Explanation Given: S = G ( D ) , where D = { ( u , v ) : u 2 + v 2 ≤ 1 , u ≥ 0 , v ≥ 0 } And G ( u , v ) = ( 2 u + 1 , u − v , 3 u + v ) from the exercise 3. Calculation: a) The surface S is given by the parametrization: ϕ ( u , v ) = ( 2 u + 1 , u − v , 3 u + v ) In Exercise 3 we computed the normal vector : n ( u , v ) = ( 4 , − 2 , − 2 ) ⇒ ‖ n ( u , v ) ‖ = 24 The surface area for u and v in the domain D = { ( u , v ) : u 2 + v 2 ≤ 1 , u ≥ 0 , v ≥ 0 } is the surface integral: A r e a ( S ) = ∬ D ‖ n ( u , v ) ‖ d u d v = ∬ D ‖ ( 4 , − 2 , − 2 ) ‖ d u d v = 24 ∬ D 1 d u d v = 24 A r e a ( D ) D is the quarter disc u 2 + v 2 ≤ 1 in the first quadrant, hence it has the area A r e a = π ⋅ 1 2 4 = π 4 Hence the area( S ) is A r e a ( S ) = 24 ⋅ π 4 = 2 6 ⋅ π 4 = π 6 2 b) To compute the surface integral of f ( x , y , z ) = x − y , we first express f in terms of the parameters u and v : f ( ϕ ( u . v ) = ( 2 u + 1 ) − ( u − v ) = u + v + 1 We now use the Theorem on surface integrals to compute the integral: ∬ S f ( x , y , z ) d S = ∬ D f ( ϕ ( u ...

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ISBN: 9781319055844

Author: FRANZOSA

Publisher: VST

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Chapter 17.4, Problem 4E

To determine

The following

a) Calculate the surface area of S.

b) Evaluate $\iint_{S} (x - y) d S$

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Chapter 17.4, Problem 3E

Chapter 17.4, Problem 5E

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The following a) Calculate the surface area of S. b) Evaluate ∬ S ( x − y ) d S

Solution Summary: The author explains how to determine the surface area of S by calculating the parametrization.

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The following

a) Calculate the surface area of S.

b) Evaluate $\iint_{S} (x - y) d S$

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Chapter 17 Solutions

The following a) Calculate the surface area of S. b) Evaluate ∬ S ( x − y ) d S

Solution Summary: The author explains how to determine the surface area of S by calculating the parametrization.

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The following a) Calculate the surface area of S. b) Evaluate ∬S(x−y) dS

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Chapter 17 Solutions

The following

a) Calculate the surface area of S.

b) Evaluate $\iint_{S} (x - y) d S$