Explanation Find the radius of the washer that is obtained by the intersection of the cylinder with the sphere. Then determine the volume of the cross-section. Given: V = π 6 h 3 . Since V can be expressed in terms of h alone, it follows that two beads of height 1 cm, one formed from a sphere the size of an orange and the other from a sphere the size of the earth would have the same volume. Calculation: The equation for the outer circle of the bead is given by x 2 + y 2 = R 2 and the inner cylinder intersects the sphere when y = ± R 2 − r 2 . Each cross-section of the bead will be a washer with outer radius given by R 2 − y 2 and inner radius r . The volume will be therefore, π ∫ − R 2 − r 2 R 2 − r 2 ( ( R 2 − y 2 ) 2 − r 2 ) d y = π ∫ − R 2 − r 2 R 2 − r 2 ( R 2 − y 2 − r 2 ) d y = π [ ( R 2 y − r 2 y − y 3 ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Question

Chapter 6.3, Problem 60E

To determine

The volume $V$ of the bead (Figure 17) in terms of $r$ and $R$ . Then show that $V = \frac{π}{6} h^{3}$ , where $h$ is the height of the bead and then explain intuitively, how the given statement is possible.

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Chapter 6.3, Problem 59E

Chapter 6.3, Problem 61E

Chapter 6 Solutions

Calculus - Standalone book

Ch. 6.1 - Prob. 1PQ Ch. 6.1 - Prob. 2PQ Ch. 6.1 - Prob. 3PQ Ch. 6.1 - Prob. 4PQ Ch. 6.1 - Prob. 5PQ Ch. 6.1 - Prob. 6PQ Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E

Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E Ch. 6.1 - Prob. 7E Ch. 6.1 - Prob. 8E Ch. 6.1 - Prob. 9E Ch. 6.1 - Prob. 10E Ch. 6.1 - Prob. 11E Ch. 6.1 - Prob. 12E Ch. 6.1 - Prob. 13E Ch. 6.1 - Prob. 14E Ch. 6.1 - Prob. 15E Ch. 6.1 - Prob. 16E Ch. 6.1 - Prob. 17E Ch. 6.1 - Prob. 18E Ch. 6.1 - Prob. 19E Ch. 6.1 - Prob. 20E Ch. 6.1 - Prob. 21E Ch. 6.1 - Prob. 22E Ch. 6.1 - Prob. 23E Ch. 6.1 - Prob. 24E Ch. 6.1 - Prob. 25E Ch. 6.1 - Prob. 26E Ch. 6.1 - Prob. 27E Ch. 6.1 - Prob. 28E Ch. 6.1 - Prob. 29E Ch. 6.1 - Prob. 30E Ch. 6.1 - Prob. 31E Ch. 6.1 - Prob. 32E Ch. 6.1 - Prob. 33E Ch. 6.1 - Prob. 34E Ch. 6.1 - Prob. 35E Ch. 6.1 - Prob. 36E Ch. 6.1 - Prob. 37E Ch. 6.1 - Prob. 38E Ch. 6.1 - Prob. 39E Ch. 6.1 - Prob. 40E Ch. 6.1 - Prob. 41E Ch. 6.1 - Prob. 42E Ch. 6.1 - Prob. 43E Ch. 6.1 - Prob. 44E Ch. 6.1 - Prob. 45E Ch. 6.1 - Prob. 46E Ch. 6.1 - Prob. 47E Ch. 6.1 - Prob. 48E Ch. 6.1 - Prob. 49E Ch. 6.1 - Prob. 50E Ch. 6.1 - Prob. 51E Ch. 6.1 - Prob. 52E Ch. 6.1 - Prob. 53E Ch. 6.1 - Prob. 54E Ch. 6.1 - 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The volume V of the bead (Figure 17) in terms of r and R . Then show that V = π 6 h 3 , where h is the height of the bead and then explain intuitively, how the given statement is possible.

Solution Summary: The author explains how the volume of the bead is given by V = 6 h 3 and how it can be explained intuitively.

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The volume $V$ of the bead (Figure 17) in terms of $r$ and $R$ . Then show that $V = \frac{π}{6} h^{3}$ , where $h$ is the height of the bead and then explain intuitively, how the given statement is possible.

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Chapter 6 Solutions

The volume V of the bead (Figure 17) in terms of r and R . Then show that V = π 6 h 3 , where h is the height of the bead and then explain intuitively, how the given statement is possible.

Solution Summary: The author explains how the volume of the bead is given by V = 6 h 3 and how it can be explained intuitively.

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The volume V of the bead (Figure 17) in terms of r and R. Then show that V=π6h3, where h is the height of the bead and then explain intuitively, how the given statement is possible.

Want to see the full answer?

Chapter 6 Solutions

The volume $V$ of the bead (Figure 17) in terms of $r$ and $R$ . Then show that $V = \frac{π}{6} h^{3}$ , where $h$ is the height of the bead and then explain intuitively, how the given statement is possible.