Hw16-4 - 16.4: 2, 8, 11, 14, 19, 24, 31, 35 2. F = M i + N...

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Unformatted text preview: 16.4: 2, 8, 11, 14, 19, 24, 31, 35 2. F = M i + N j = y i , and the path C is r = ( a cos t ) i + ( a sin t ) j , 0 t 2 . I C F T ds = I C y dx = Z 2 ( a sin t )(- a sin tdt ) =- a 2 Z 2 sin 2 tdt =- a 2 By now you have seen many times how to integrate sin 2 t . ZZ R N x- M y dxdy = ZZ R (- 1) dxdy =- a 2 since thats the area of the circle. For the flux, we have I C F n ds = I C y dy = Z 2 ( a sin t )( a cos tdt ) = 1 2 a 2 sin 2 t 2 = 0 ZZ R M x + N y dxdy = ZZ R dxdy = 0 8. F = ( x + y ) i- ( x 2 + y 2 ) j , and region R bounded by C is the triangle bounded by y = 0, x = 1, and y = x . circ = ZZ R N x- M y dxdy = ZZ R (- 2 x- 1) dxdy = Z 1 Z 1 y (- 2 x- 1) dxdy = Z 1 [- x 2- x ] 1 y dy = Z 1 (- 2 + y 2 + y ) dy =- 2 y + y 3 3 + y 2 2 1 =- 2 + 1 3 + 1 2 =- 7 6 flux = ZZ R M x + N y dxdy = ZZ R (1- 2 y ) dxdy = Z 1 Z 1 y (1- 2 y ) dxdy = Z 1 (1- 2 y ) x 1 y dy = Z 1 (1- 2 y- y + 2 y 2 ) dy = y- 3 y 2 2 +...
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This note was uploaded on 06/11/2011 for the course MATH 1920 at Cornell University (Engineering School).

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Hw16-4 - 16.4: 2, 8, 11, 14, 19, 24, 31, 35 2. F = M i + N...

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