ps9sol - 18.02 Fall 2008 Problem Set 9, Part B Solutions 1....

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Unformatted text preview: 18.02 Fall 2008 Problem Set 9, Part B Solutions 1. a) The normal vector to C at ( x,y ) is n = x + y . Therefore vector F 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 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 n ds = integraltext C Qdx + P dy . Parametrizing the circle by x = cos , y = sin , dx = sin d , dy = cos d , we calculate integraldisplay C y 2 dx + xy dy = integraldisplay C cos 2 ( cos d ) + cos sin (cos d ) = integraldisplay 2 (sin 2 + cos 2 ) sin d = integraldisplay 2 sin d = 0 . (or, using (a) and observing that ds = d , integraltext C vector F n ds = integraltext C y ds = integraltext 2 sin d = 0) We get zero because, as seen in (a), vector F n = y , so the (negative) flux through the lower...
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