Explanation Given: F → = 〈 e y − z , 0 , 0 〉 , the square with the vertices ( 1 , 0 , 1 ) , ( 1 , 1 , 1 ) , ( 0 , 1 , 1 ) , a n d ( 0 , 0 , 1 ) . 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 Calculation: From the given data we know that F → = 〈 e y − z , 0 , 0 〉 , the square with the vertices ( 1 , 0 , 1 ) , ( 1 , 1 , 1 ) , ( 0 , 1 , 1 ) , a n d ( 0 , 0 , 1 ) . Closed loop line intergral is given by- ∮ F → . d r → = ∫ A B F → . d r → + ∫ B C F → . d r → + ∫ C D F → . d r → + ∫ D A F → . d r → For A → B r → = 〈 1 , t , 1 〉 W h e r e 0 ≤ t ≤ 1 d r → = 〈 0 , d t , 0 〉 ∫ A B F → . d r → = ∫ A B 〈 e y − z , 0 , 0 〉 . 〈 0 , d t , 0 〉 = 0 For B → C r → = 〈 ( 1 − t ) , 1 , 1 〉 d r → = 〈 − d t , 0 , 0 〉 ∫ B C F → . d r → = ∫ B C 〈 e y − z , 0 , 0 〉 . 〈 − d t , 0 , 0 〉 = ∫ t = 0 t = − 1 e 0 ( − d t ) = − 1 For C → D r → = 〈 ( 0 , 1 − t ) , 1 〉 d r → = 〈 0 , − d t , 0 〉 ∫ C D F → ...

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Chapter 18.2, Problem 1E

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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 5PQ

Chapter 18.2, Problem 2E

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Chapter 18 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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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