Explanation Given: F is a vector field such that ∮ C 1 F . d r = 9 and ∂ F 2 ∂ x − ∂ F 1 ∂ y = x 2 + y 2 for ( x , y ) in the annulus 1 ≤ x 2 + y 2 ≤ 4 . Formula used: Green’s Theorem: ∫ C 2 F d r − ∫ C 1 F d r = ∬ D ( ∂ F 2 ∂ x − ∂ F 1 ∂ y ) d x d y Calculation: We have to find the value of ∮ C 2 F . d r . The annulus D between the circle C 1 and C 2 oriented.’ Green’s Theorem: ∫ C 2 F d r − ∫ C 1 F d r = ∬ D ( ∂ F 2 ∂ x − ∂ F 1 ∂ y ) d x d y … (1) Substitute the value of ∮ C 1 F . d r = 9 and ∂ F 2 ∂ x − ∂ F 1 ∂ y = x 2 + y 2 in equation (1). ∫ C 2 F d r − ∫ C 1 F d r = ∬ D ( ∂ F 2 ∂ x − ∂ F 1 ∂ y ) d x d y ∫ C 2 F d r − 9 = ∬ D ( x 2 + y 2 ) d x d y … (2) Solve the integral by using polar coordinates. Let x = r cos θ , y = r sin θ . Now, find the value of x 2 + y 2 . x 2 + y 2 = ( r cos θ ) 2 + ( r sin θ ) 2 = r 2 cos 2 θ + r 2 sin 2 θ = r 2 ( cos 2 θ + sin 2 θ ) = r 2 The region becomes D = { ( r , θ ) | 1 ≤ r ≤ 2 , 0 ≤ θ ≤ 2 π } ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 18.1, Problem 27E

To determine

Use the general form of Green’s Theorem to determine $\oint_{C_{2}} F . d r$ .

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Chapter 18.1, Problem 26E

Chapter 18.1, Problem 28E

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Chapter 18 Solutions

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