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# notes001 (81) - =& 2 3 The&ux of F across C = C 1 C...

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MATH 2203 - Quiz 5 (Version 1) Solution April 25, 2011 NAME______________________ 1. Find the outward °ux of the vector ±eld F ( x; y ) = ( x + y ) i ° ° x 2 + y 2 ± j across the triangle with vertices at (1 ; 0) , (0 ; 1) , and ( ° 1 ; 0) . Assume that the boundary curve of the triangle has counterclockwise orienta- tion (as usual). Include a picture of this triangle. Do this in two ways: a) directly (without using Green²s Theorem) b) using Green²s Theorem All details of your work must be shown. If you do the problem correctly in just one way, then you can get up to 16 points. Solution: Here is a picture of the triangle. 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 x y 1

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For the curve, C 1 , on the bottom of the triangle, we have x = x dx = dx y = 0 dy = 0 . For the curve, C 2 , on the right of the triangle, we have x = x dx = dx y = 1 ° x dy = ° dx . For the curve, C 3 , on the left of the triangle, we have x = x dx = dx y = 1 + x dy = dx . We thus obtain I C 1 M dy ° N dx = I C 1 ( x + y ) dy + ° x 2 + y 2 ± dx = I C 1 ° x 2 + y 2 ± dx = Z 1 ° 1 x 2 dx = 2 3 and I C 2 M dy ° N dx = I C 2 ( x + y ) dy + ° x 2 + y 2 ± dx = I C 2 1 ( ° 1) dx + ° x 2 + (1 ° x ) 2 ± dx = Z 0 1 ° 2 x 2 ° 2 x ± dx = 1 3 2
and I C 3 M dy ° N dx = I C 3 ( x + y ) dy + ° x 2 + y 2 ± dx = I C 3 (2 x + 1) dx + ° x 2 + (1 + x ) 2 ± dx = Z ° 1 0 ° 2 x 2 + 4 x + 2 ± dx =

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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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notes001 (81) - =& 2 3 The&ux of F across C = C 1 C...

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