15_6_1 - 1 F = xi zj The surface of the tetrahedron has...

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Unformatted text preview: 1. F = xi + zj. The surface of the tetrahedron has four faces: ^ ^ On 1 , x = 0, N = -i, F N = 0. ^ ^ On 2 , y = 0, N = -j, F N = -z, d S = d x dz. ^ = -k, F N = 0. ^ On 3 , z = 0, N i + 2j + 3k x + 2z d x dy 14 ^ ^ ,FN = , dS = = On 4 , x + 2y + 3z = 6, N = d x dz. ^ 2 14 14 |N j| We have 1 2 ^ F N dS = - =- 4 ^ F N dS = 6-3z 2 14 1 (x + 2z) d x dz 2 14 0 0 1 2 (6 - 3z)2 + 2z(6 - 3z) dz = 2 0 2 1 2 = (6 - 3z)(6 + z) dz 4 0 2 1 = (36z - 6z 2 - z 3 ) = 10. 4 0 The flux of F out of the tetrahedron is ^ F N d S = 0 - 4 + 0 + 10 = 6. 2 4 x Fig. 6-1 6 3 ^ F N dS = 3 ^ F N dS = 0 2 6-3z z dz 0 2 0 0 dx (6z - 3z 2 ) dz = -4 z 2 1 3 y ...
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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