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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.15
Chapter 6, Problem 6.PE.15

Volumes
Volume of a solid sphere hole A round hole of radius √3 ft is bored through the center of a solid sphere of radius 2 ft. Find the volume of material removed from the sphere.

Verified step by step guidance
1
Identify the problem: A cylindrical hole of radius \(\sqrt{3}\) ft is bored through the center of a solid sphere of radius 2 ft. We need to find the volume of the material removed, which is the volume of the cylindrical hole including the spherical caps removed at the ends.
Set up the coordinate system: Place the sphere centered at the origin with radius \(R = 2\) ft. The hole is drilled along the \(z\)-axis, creating a cylindrical hole of radius \(r = \sqrt{3}\) ft.
Express the volume of the removed material as the volume of the cylinder minus the volume of the spherical caps. Alternatively, use the method of slicing: For each \(z\) between \(-h\) and \(h\), find the cross-sectional area of the hole and integrate.
Determine the height \(h\) of the hole inside the sphere using the Pythagorean theorem: Since the hole radius is \(r = \sqrt{3}\), and the sphere radius is \(R = 2\), the half-height \(h\) satisfies \(h = \sqrt{R^2 - r^2}\). This gives the limits of integration for \(z\) from \(-h\) to \(h\).
Set up the integral for the volume of the removed material as \(V = \int_{-h}^{h} \pi r^2 \, dz = \pi r^2 (2h)\), which is the volume of the cylindrical hole. Since the hole removes material from the sphere, this volume corresponds to the volume of the material removed.

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Key Concepts

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

Volume of a Sphere

The volume of a sphere is given by the formula V = (4/3)πr³, where r is the radius. This formula helps determine the total volume of the original solid sphere before any material is removed.
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Volume of a Cylindrical Hole

A cylindrical hole bored through a sphere can be modeled as a cylinder with radius equal to the hole's radius and height equal to the chord length through the sphere. Calculating this volume involves understanding the geometry of the intersection.
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Using the Pythagorean Theorem in Solids of Revolution

The Pythagorean theorem helps relate the radius of the sphere, the radius of the hole, and the height of the cylindrical hole. This relationship is essential to find the length of the hole and thus compute the volume of the removed material.
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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 between the x-axis and curve y = x² ―2x about

b. the line y = ―1

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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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.

1
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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 .

1
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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 = 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.

1
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