Explanation Given information: The given function is y = 1 4 x 2 − 1 2 ln x and the interval is [ 1 , e ] . 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 indefinite integral can be calculated as shown below. ∫ ( 1 4 x 2 − 1 2 ln x ) ( x 2 + 1 2 x ) d x = 1 32 ( x 4 + 4 x 2 − 4 x 2 ln x − 4 ln 2 ( x ) ) + C Calculation: Consider the given function. y = f ( x ) = 1 4 x 2 − 1 2 ln x Differentiate the above function with respect to x . f ′ ( x ) = 1 4 × 2 x − 1 2 × 1 x = x 2 − 1 2 x = x 2 − 1 2 x The given interval is [ 1 , e ] . Here, a = 1 , b = e For 1 ≤ x ≤ e , x ≥ 0 and f ′ exists and continuous on [ 1 , e ] . Now calculate the surface area of revolution about x -axis over the interval [ 1 , e ] as shown below

ACHIEVE FOR CALCULUS 4 TERM >CSI CUSTOM<

21st Edition

ISBN: 9781319438333

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 8.2, Problem 45E

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Thesurface area of revolution about the x -axis over the interval [1,e] for the given function.

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Chapter 8.2, Problem 44E

Chapter 8.2, Problem 46E

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ACHIEVE FOR CALCULUS 4 TERM >CSI CUSTOM<

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Thesurface area of revolution about the x -axis over the interval [ 1 , e ] 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 [1,e] for the given function.

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Chapter 8 Solutions

Thesurface area of revolution about the x -axis over the interval [ 1 , e ] 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 [1,e] for the given function.

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Chapter 8 Solutions

To compute:

Thesurface area of revolution about the x -axis over the interval [1,e] for the given function.