Explanation Given information: Given surface is y = 4 − z 2 , 0 ≤ x ≤ 2 , 0 ≤ z ≤ 2 A function f ( x , y , z ) = 3 z Formula used: For any function f ( x , y , z ) , if the surface S = F ( x , y , z ) is given such that z = g ( x , y ) And d S = 1 + g x 2 ( x , y ) + g y 2 ( x , y ) d x d y The integral ∬ S f ( x , y , z ) d S = ∫ a b ∫ c d p ( x , y ) 1 + g x 2 ( x , y ) + g y 2 ( x , y ) d x d y Where c ≤ x ≤ d , a ≤ y ≤ b Calculation: Let us consider the surface S given by y = 4 − z 2 . z = g ( x , y ) = ± 4 − y So, g x ( x , y ) = 0 g y ( x , y ) = ∓ 1 2 4 − y Again, 0 ≤ z ≤ 2 gives 0 ≤ y ≤ 4 . The value of the surface integral can be calculated as shown below. ∬ S f ( x , y , z ) d S = ∫ y = 0 4 ∫ x = 0 2 3 z 1 + g x 2 ( x , y ) + g y 2 ( x , y ) d x d y = ± ∫ y = 0 4 ∫ x = 0 2 3 4

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 16.4, Problem 15E

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The integral ∬Sf(x,y,z)dS for the given surface and function.

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Chapter 16.4, Problem 14E

Chapter 16.4, Problem 16E

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Chapter 16 Solutions

Calculus: Early Transcendentals

Ch. 16.1 - Prob. 1PQ Ch. 16.1 - Prob. 2PQ Ch. 16.1 - Prob. 3PQ Ch. 16.1 - Prob. 4PQ Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E Ch. 16.1 - Prob. 5E Ch. 16.1 - Prob. 6E

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The integral ∬ S f ( x , y , z ) d S for the given surface and function.

Solution Summary: The author calculates the integral displaystyleiint_Sf(x,y,z).

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To calculate:

The integral ∬Sf(x,y,z)dS for the given surface and function.

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Chapter 16 Solutions

The integral ∬ S f ( x , y , z ) d S for the given surface and function.

Solution Summary: The author calculates the integral displaystyleiint_Sf(x,y,z).

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To calculate: The integral ∬Sf(x,y,z)dS for the given surface and function.

Want to see the full answer?

Chapter 16 Solutions

To calculate:

The integral ∬Sf(x,y,z)dS for the given surface and function.