Explanation Given: F → = 〈 y z , 0 , x 〉 , the portion of the plane x 2 + y 3 + z = 1 where x , y , z ≥ 0 . Formula used: Stokes’ theorem is given by- ∮ c F → . d r → = ∬ s ( ∇ → × F → ) . d s → Where, d r → is the length element on the closed curve c d s → is the surface normal to the surface s enclosed by curve c From the given data we know that F → = 〈 y z , 0 , x 〉 and portion of the plane x 2 + y 3 + z = 1 where x , y , z ≥ 0 . From A → B z = 0 x 2 + y 3 + z = 1 i.e. r → = 〈 x , 3 ( 1 − x 2 ) , 0 〉 W h e r e , 0 ≤ x ≤ 2 d r → = 〈 d x , − 3 2 d x , 0 〉 ∫ A B F → . d r → = ∫ x = 2 x = 0 〈 y z , 0 , x 〉 . 〈 d x , − 3 2 d x , 0 〉 = ∫ x = 2 x = 0 〈 0 , 0 , x 〉 . 〈 d x , − 3 2 d x , 0 〉 = 0 From B → C r → = 〈 0 , y , ( 1 − y 3 ) 〉 W h e r e , 0 ≤ y ≤ 3 d r → = 〈 0 , d y , − d y 3 〉 ∫ B C F → . d r → = ∫ y = 3 y = 0 〈 y z , 0 , 0 〉 . 〈 0 , d y , − d y 3 〉 = 0 From C → A r → = 〈 x , 0 , ( 1 − x 2 ) 〉 W h e r e , 0 ≤ x ≤ 2 d r → = 〈 d x , 0 , − d x 2 〉 ∫ A C F → . d r → = ∫ x = 2 x = 0 〈 0 , 0 , x 〉 ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 18.2, Problem 3E

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The validity of Stokes’ theorem for the given vector field and surface, oriented with an upward-pointing normal.

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Chapter 18.2, Problem 2E

Chapter 18.2, Problem 4E

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The validity of Stokes’ theorem for the given vector field and surface, oriented with an upward-pointing normal.

Solution Summary: The author explains how Stokes' theorem is verified for the given vector field and surface, oriented with an upward-pointing normal.

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Chapter 18 Solutions