Let s1 s2 s3 and s4 be the four faces of the

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Unformatted text preview: x ≤ 1, −1 ≤ y ≤ 1, −1 ≤ z ≤ 1} div F = ∂ ∂ ∂ 3y 2 z 3 + 9x2 yz 2 + −4xy 2 = 9x2 z 2 ∂x ∂y ∂z 1 1 1 9x2 z 2 dx dy dz = 8 F • dS = −1 S 1 −1 −1 z z x x y y Figure 41a (b) Figure 41b i. Let S1 , S2 , S3 , and S4 be the four faces of the tetrahedron. (Figure 42a and 42b) Top Left Back Bottom (yellow) (green) (blue) (pink) S1 : S2 : S3 : S4 : r (x, y ) = [x, y, 1 − x − y ] r (x, z ) = [x, 0, z ] r (y, z ) = [0, y, z ] r (x, y ) = [x, y, 0] 1−x 1 F • dS = 0 0 with with with with 0≤x≤1 0≤x≤1 0≤y≤1 0≤x≤1 and and and and 0≤y ≤1−x 0≤z ≤1−x 0≤z ≤1−y 0≤y ≤1−x 3xy, y 2 , −x2 y 4 • ([1, 0, −1] × [0, 1, −1]) dy dx = 29 140 S1 1−x 1 F • dS = 0 0 [0, 0, 0] • ([1, 0, 0] × [0, 0, 1]) dz dx = 0 S2 1−y 1 F • dS = 0 0 0, y 2 , 0 • ([0, 0, 1] × [0, 1, 0]) dz dy = 0 S3 1−x 1 F • dS = 0 0 3xy, y 2 , −x2 y 4 • ([0, 1, 0] × [1, 0, 0]) dy dx = 1 840 S4 Therefore, F • dS = S F • dS + S1 F • dS + S2 F • dS + S3 F • dS = S4 ii. E = {(x, y, z ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x, 0 ≤ z ≤ 1 − x − y } div F = 3y + 2y = 5y 1−x 1 F • dS = 0 S 2 0 1−x−y 0 5y dz dy dx = 5 24 5 . 24 z z y y x x Figure 42a Figure 42b 2. (a) E = {(x, y, z ) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 2, 0 ≤ z ≤ 1} div F = 2xy + 2yz 3 2 1 F • dS = (2xy + 2yz ) dx dy dz = 24 0 0 0 S (b) E = {(x, y, z ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 − 2x, 0 ≤ z ...
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This document was uploaded on 01/23/2014.

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