Explanation Given: The surface of a trough is z = x 2 for − 2 ≤ x ≤ 2 , 0 ≤ y ≤ 4 . Theorem used: Evaluation of surface Integrals of Scalar-Valued Functions on Explicitly Defined Surfaces “Let f be a continuous function on a smooth surface S given by z = g ( x , y ) , for ( x , y ) in a region R . The surface integral of f over S is ∬ S f ( x , y , z ) d S = ∬ R f ( x , y , g ( x , y ) ) z x 2 + z y 2 + 1 d A . If f ( x , y , z ) = 1 , the surface integral equals to the area of the surface”. Formula used: ∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln | x + x 2 + a 2 a | . Calculation: Obtain the value of z x 2 + z y 2 + 1 . From the given z = x 2 , Compute the following. Differentiate z with respect to x . d z = ( 2 x ) d x d z d x = 2 x z x = 2 x Differentiate z with respect to y . d z = ( 0 ) d y d z d y = 0 z y = 0 Compute the value of z x 2 + z y 2 + 1 . z x 2 + z y 2 + 1 = ( 2 x ) 2 + 0 + 1 = 4 x 2 + 1 Thus, the surface area of the integral becomes ∬ S d S = ∫ 0 4 ∫ − 2 2 4 x 2 + 1 d x d y . Integrate the above integral with respect to x and apply the limit. ∬ S d S = ∫ 0 4 ∫ − 2 2 4 x 2 + 1 d x d y = 1 2 ∫ 0 4 [ 2 x 2 4 x 2 + 1 + 1 2 ln | 2 x + 4 x 2 + 1 1 | ] − 2 2 d y = 1 2 ∫ 0 4 { [

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ISBN: 9781323178522

Author: Briggs

Publisher: PEARSON

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Chapter 14.6, Problem 33E

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To find: The surface area with the help of explicit description.

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Chapter 14.6, Problem 32E

Chapter 14.6, Problem 34E

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The surface area with the help of explicit description.

Solution Summary: The author explains the surface area with the help of explicit description. The surface of a trough is z=x2.

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