Explanation Given information: The given function is y = sin x and the interval is [ 0 , π ] . Formula used: Assume that f ( x ) ≥ 0 and f ′ exists and continuous on the interval [ a , b ] . The surface area S of revolution about the x -axis over the interval is given by S = 2 π ∫ a b f ( x ) 1 + [ f ′ ( x ) ] 2 d x The following integration can be calculated as shown below. ∫ 1 + x 2 d x = 1 2 x 1 + x 2 + 1 2 ln | x + 1 + x 2 | + C Calculation: Consider the given function. y = f ( x ) = sin x Differentiate the above function with respect to x . f ′ ( x ) = cos x The given interval is [ 0 , π ] . Here, a = 0 , b = π For 0 ≤ x ≤ π , x ≥ 0 and f ′ exists and continuous on [ 0 , π ] . Now calculate the surface area of revolution about x -axis over the interval [ 0 , π ] as shown below. S = 2 π ∫ a b f ( x ) 1 + [ f ′ ( x ) ] 2 d x = 2 π ∫ 0 π sin x 1 + [ cos x ] 2 d x = 2 π ∫ 0 π

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 8.2, Problem 46E

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Thesurface area of revolution about the x -axis over the interval [0,π] for the given function.

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Chapter 8.2, Problem 45E

Chapter 8.2, Problem 47E

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Chapter 8 Solutions

Calculus: Early Transcendentals

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Thesurface area of revolution about the x -axis over the interval [ 0 , π ] for the given function.

Solution Summary: The author calculates the surface area of revolution about the x -axis over the interval.

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Thesurface area of revolution about the x -axis over the interval [0,π] for the given function.

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Chapter 8 Solutions

Thesurface area of revolution about the x -axis over the interval [ 0 , π ] for the given function.

Solution Summary: The author calculates the surface area of revolution about the x -axis over the interval.

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To compute: Thesurface area of revolution about the x -axis over the interval [0,π] for the given function.

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Chapter 8 Solutions

To compute:

Thesurface area of revolution about the x -axis over the interval [0,π] for the given function.