Explanation Place the centre of the sphere at the origin in three-dimensional space Let the cylinder be of radius y and half-height x The Pythagorean Theorem state x 2 + y 2 = R 2 y 2 = R 2 − x 2 The volume of the cylinder is V ( x ) = π y 2 ( 2 x ) = 2 π ( R 2 − x 2 ) x = 2 π ( R 2 − x 2 ) x = 2 π x 3 Allowing for degenerate cylinders we have 0 ≤ x ≤ R solve V ' ( x ) = 2 π R 2 − 6 π x 2 = 0 for

4th Edition

ISBN: 9781319434540

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.6, Problem 33E

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In his work nova published in 1615 astronomer Johannes Kepler stated and solved the following problem a sphere of radius R scribed cylinder has volume 2πx(R2−x2) where x is one-half height of the cylinder

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Chapter 4.6, Problem 32E

Chapter 4.6, Problem 34E

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Chapter 4 Solutions

ACHIEVE STANDALONE ACCESS F/ CALC 4E

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

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In his work nova published in 1615 astronomer Johannes Kepler stated and solved the following problem a sphere of radius R scribed cylinder has volume 2 π x ( R 2 − x 2 ) where x is one-half height of the cylinder

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In his work nova published in 1615 astronomer Johannes Kepler stated and solved the following problem a sphere of radius R scribed cylinder has volume 2πx(R2−x2) where x is one-half height of the cylinder

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Chapter 4 Solutions