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

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Unformatted text preview: 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 Greens theorem, I C P dx + Q dy = ZZ R ( Q x- P y ) dA = ZZ R ( y- 2 y ) dA =- ZZ 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 Greens theorem, Z C P dx + Q dy = ZZ R Q x- P y dA, take P = 0 and Q = x 2 to get Z C x 2 dy = ZZ R 2 x dA. On the other hand, by definition, x = RR R x dA Area( R ) = 1 2 RR R 2 x dA A = 1 2 A Z C x 2 dy. Similarly, taking P =- y 2 and Q = 0 in Greens theorem yields- Z C y 2 dx = ZZ R 2 y dA, 1 so y = RR R y dA Area( R ) = 1 2 RR R 2 y dA A =- 1 2 A Z C y 2 dx. Solution. (to p. 1045, 38) (a) The curve C is parametrized by ( x ( t ) , y ( t )) := h x 1 , y 1 i + t h x 2- x 1 , y 2- y 1 i = h (1- t ) x 1 + tx 2 , (1- t ) y 1 + ty 2 i . 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 Z C x dy- y dx = Z 1 (((1- t ) x 1 + tx 2 )( y 2- y 1 )- ((1- t ) y 1 + ty 2 )( x 2- x 1 )) dt = Z 1 ( 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), I C x dy- y dx = (0 y 1- x 1 0) + ( x 1 y 2- x 2 y 1 ) + ( x 2- y 2 ) = x 1 y 2- x 2 y 1 . On the other hand, by Greens theorem with P :=- y and Q := x , I C x dy- y dx = ZZ triangle Q x- P y dA = ZZ 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 Greens theorem with P :=- x 2 y and Q := xy 2 , I C- x 2 y dx + xy 2 dy = ZZ R Q x- P y dA = ZZ R ( y 2 + x 2 ) dA > , because the integrand x 2 + y 2 is always nonnegative, and it is 0 zero only at (0...
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This note was uploaded on 09/14/2011 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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

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