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# h6s - 18.02 HOMEWORK#6 DUE NOVEMBER 4 2010 BJORN POONEN 1...

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18.02 HOMEWORK #6, DUE NOVEMBER 4, 2010 BJORN POONEN 1. Part A (After Oct. 28) p. 1045: 9, 29* (8 pts.), 38* (8 pts.) Solution. (to p. 1045, 9) Let R be the interior of the ellipse. By Green’s theorem, C P dx + Q dy = R ( Q x - P y ) dA = R ( y - 2 y ) dA = - R y dA = 0 by symmetry (for every piece of area inside the ellipse, there is a corresponding piece of area on which y takes the opposite value). Solution. (to p. 1045, 29) In Green’s theorem, C P dx + Q dy = R ∂Q ∂x - ∂P ∂y dA, take P = 0 and Q = x 2 to get C x 2 dy = R 2 x dA. On the other hand, by definition, ¯ x = R x dA Area( R ) = 1 2 R 2 x dA A = 1 2 A C x 2 dy. Similarly, taking P = - y 2 and Q = 0 in Green’s theorem yields - C y 2 dx = R 2 y dA, 1

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so ¯ y = R y dA Area( R ) = 1 2 R 2 y dA A = - 1 2 A C y 2 dx. Solution. (to p. 1045, 38) (a) The curve C is parametrized by ( x ( t ) , y ( t )) := x 1 , y 1 + t x 2 - x 1 , y 2 - y 1 = (1 - t ) x 1 + tx 2 , (1 - t ) y 1 + ty 2 . for t [0 , 1]. In terms of the parameter t , we have dx = ( x 2 - x 1 ) dt dy = ( y 2 - y 1 ) dt Then C x dy - y dx = 1 0 (((1 - t ) x 1 + tx 2 )( y 2 - y 1 ) - ((1 - t ) y 1 + ty 2 )( x 2 - x 1 )) dt = 1 0 ( x 1 ( y 2 - y 1 ) - y 1 ( x 2 - x 1 )) dt = x 1 ( y 2 - y 1 ) - y 1 ( x 2 - x 1 ) = x 1 y 2 - x 2 y 1 . (b) Let C be the boundary of the triangle traversed counterclockwise. By part (a), C x dy - y dx = (0 y 1 - x 1 0) + ( x 1 y 2 - x 2 y 1 ) + ( x 2 0 - 0 y 2 ) = x 1 y 2 - x 2 y 1 . On the other hand, by Green’s theorem with P := - y and Q := x , C x dy - y dx = triangle ∂Q ∂x - ∂P ∂y dA = triangle 1 - ( - 1) dA = 2 A. Comparing shows that A = 1 2 ( x 1 y 2 - x 2 y 1 ) . 2
(After Oct. 28) 4D-1c, 4D-3, 4D-6* (5 pts.) (assume that C is positively oriented), 4D-7* (5 pts.) Solution. (to 4D-6*) Let R be the region enclosed by C . By Green’s theorem with P := - x 2 y and Q := xy 2 , C - x 2 y dx + xy 2 dy = R ∂Q ∂x - ∂P ∂y dA = R ( y 2 + x 2 ) dA > 0 , because the integrand x 2 + y 2 is always nonnegative, and it is 0 zero only at (0 , 0), but R consists of more than just the point (0 , 0). Solution. (to 4D-7*) Let C be the boundary of the triangle, and let R be the region inside.

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h6s - 18.02 HOMEWORK#6 DUE NOVEMBER 4 2010 BJORN POONEN 1...

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