Explanation Given: The two graphs f ( x ) and g ( x ) such that 0 ≤ f ( x ) ≤ g ( x ) , both rotated about the x-axis in the interval [ a , b ] Formula Used: S = 2 π ∫ a b f ( x ) 1 + ( d f ( x ) d x ) 2 d x Calculation: Yes, g ( x ) can have less surface area when rotated about x-axis over the interval [ a , b ] , even if 0 ≤ f ( x ) ≤ g ( x ) . Example: Let f ( x ) = x − 2 , g ( x ) = x over the interval [ 0 , 1 ] , then 0 ≤ f ( x ) ≤ g ( x ) . Now the surface area of f ( x ) is given by S = 2 π ∫ 0 1 ( x − 2 ) 1 + ( d ( x − 2 ) d x ) 2 d x S = 2 π ∫ 0 1 ( x − 2 ) 1 + ( 1 − 0 ) 2 d x S = 2 π ∫ 0 1 ( x − 2 ) 1 + 1 d x

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 9.1, Problem 3PQ

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Chapter 9 Solutions

Calculus - Standalone book

Ch. 9.1 - Prob. 1PQ Ch. 9.1 - Prob. 2PQ Ch. 9.1 - Prob. 3PQ Ch. 9.1 - Prob. 4PQ Ch. 9.1 - Prob. 5PQ Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E

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The difference between surface areas of the two graphs related.

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