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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
All textbooksHass 15th EditionCh. 6 - Applications of Definite IntegralsProblem 6.PE.3
Chapter 6, Problem 6.PE.3

Volumes
Find the volumes of the solids in Exercises 1–18.
The solid lies between planes perpendicular to the x-axis at x = π/4 and x = 5π/4. The cross-sections between these planes are circular disks whose diameters run from the curve y = 2 cos x to the curve y = 2 sin x.

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1
Identify the interval over which the volume is to be found: from \(x = \frac{\pi}{4}\) to \(x = \frac{5\pi}{4}\).
Determine the diameter of each circular cross-section at a given \(x\). The diameter is the vertical distance between the two curves, so calculate \(D(x) = |2 \sin x - 2 \cos x|\).
Express the radius of the circular cross-section as half the diameter: \(r(x) = \frac{D(x)}{2} = \frac{|2 \sin x - 2 \cos x|}{2} = |\sin x - \cos x|\).
Write the area of each circular cross-section as a function of \(x\): \(A(x) = \pi [r(x)]^2 = \pi (\sin x - \cos x)^2\) (since squaring removes the absolute value).
Set up the volume integral by integrating the cross-sectional area along the \(x\)-axis between the given bounds: \(V = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \pi (\sin x - \cos x)^2 \, dx\).

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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 its cross-sections perpendicular to an axis. When cross-sections are circular disks, the area is π times the radius squared. The volume is obtained by integrating these areas over the given interval along the axis.
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Introduction to Cross Sections

Determining the Radius from Two Curves

The radius of each circular cross-section is half the distance between the two curves defining the diameter. Here, the diameter is the vertical distance between y = 2 cos x and y = 2 sin x, so the radius is half the absolute difference of these functions at each x.
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Definite Integration with Trigonometric Functions

To find the volume, you integrate the area function involving trigonometric expressions over the interval [π/4, 5π/4]. Understanding how to integrate squares of sine and cosine functions, possibly using identities, is essential for evaluating the integral correctly.
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Definition of the Definite Integral
Related Practice
Textbook Question

Work


Assume that a spring does not follow Hooke’s Law. Instead, the force required to stretch the spring x ft from its natural length is ƒ(𝓍) = 10𝓍³/² lb . How much work does it take to

a. stretch the spring 4 ft from its natural length?

Textbook Question

Volumes

Find the volume of the solid generated by revolving the region bounded by the curve y = sin x and the lines x = 0, x = π and y = 2 about the line y = 2.

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Textbook Question

Volumes

Find the volume of the solid generated by revolving the region bounded on the left by the parabola x = y² + 1 and on the right by the line x = 5 about

a. the x-axis

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Textbook Question

Volumes

Find the volume of the solid generated by revolving the region bounded by the x-axis, the curve y = 3x⁴ , and the lines x = 1 and x = ―1 about

c. the line x = 1

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Textbook Question

Volumes

Find the volume of the solid generated by revolving the region between the x-axis and curve y = x² ―2x about

b. the line y = ―1

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Textbook Question

Find the lengths of the curves in Exercises 19–22.

y = (5/12) x⁶/⁵ ― (5/8)x⁴/⁵ , 1 ≤ x ≤ 32

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