Explanation Given: S is the part of the graph z = g ( x , y ) lying over a domain D in the x y -plane and ϕ = ϕ ( x , y ) be the angle between the normal to S and the vertical. Formula used: Normal vector to a surface z = g ( x , y ) at point ( x , y ) in the upward direction is given by n → = 〈 − ∂ g ∂ x , − ∂ g ∂ y , 1 〉 Calculation: d r → = 〈 d x , d y , d z 〉 When x is constant d x = 0 this gives d r → y = 〈 0 , d y , ∂ z ∂ y d y 〉 When y is constant d y = 0 this gives d r → x = 〈 d x , 0 , ∂ z ∂ x d x 〉 d r → x × d r → y = 〈 d x , 0 , ∂ z ∂ x d x 〉 × 〈 0 , d y , ∂ z ∂ y d y 〉 = 〈 − ∂ z ∂ x d x d y , − ∂ z ∂ y d x d y , d x d y 〉 | d r → x × d ...

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.4, Problem 51E

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The proof that $A r e a (S) = \iint_{D} \frac{d A}{| \cos ϕ |}$

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Chapter 17.4, Problem 50E

Chapter 17.5, Problem 1PQ

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Chapter 17 Solutions

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The proof that A r e a ( S ) = ∬ D d A | cos ϕ |

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The proof that $A r e a (S) = \iint_{D} \frac{d A}{| \cos ϕ |}$

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Chapter 17 Solutions

The proof that A r e a ( S ) = ∬ D d A | cos ϕ |

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The proof that $A r e a (S) = \iint_{D} \frac{d A}{| \cos ϕ |}$