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Ch. 6 - Applications of Definite Integrals
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. 6 - Applications of Definite IntegralsProblema 6.PE.15
Capitolo 6, Problema 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.

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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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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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Finding Volume Using Disks

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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Finding Volume Using Disks

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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Percorso guidato
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Fundamental Theorem of Calculus Part 1
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Work


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

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Volumes

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

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

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Volumes

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

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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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Volumes

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

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