Explanation Formula Used: Gradient Theorem, ∫ C ∇ f ⋅ d r = f ( B ) − f ( A ) Green’s Theorem – I = ∮ C F ⋅ d r = ∬ D ( ∂ Q ∂ y − ∂ P ∂ x ) d A Given: I = ∮ C F ⋅ d r F ( x , y ) = 〈 2 x e y , x + x 2 e y 〉 F = G + ∇ f , where G = 〈 0 , x 〉 Calculation: (a) First we have to find function f ( x , y ) . We have been given, F ( x , y ) = 〈 2 x e y , x + x 2 e y 〉 , F = G + ∇ f , where G = 〈 0 , x 〉 F = G + ∇ f ∇ f = F − G ∇ f = 〈 2 x e y , x + x 2 e y 〉 − 〈 0 , x 〉 ∇ f = 〈 2 x e y , x 2 e y 〉 Gradient vector, ∇ f = 〈 f x , f y 〉 Therefore, f ( x , y ) = ∫ f x d x = ∫ f y d y f ( x , y ) = ∫ 2 x e y d x = ∫ x 2 e y d y = x 2 e y Therefore we have, f ( x , y ) = x 2 e y (b) Now we will find line integral of G along O A ¯ and O B ¯ Now we have to parameterize G = 〈 0 , x 〉 along O A ¯ x = t 0 ≤ t ≤ 4 and y = 0 Therefore, d x = d t d x d t = 1 d y = 0 Line integral of G along O A ¯ is given by, ∫ O A ¯ G ⋅ d r Now, G = 〈 0 , t 〉 d r = 〈 d x d t , d y d t 〉 d t = 〈 1 , 0 〉 d t Therefore, ∫ O A ¯ G ⋅ d r = ∫ 0 4 〈 0 , t 〉 ⋅ 〈 1 , 0 〉 d t ∫ O A ¯ G ⋅ d r = ∫ 0 4 0 ( 1 ) + t ( 0 ) d t ∫ O A ¯ G ⋅ d r = ∫ 0 4 0 d t ∫ O A ¯ G ⋅ d r = 0 Now we have to parameterize G = 〈 0 , x 〉 along O B ¯ x = 0 and y = t 0 ≤ t ≤ 4 Therefore, d x = 0 d y = d t d y d t = 1 Line integral of G along O B ¯ is given by, ∫ O B ¯ G ⋅ d r Now, G = 〈 0 , 0 〉 d r = 〈 d x d t , d y d t 〉 d t = 〈 0 , 1 〉 ∫ O B ¯ G ⋅ d r = ∫ 0 4 〈 0 , 0 〉 ⋅ 〈 0 , 1 〉 ∫ O B ¯ G ⋅ d r = ∫ 0 4 0 ( 0 ) + 0 ( 1 ) d t ∫ O B ¯ G ⋅ d r = 0 Therefore, we have Line integrals of G along the segments O A ¯ and O B ¯ are zero...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 18.1, Problem 11E

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To evaluate: $I = \oint_{C} F \cdot d r$

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Chapter 18.1, Problem 10E

Chapter 18.1, Problem 12E

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Chapter 18 Solutions

Calculus - Standalone book

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I = ∮ C F ⋅ d r

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Chapter 18 Solutions

To evaluate: $I = \oint_{C} F \cdot d r$