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Ch. 6 - Applications of Definite Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 6, Problem 6.PE.1

Volumes
Find the volumes of the solids in Exercises 1–18.
The solid lies between planes perpendicular to the x-axis at x = 0 and x = 1. The cross-sections perpendicular to the x-axis between these planes are circular disks whose diameters run from the parabola y = x² to the parabola y = √x.

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Identify the region bounded by the two curves: the parabola \(y = x^{2}\) and the curve \(y = \sqrt{x}\), between \(x = 0\) and \(x = 1\).
Since the cross-sections perpendicular to the x-axis are circular disks with diameters running from \(y = x^{2}\) to \(y = \sqrt{x}\), find the length of the diameter at a general point \(x\) by subtracting the lower curve from the upper curve: \(D(x) = \sqrt{x} - x^{2}\).
Calculate the radius of each circular cross-section as half the diameter: \(r(x) = \frac{D(x)}{2} = \frac{\sqrt{x} - x^{2}}{2}\).
Write the area of the circular cross-section as a function of \(x\): \(A(x) = \pi [r(x)]^{2} = \pi \left( \frac{\sqrt{x} - x^{2}}{2} \right)^{2}\).
Set up the volume integral by integrating the cross-sectional area from \(x = 0\) to \(x = 1\): \(V = \int_{0}^{1} A(x) \, dx = \int_{0}^{1} \pi \left( \frac{\sqrt{x} - x^{2}}{2} \right)^{2} \, dx\). This integral will give the volume of the solid.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Volume of Solids with Known Cross-Sections

This concept involves finding the volume of a solid by integrating the area of cross-sectional slices perpendicular to an axis. Each cross-section's area is expressed as a function of the variable of integration, and integrating these areas over the given interval yields the total volume.
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Introduction to Cross Sections

Area of a Circular Cross-Section

When cross-sections are circular disks, their area is calculated using the formula A = πr², where r is the radius. In this problem, the diameter is given by the vertical distance between two curves, so the radius is half that distance, which must be expressed as a function of x.
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Introduction to Cross Sections Example 2

Using Functions to Determine Cross-Section Dimensions

The diameters of the circular cross-sections are determined by the vertical distance between two curves, y = x² and y = √x. Understanding how to find this distance at each x-value is essential to express the diameter (and thus radius) as a function for integration.
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Introduction to Cross Sections
Related Practice
Textbook Question

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Textbook Question

Volumes

Find the volumes of the solids in Exercises 1–18.

The solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. The cross-sections of the solid perpendicular to the x-axis between these planes are circular disks whose diameters run from the curve x² = 4y to the curve y² = 4x.

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Textbook Question

Centers of Mass and Centroids

Find the centroid of a thin, flat plate covering the region enclosed by the parabolas 𝔂 = 2𝓍² and 𝔂 = 3 ― 𝓍² .

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Textbook Question

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In Exercises 23–26, find the areas of the surfaces generated by revolving the curves about the given axes.

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y = √2x + 1 , 0 ≤ x ≤ 3 ; x-axis"

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Textbook Question

Work

Pumping a conical tank A right-circular conical tank, point down, with top radius 5 ft and height 10 ft, is filled with a liquid whose weight-density is 60lb/ft³. How much work does it take to pump the liquid to a point 2 ft above the tank? If the pump is driven by a motor rated at 275ft-lb/sec (1/2 hp), how long will it take to empty the tank? 

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Textbook Question

Centers of Mass and Centroids

Find the center of mass of a thin, flat plate covering the region enclosed by the parabola 𝔂² = 𝓍 and the line 𝓍 = 2𝔂 if the density function is δ(𝔂) = 1 + 𝔂. (Use horizontal strips.)

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