Explanation Given Info: Given z = 4 − x 2 − y 2 , 0 ≤ z ≤ 3 . f ( x , y , z ) = x 2 4 − z . Formula Used: Let z = g ( x , y ) . Then let g x = ∂ g ∂ x , g y = ∂ g ∂ y . d S = 1 + g x 2 + g y 2 d x d y . Calculation: Here g ( x , y ) = 4 − x 2 − y 2 . f ( x , y , z ) = x 2 4 − z = x 2 x 2 + y 2 . z = 4 − x 2 − y 2 , 0 ≤ z ≤ 3 ⇒ 1 ≤ x 2 + y 2 ≤ 4 g x = ∂ ( 4 − x 2 − y 2 ) ∂ x = − 2 x g y = ∂ ( 4 − x 2 − y 2 ) ∂ y = − 2 y ∬ S f ( x , y , z ) d S = ∬ S f ( x , y , z ) 1 + g x 2 + g y 2 d x d y ∬ S f ( x , y , z ) d S = ∬ 1 ≤ x 2 + y 2 ≤ 4 x 2 x 2 + y 2 1 + ( − 2 x ) 2 + ( − 2 y ) 2 d x d y ∬ S f ( x , y , z ) d S = ∬ 1 ≤ x 2 + y 2 ≤ 4 x 2 x 2 + y 2 1 + 4 ( x 2 + y 2 ) d x d y ∬ S f ( x , y , z ) d S = ∫ 0 2 π ∫ 1 2 r 2 cos 2 θ r 2 1 + 4 r 2 r d r d θ ∬ S f ( x , y ,

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.4, Problem 18E

To determine

To calculate: $\iint_{S} f (x, y, z) d S$ for the given surface and function.

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Chapter 17.4, Problem 17E

Chapter 17.4, Problem 19E

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Chapter 17 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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

Solution Summary: The author calculates S f ( x, y ) d S for the given surface and function.

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To calculate: $\iint_{S} f (x, y, z) d S$ for the given surface and function.

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Chapter 17 Solutions

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

Solution Summary: The author calculates S f ( x, y ) d S for the given surface and function.

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

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Chapter 17 Solutions

To calculate: $\iint_{S} f (x, y, z) d S$ for the given surface and function.