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Hw16-4

# Hw16-4 - 16.4 2 8 11 14 19 24 31 35 2 F = M i N j = y i and...

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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 π . I C F · T ds = I C y dx = Z 2 π 0 ( a sin t )( - a sin t dt ) = - a 2 Z 2 π 0 sin 2 t dt = - πa 2 By now you have seen many times how to integrate sin 2 t . ZZ R ∂N ∂x - ∂M ∂y dx dy = ZZ R ( - 1) dx dy = - πa 2 since that’s the area of the circle. For the flux, we have I C F · n ds = I C y dy = Z 2 π 0 ( a sin t )( a cos t dt ) = 1 2 a 2 sin 2 t 2 π 0 = 0 ZZ R ∂M ∂x + ∂N ∂y dx dy = ZZ R 0 dx dy = 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 dx dy = ZZ R ( - 2 x - 1) dx dy = Z 1 0 Z 1 y ( - 2 x - 1) dx dy = Z 1 0 [ - x 2 - x ] 1 y dy = Z 1 0 ( - 2 + y 2 + y ) dy = - 2 y + y 3 3 + y 2 2 1 0 = - 2 + 1 3 + 1 2 = - 7 6 flux = ZZ R ∂M ∂x + ∂N ∂y dx dy = ZZ R (1 - 2 y ) dx dy = Z 1 0 Z 1 y (1 - 2 y ) dx dy = Z 1 0 (1 - 2 y ) x 1 y dy = Z 1 0 (1 - 2 y - y + 2 y 2 ) dy = y - 3 y 2 2 + 2 y 3 3 1 0 = 1 6 11. F = x 3 y 2 i + 1 2 x 4 y j , and the region R is bounded below by y = x 2 - x and bounded above by y = x ( x goes from 0 to 2).

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Hw16-4 - 16.4 2 8 11 14 19 24 31 35 2 F = M i N j = y i and...

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