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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 8 - Techniques of IntegrationProblema 8.8.90c
Capitolo 8, Problema 8.8.90c

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 (ii) about the y-axis.

Guida verificata passo dopo passo
1
First, 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 \( x = 1 \), above the x-axis and below the curve \( y = \frac{1}{\sqrt{x}} \).
Since the solid is formed by revolving this region about the y-axis, we will use the method of cylindrical shells. The formula for the volume using cylindrical shells when revolving around the y-axis is: \(\n\[\n\)\[ V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx \]\(\n\]\nwhere\) the radius is the distance from the y-axis (which is \( x \)) and the height is the function value \( y = \frac{1}{\sqrt{x}} \).
Set up the integral with the limits of integration from \( x = 0 \) to \( x = 1 \): \(\n\[\n\)\[ V = 2\pi \int_{0}^{1} x \cdot \frac{1}{\sqrt{x}} \, dx \]\(\n\]\nSimplify\) the integrand before integrating.
Simplify the integrand \( x \cdot \frac{1}{\sqrt{x}} = x \cdot x^{-1/2} = x^{1 - \frac{1}{2}} = x^{\frac{1}{2}} \). So the integral becomes \(\n\)\(\n\)\[ V = 2\pi \int_{0}^{1} x^{\frac{1}{2}} \, dx \]
Evaluate the integral \( \int_{0}^{1} x^{\frac{1}{2}} \, dx \) using the power rule for integration: \(\n\[\n\)\[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \]\(\n\]\nApply\) the limits from 0 to 1 and multiply by \( 2\pi \) to express the volume.

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Concetti chiave

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Setting up the region bounded by curves

Understanding the region involves identifying the area enclosed by the given curves y = 1/√x, y = 0, x = 0, and x = 1 in the first quadrant. This means recognizing the limits of integration and the shape formed, which is essential before applying volume formulas.
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Finding Area Between Curves on a Given Interval

Method of cylindrical shells for volume

When revolving a region around the y-axis, the cylindrical shells method is often used. It involves integrating the volume of thin cylindrical shells with radius equal to the x-value, height given by the function, and thickness dx, allowing calculation of the solid's volume.
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Finding Volume Using Disks

Integration with respect to x

Since the region is bounded between x = 0 and x = 1, and the function is given as y in terms of x, the volume integral is set up with respect to x. Proper integration techniques must be applied to evaluate the integral and find the exact volume.
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Integrals of Natural Exponential Functions (e^x)
Pratica correlata
Domanda del libro di testo

88. The region in Exercise 87 is revolved about the x-axis to generate a solid.

b. Show that the inner and outer surfaces of the solid have infinite area.

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Finding area

Find the area of the region enclosed by the curve y = x sin(x) and the x-axis (see the accompanying figure) for:

c. 2π ≤ x ≤ 3π.

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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.)

∫ from 1 to 2 of x dx

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Using different substitutions

Show that the integral

∫((x² - 1)(x + 1))^(-2/3) dx

can be evaluated with any of the following substitutions.

c. u = arctan x

What is the value of the integral?

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Centroid:

Find the centroid of the region cut from the first quadrant by the curve

y = 1/√(x + 1) and the line x = 3.

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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 (ii) about the y-axis.

19
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