Explanation Given: The surface is given by f ( x , y ) = x 2 + y 2 , above the region R , represented by the limit shown R = { ( x , y ) : 0 ≤ f ( x , y ) ≤ 1 } 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 The equation of circle x 2 + y 2 = r 2 , where r is the radius Calculation: The function given is f ( x , y ) = x 2 + y 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 ( x 2 + y 2 ) = 2 x 2 x 2 + y 2 = x x 2 + y 2 Now, with respect to y f y ( x , y ) = d d x ( x 2 + y 2 ) = 2 y 2 x 2 + y 2 = y x 2 + y 2 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 13E

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To calculate: The area of the surface given by z=f(x,y)=x2+y2 which lies above the region R, represented by the limit shown R={(x,y): 0≤f(x,y)≤1}

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Chapter 14.5, Problem 12E

Chapter 14.5, Problem 14E

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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 ) = x 2 + y 2 which lies above the region R , represented by the limit shown R = { ( x , y ) : 0 ≤ f ( x , y ) ≤ 1 }

Solution Summary: The author calculates the surface area of the region R by z=f(x,y)=sqrtx2+y

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To calculate: The area of the surface given by z=f(x,y)=x2+y2 which lies above the region R, represented by the limit shown R={(x,y): 0≤f(x,y)≤1}

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Chapter 14 Solutions