x = 3.
추천 영상:
05:23
Finding Area Between Curves on a Given Interval

Properties of the Function y = (ln x) / x²

Understanding the behavior of y = (ln x) / x² is crucial for setting correct bounds and ensuring the function is positive over the interval. The natural logarithm ln x is positive for x > 1, and dividing by x² affects the shape, so the region lies above y=0 between x=1 and x=3.
추천 영상:
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06:21
Properties of Functions
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