Explanation Given: F = 〈 z , 0 , y 〉 S is the oriented surface parameterized by G ( u , v ) = ( u 2 − v , u , v 2 ) for 0 ≤ u ≤ 2 , − 1 ≤ v ≤ 4 Calculation: (a) Consider G ( u , v ) = ( u 2 − v , u , v 2 ) Differentiating above relation with respect to u and v , we get the partial derivatives as T u = ∂ G ∂ u = 〈 2 u , 1 , 0 〉 T v = ∂ G ∂ v = 〈 − 1 , 0 , 2 v 〉 Therefore, the normal is determined as N = T u × T v = | i j k 2 u 1 0 − 1 0 2 v | = i ( 2 v .1 − 0.0 ) − j ( 2 u .2 v − ( − 1 ) .0 ) + k ( 2 u .0 − ( − 1 ) .1 ) = 2 v i − 4 u v j + k = 〈 2 v , − 4 u v , 1 〉 Thus, n = ( 2 v , − 4 u v , 1 ) Consider F = 〈 z , 0 , y 〉 Therefore, F ( G ( u , v ) ) = F ( u 2 − v , u , v 2 ) = 〈 v 2 , u 2 − v , u 〉 F . N is determined as F ( G ( u , v ) ) . N ( u , v ) = 〈 v 2 , u 2 − v , u 〉 . 〈 2 v , − 4 u v , 1 〉 = 〈 v 2 .2 v , ( u 2 − v ) ( − 4 u v ) , u .1 〉 = 2 v 3 − 4 u 3 v + 4 u v 2 + u Thus, F . N = 2 v 3 − 4 u 3 v + 4 u v 2 + u (b) The normal component of F to the surface is F . e n = F . n ‖ n ‖ = F . n ‖ n ‖ = ( 2 v 3 − 4 u 3 v + 4 u v 2 + u ) ( 2 v ) 2 + ( − 4 u v ) 2 + ( 1 ) 2 = ( 2 v 3 − 4 u 3 v + 4 u v 2 + u ) 4 v 2 + 16 u 2 v 2 + 1 At P = ( 3 , 2 , 1 ) = G ( 2 , 1 ) F ( p ) . e n = 2 ( 1 ) 3 − 4 ( 2 ) 3 ( 1 ) + 4 ( 2 ) ( 1 ) 2 + ( 2 ) 4 ( 1 ) 2 + 16 ( 2 ) 2 ( 1 ) 2 + 1 = 2 − 32 + 8 + 2 4 + 64 + 1 = ( − 20 ) 69 = − 20 69 Thus, F ( p ) ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 17.5, Problem 1E

To determine

To calculate:

(a) $N$ and $F . N$ as functions of $u$ and $v$ .

(b) The normal component of $F$ to the surface at $P = (3, 2, 1) = G (2, 1)$

(c) $\int \int_{S} F . d S$

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Chapter 17.5, Problem 7PQ

Chapter 17.5, Problem 2E

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Chapter 17 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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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(a) N and F . N as functions of u and v . (b) The normal component of F to the surface at P = ( 3 , 2 , 1 ) = G ( 2 , 1 ) (c) ∫ ∫ S F . d S

Solution Summary: The author calculates N and F as functions of u and v. The normal component of F is the oriented surface parameterized by G.

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To calculate:

(a) $N$ and $F . N$ as functions of $u$ and $v$ .

(b) The normal component of $F$ to the surface at $P = (3, 2, 1) = G (2, 1)$

(c) $\int \int_{S} F . d S$

Want to see the full answer?

Chapter 17 Solutions

(a) N and F . N as functions of u and v . (b) The normal component of F to the surface at P = ( 3 , 2 , 1 ) = G ( 2 , 1 ) (c) ∫ ∫ S F . d S

Solution Summary: The author calculates N and F as functions of u and v. The normal component of F is the oriented surface parameterized by G.

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To calculate: (a) N and F.N as functions of u and v. (b) The normal component of F to the surface at P=(3,2,1)=G(2,1) (c) ∫∫SF.dS

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Chapter 17 Solutions

To calculate:

(a) $N$ and $F . N$ as functions of $u$ and $v$ .

(b) The normal component of $F$ to the surface at $P = (3, 2, 1) = G (2, 1)$

(c) $\int \int_{S} F . d S$