Unformatted text preview: 18.02 Fall 2008 – Problem Set 9, Part B Solutions 1. a) The normal vector to C at ( x,y ) is ˆ n = x ˆ ı + y ˆ . Therefore vector F · ˆ n = ( xy,y 2 )·( x,y ) = x 2 y + y 3 = ( x 2 + y 2 ) y = y . The contribution to flux is positive when vector F · ˆ n = y > 0, which corresponds to the upper half-circle, and negative for the lower half-circle. b) When vector F = ( P,Q ) , integraltext C vector F · ˆ n ds = integraltext C − Qdx + P dy . Parametrizing the circle by x = cos θ , y = sin θ , dx = − sin θ dθ , dy = cos θ dθ , we calculate integraldisplay C − y 2 dx + xy dy = integraldisplay C − cos 2 θ ( − cos θ dθ ) + cos θ sin θ (cos θ dθ ) = integraldisplay 2 π (sin 2 θ + cos 2 θ ) sin θ dθ = integraldisplay 2 π sin θ dθ = 0 . (or, using (a) and observing that ds = dθ , integraltext C vector F · ˆ n ds = integraltext C y ds = integraltext 2 π sin θ dθ = 0) We get zero because, as seen in (a), vector F · ˆ n = y , so the (negative) flux through the lower...
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This note was uploaded on 04/28/2009 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.
- Fall '08