Explanation Given: The surface is given by f ( x , y ) = 2 + 2 3 y 3 2 , above the region R , which is represented by R = { ( x , y ) : 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 − x } . 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 ) = 2 + 2 3 y 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 ( 2 + 2 3 y 3 2 ) = 0 Now, with respect to y f y ( x , y ) = d d y ( 2 + 2 3 y 3 2 ) = 0 + 2 3 ( 3 2 ) y = y Substitute in the formula S = ∬ R 1 + [ f x ( x , y ) ] 2 + [ f y ( x , y ) ] 2 d A . S = ∬ R 1 + [ 0 ] 2 + [ y ] 2 d A = ∬ R 1 + 0 + y d A = ∬ R 1 + y d A Plot the provided points R = { ( x , y ) : 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 − x } on the graph

11th Edition

ISBN: 9781337275347

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 14.5, Problem 10E

To determine

To Calculate: The area of the surface given by $z=f(x,y)=2+23y32$ which lies above the region R, which is represented by $R={(x,y): 0≤x≤2,0≤y≤2−x}$ .

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Chapter 14 Solutions

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The area of the surface given by z = f ( x , y ) = 2 + 2 3 y 3 2 which lies above the region R , which is represented by R = { ( x , y ) : 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 − x } .

Solution Summary: The author calculates the area of the surface given by z=f(x,y)=2+23y

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To Calculate: The area of the surface given by $z=f(x,y)=2+23y32$ which lies above the region R, which is represented by $R={(x,y): 0≤x≤2,0≤y≤2−x}$ .

Chapter 14 Solutions