Explanation Given: C is any Jordan curve whose interior doesn't contain the point (1,0) traversed counterclockwise Calculation: Given that ∮ C [ − y d x + ( x − 1 ) d y ( x − 1 ) 2 + y 2 ] = ∮ C [ − y ( x − 1 ) 2 + y 2 d x + x − 1 ( x − 1 ) 2 + y 2 d y ] So, M = − y ( x − 1 ) 2 + y 2 ⇒ ∂ M ∂ y = − − 1 ( x − 1 ) 2 + y 2 − 2 y 2 ( ( x − 1 ) 2 + y 2 ) 2 ⇒ ∂ M ∂ y = − x 2 + 2 x − 1 + y 2 ( ( x − 1 ) 2 + y 2 ) 2 N = x − 1 ( x − 1 ) 2 + y 2 ⇒ ∂ N ∂ x = 1

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 13.4, Problem 46PS

To determine

To find: the integral ∮C[ −ydx+( x−1 )dy ( x−1 ) 2 + y 2 ]

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Chapter 13.4, Problem 45PS

Chapter 13.4, Problem 47PS

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the integral ∮ C [ − y d x + ( x − 1 ) d y ( x − 1 ) 2 + y 2 ]

Solution Summary: The author explains how to find the integral ldisplaystyleundersetCoint, whose interior doesn't contain the point (1,0) traverse

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To find: the integral ∮C[ −ydx+( x−1 )dy ( x−1 ) 2 + y 2 ]

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Chapter 13 Solutions