Explanation Given: The surface: x 2 + y 2 + z 2 = 4 , 1 ≤ z ≤ 2 And, the function: f ( x , y , z ) = z 2 ( x 2 + y 2 + z 2 ) − 1 Calculation: Consider the surface: x 2 + y 2 + z 2 = 4 , 1 ≤ z ≤ 2 Parameterize the surface: G ( u , v ) = ( 2 cos u sin v , 2 cos u sin v , 2 cos v ) , 0 ≤ u ≤ 2 π , 0 ≤ v ≤ π 3 And, the tangent vectors are: T u = 〈 − 2 sin u sin v , 2 cos u sin v , 0 〉 , T v = 〈 2 cos u cos v , 2 sin u cos v , − 2 sin v 〉 Now, the Normal vector is: N = T u × T v = i j k − 2 sin u sin v 2 cos u sin v 0 2 cos u cos v 2 sin u cos v − 2 sin v = − 4 cos u sin 2 v i − 4 sin u sin 2 v j + [ − 4 sin 2 u sin v cos v − 4 cos 2 u sin v cos v ] k = − 4 cos u sin 2 v i − 4 sin u sin 2 v j − 4 sin v cos v k Now, the norm of the normal vector is: ‖ N ‖ = ( − 4 cos u sin 2 v ) 2 + ( − 4 sin u sin 2 v ) 2 + ( − 4 sin v cos v ) 2 = 16 ( cos 2 u sin 4 v + sin 2 u sin 4 v + sin 2 v cos 2 v ) = 4 sin 4 v + sin 2 v cos 2 v

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 17.4, Problem 23E

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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 17.4, Problem 22E

Chapter 17.4, Problem 24E

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Chapter 17 Solutions

CALCULUS (CLOTH)

Ch. 17.1 - Prob. 1PQ Ch. 17.1 - Prob. 2PQ Ch. 17.1 - Prob. 3PQ Ch. 17.1 - Prob. 4PQ Ch. 17.1 - Prob. 1E Ch. 17.1 - Prob. 2E Ch. 17.1 - Prob. 3E Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E Ch. 17.1 - Prob. 6E

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

Solution Summary: The author explains how to calculate S f (x, y, and z) 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 explains how to calculate S f (x, y, and z) 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.

Want to see the full answer?

Chapter 17 Solutions

To calculate:

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