Explanation Given information: The given function is y = ( 4 − x 2 3 ) 3 2 and the interval is [ 0 , 8 ] . 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 power rule of indefinite integrals. ∫ x n d x = x n n + 1 + C Calculation: Consider the given function. y = f ( x ) = ( 4 − x 2 3 ) 3 2 Differentiate the above function with respect to x . f ′ ( x ) = 3 2 ( 4 − x 2 3 ) 3 2 − 1 d d x ( x 2 3 ) = 3 2 × 2 3 x 2 3 − 1 ( 4 − x 2 3 ) 1 2 = x − 1 3 ( 4 − x 2 3 ) 1 2 The given interval is [ 0 , 8 ] . Here, a = 0 , b = 8 For 0 ≤ x ≤ 8 , x ≥ 0 and f ′ exists and continuous on [ 0 , 8 ] . Now calculate the surface area of revolution about x -axis over the interval [ 0 , 8 ] 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 43E

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Thesurface area of revolution about the x -axis over the interval [0,8] for the given function.

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Chapter 8.2, Problem 42E

Chapter 8.2, Problem 44E

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Chapter 8 Solutions

ACHIEVE FOR CALCULUS 4 TERM >CSI CUSTOM<

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Thesurface area of revolution about the x -axis over the interval [ 0 , 8 ] for the given function.

Solution Summary: The author explains how to compute 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 [0,8] for the given function.

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Chapter 8 Solutions

Thesurface area of revolution about the x -axis over the interval [ 0 , 8 ] for the given function.

Solution Summary: The author explains how to compute 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 [0,8] for the given function.

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Chapter 8 Solutions

To compute:

Thesurface area of revolution about the x -axis over the interval [0,8] for the given function.