Explanation Formula Used: Green’s Theorem – I = ∮ C F ⋅ d r = ∬ D ( ∂ Q ∂ x − ∂ P ∂ y ) d A Given: I = ∫ C ( sin x + y ) d x + ( 3 x + y ) d y Nonclosed path ABCD Calculation: We have been given, I = ∫ C ( sin x + y ) d x + ( 3 x + y ) d y Let, F = 〈 ( sin x + y ) , ( 3 x + y ) 〉 Therefore, P = sin x + y and Q = 3 x + y In the above figure, we see the region D enclosed by the path C 1 , which is determined by the nonclosed path C and the line segment D A ¯ . Now, we will use Green’s Theorem, ∫ C 1 P d x + Q d y = ∬ D ( ∂ Q ∂ x − ∂ P ∂ y ) d A Now, ∂ Q ∂ x = ∂ ∂ x ( 3 x + y ) = 3 ∂ P ∂ y = ∂ ∂ y ( sin x + y ) = 1 Therefore, ∫ C 1 P d x + Q d y = ∬ D ( ∂ Q ∂ x − ∂ P ∂ y ) d A = ∬ D ( 3 − 1 ) d A = ∬ D 2 d A = 2 ∬ D d A We know are of closed region D is given by, A r e a ( D ) = ∬ D d A Therefore, ∫ C 1 P d x + Q d y = 2 ( A r e a ( D ) ) -----------------(1) From the above figure, we can see the D region is a trapezium. Therefore, A r e a ( D ) = ( s u m o f p a r a l l e l s i d e s ) h e i g h t 2 A r e a ( D ) = ( B C + D A ) h 2 A r e a ( D ) = ( 2 + 6 ) 2 2 = 8 Let’s substitute value of A r e a ( D ) in (1) Therefore, ∫ C 1 P d x + Q d y = 2 ( 8 ) = 16 Now, using line integral we can write, ∫ C 1 P d x + Q d y = ∫ C P d x + Q d y + ∫ D A ¯ P d x + Q d y ∫ C P d x + Q d y + ∫ D A ¯ P d x + Q d y = 16 -----------------------(2) Now we have to calculate the line integral of D A ¯ , using parameterization...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 18.1, Problem 15E

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To evaluate: $I = \int_{C} (\sin x + y) d x + (3 x + y) d y$

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Chapter 18.1, Problem 14E

Chapter 18.1, Problem 16E

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Chapter 18 Solutions

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I = ∫ C ( sin x + y ) d x + ( 3 x + y ) d y

Solution Summary: The author explains that the region D enclosed by the nonclosed path C and the line segment D A is determined by a Green’s Theorem.

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To evaluate: I=∫C(sinx+y)dx+(3x+y)dy

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Chapter 18 Solutions

To evaluate: $I = \int_{C} (\sin x + y) d x + (3 x + y) d y$