Explanation Given: The surface is given by f ( x , y ) = 3 + 2 x 3 2 , above the region R , which is rectangular with vertices ( 0 , 0 ) , ( 0 , 4 ) , ( 1 , 4 ) and ( 1 , 0 ) . Formula used: The surface area can be calculated of the region R by, S = ∬ R 1 + [ f x ( x , y ) ] 2 + [ f y ( x , y ) ] 2 d A Differentiation formula d d x ( x n ) = n x n − 1 , d d x ( constant ) = 0 Calculation: The function given is f ( x , y ) = 3 + 2 x 3 2 . Now partially differentiating it with respect to x , use d d x ( x n ) = n x n − 1 , d d x ( constant ) = 0 f x ( x , y ) = d d x ( 3 + 2 x 3 2 ) = 2 ( 3 2 ) x 3 − 2 2 = 3 x 1 2 Now, with respect to y . f y ( x , y ) = d d y ( 3 + 2 x 3 2 ) = 0 Substitute in the formula S = ∬ R 1 + [ f x ( x , y ) ] 2 + [ f y ( x , y ) ] 2 d A

Calculus: Early Transcendental Functions

7th Edition

ISBN: 9781337670388

Author: Larson

Publisher: Cengage

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Chapter 14.5, Problem 9E

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To calculate: The area of the surface given by z=f(x,y)=3+2x32 which lies above the region R, which is rectangular with vertices (0,0),(0,4),(1,4) and (1,0).

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Chapter 14.5, Problem 8E

Chapter 14.5, Problem 10E

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Chapter 14 Solutions

Calculus: Early Transcendental Functions

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The area of the surface given by z = f ( x , y ) = 3 + 2 x 3 2 which lies above the region R , which is rectangular with vertices ( 0 , 0 ) , ( 0 , 4 ) , ( 1 , 4 ) and ( 1 , 0 ) .

Solution Summary: The author calculates the surface area of the region R by z=f(x,y)=3+2x32

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To calculate: The area of the surface given by z=f(x,y)=3+2x32 which lies above the region R, which is rectangular with vertices (0,0),(0,4),(1,4) and (1,0).

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Chapter 14 Solutions