Domanda del libro di testo
Consider the region bounded by the graphs of
y = arctan(x), y = 0, and x = 1.
b. Find the volume of the solid formed by revolving this region about the y-axis.
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Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 8, Problema 8.8.90b
90. Consider the infinite region in the first quadrant bounded by the graphs of
y = 1 / √x, y = 0, x = 0, and x = 1.
b. Find the volume of the solid formed by revolving the region (i) about the x-axis
Guida verificata passo dopo passo1
Identify the region bounded by the curves: \(y = \frac{1}{\sqrt{x}}\), \(y = 0\), \(x = 0\), and \(x = 1\). This region lies in the first quadrant between \(x=0\) and \(x=1\) under the curve \(y = \frac{1}{\sqrt{x}}\).
Since the region is revolved about the x-axis, use the disk method to find the volume. The volume \(V\) is given by the integral \(V = \pi \int_{a}^{b} [f(x)]^{2} \, dx\), where \(f(x)\) is the function representing the radius of the disks.
Here, the radius of each disk is \(y = \frac{1}{\sqrt{x}}\). So, the volume integral becomes \(V = \pi \int_{0}^{1} \left( \frac{1}{\sqrt{x}} \right)^{2} \, dx\).
Simplify the integrand: \(\left( \frac{1}{\sqrt{x}} \right)^{2} = \frac{1}{x}\). So the integral is \(V = \pi \int_{0}^{1} \frac{1}{x} \, dx\).
Evaluate the integral \(\int_{0}^{1} \frac{1}{x} \, dx\). Note that this integral is improper at \(x=0\), so consider it as a limit: \(\lim_{t \to 0^{+}} \int_{t}^{1} \frac{1}{x} \, dx\). Set up this limit to analyze the behavior of the volume.
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Definite Integrals for Area and Volume
Definite integrals calculate the accumulation of quantities, such as area under a curve or volume of a solid. In this problem, the integral bounds are from 0 to 1, and the integral will be used to find the volume generated by revolving a region around an axis.
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Percorso guidato
05:43
Definition of the Definite Integral
Volume of Solids of Revolution (Disk/Washer Method)
When a region is revolved around an axis, the volume of the resulting solid can be found using the disk or washer method. This involves integrating the area of circular cross-sections perpendicular to the axis of revolution, typically using the formula V = π∫[R(x)]² dx.
Video consigliato:
04:48Finding Volume Using Disks
Understanding the Region Bounded by Curves
Identifying the region bounded by y = 1/√x, y = 0, x = 0, and x = 1 is crucial. This region lies in the first quadrant and defines the shape to be revolved. Correctly interpreting these boundaries ensures the integral is set up with proper limits and functions.
Video consigliato:
05:06Finding Area When Bounds Are Not Given
Pratica correlata
Domanda del libro di testo
89. Consider the infinite region in the first quadrant bounded by the graphs of
y = 1 / x², y = 0, and x = 1.
b. Find the volume of the solid formed by revolving the region (i) about the x-axis.
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Domanda del libro di testo
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Find the area of the region enclosed by the curve y = x sin(x) and the x-axis (see the accompanying figure) for:
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Domanda del libro di testo
Finding volume: Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes and the curve y = cos(x), 0 ≤ x ≤ π/2, about
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60
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Domanda del libro di testo
In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)
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