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Unformatted text preview: = & 2 3 . The &ux of F across C = C 1 [ C 2 [ C 3 is thus I C M dy & N dx = 2 3 + 1 3 & 2 3 = 1 3 . We will now do the problem using Green±s Theorem (&ux²divergence form) which says that I C M dy & N dx = ZZ D ² @M @x + @N @y ³ dA . Since the region D in this case is the union of two Type I regions, we will compute two double integrals and then add the results. For the region D 1 which is the left half of the triangle, we have ZZ D 1 ² @M @x + @N @y ³ dA = Z & 1 Z x +1 (1 & 2 y ) dy dx = Z & 1 & y & y 2 ±´ ´ y = x +1 y =0 dx = Z & 1 & ( x + 1) & ( x + 1) 2 ± dx = Z & 1 & & x & x 2 ± dx = 1 6 . 3 For the region D 2 which is the right half of the triangle, we have ZZ D 2 & @M @x + @N @y ± dA = Z 1 Z 1 & x (1 & 2 y ) dy dx = 1 6 . This gives ZZ D & @M @x + @N @y ± dA = 1 6 + 1 6 = 1 3 . 4...
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 Fall '10
 Ellermeyer
 Math, Trigraph, dx, Bryan Singer

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