This problem will illustrate the divergence theorem by computing the outward flux of the vector field...
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This problem will illustrate the divergence theorem by computing the outward flux of the vector field F(x, y, 2) = 4ri + 5yj +50k across the boundary of the right rectangular prism: -2 ≤ x ≤ 3,-3 ≤ y ≤ 5,-2 ≤ z ≤ 3 oriented outwards using a surface integral and a triple integral over the solid bounded by rectangular prism. Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the prism to be positive. Part 1 - Using a Surface Integral First we parameterize the six faces using 0 ≤ s≤ 1 and 0 ≤t≤ 1: The face with z = -2: 0₁ = (x₁(s), y₁(t), z₁(s, t)) *1(s) = yı(t) = z1(s, t) -2 The face with z = 3: 02 = (x2(8), y2(t), z2(s, t)) x2 (s) = y2(t) = z2(s, t) = 3 The face with x = -2:03 = 23(s, t), y3(8), z3(t)) 23(s, t) = -2 y3(s) = z3(t) = The face with x = 3: 04 (4(s, t), y4(8), 24(t)) = 24 (s, t) = 3 Y4 (8) = z4(t) = The face with y = -3: 05 = x5 (8) = y5(s, t) =-3 25 (t) = The face with y = 5: 06 = = (x6(3), y6(s, t), 26(t)) 26 (8) = y6 (s, t) = 5 z6(t) = = (r5(s), ys(s, t), 25(t)) Then (mind the orientation) X √ [F.nds = [["²" F(0₁) - (30 + S₁² S² F (0₂) · ( 202 ds dt 802 მა Ət 0 0 доз 803 Ət Əs 804 მs 805 + S₁ S² F (03). 0 + S² S² F (0₂). (² 0 0 + 0 0 + S₁ S² P (0₂). (² + Sª S² F (06). (² 0 II X X X X 20₁ Ət 205 მა Ət 806 806 X Ət Əs + ds dt ds dt ds dt ds dt ὃσι 301) ds d + ds 1 206 [² [² F (06). (20 Ət 0 0 = So || + F.ndS= ·n d + = dt 206 Əs Part 2 - Using the Divergence Theorem [[² = [ /[di G =[ ] , , □ + ds dt + divF dV = + dx dy dz This problem will illustrate the divergence theorem by computing the outward flux of the vector field F(x, y, 2) = 4ri + 5yj +50k across the boundary of the right rectangular prism: -2 ≤ x ≤ 3,-3 ≤ y ≤ 5,-2 ≤ z ≤ 3 oriented outwards using a surface integral and a triple integral over the solid bounded by rectangular prism. Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the prism to be positive. Part 1 - Using a Surface Integral First we parameterize the six faces using 0 ≤ s≤ 1 and 0 ≤t≤ 1: The face with z = -2: 0₁ = (x₁(s), y₁(t), z₁(s, t)) *1(s) = yı(t) = z1(s, t) -2 The face with z = 3: 02 = (x2(8), y2(t), z2(s, t)) x2 (s) = y2(t) = z2(s, t) = 3 The face with x = -2:03 = 23(s, t), y3(8), z3(t)) 23(s, t) = -2 y3(s) = z3(t) = The face with x = 3: 04 (4(s, t), y4(8), 24(t)) = 24 (s, t) = 3 Y4 (8) = z4(t) = The face with y = -3: 05 = x5 (8) = y5(s, t) =-3 25 (t) = The face with y = 5: 06 = = (x6(3), y6(s, t), 26(t)) 26 (8) = y6 (s, t) = 5 z6(t) = = (r5(s), ys(s, t), 25(t)) Then (mind the orientation) X √ [F.nds = [["²" F(0₁) - (30 + S₁² S² F (0₂) · ( 202 ds dt 802 მა Ət 0 0 доз 803 Ət Əs 804 მs 805 + S₁ S² F (03). 0 + S² S² F (0₂). (² 0 0 + 0 0 + S₁ S² P (0₂). (² + Sª S² F (06). (² 0 II X X X X 20₁ Ət 205 მა Ət 806 806 X Ət Əs + ds dt ds dt ds dt ds dt ὃσι 301) ds d + ds 1 206 [² [² F (06). (20 Ət 0 0 = So || + F.ndS= ·n d + = dt 206 Əs Part 2 - Using the Divergence Theorem [[² = [ /[di G =[ ] , , □ + ds dt + divF dV = + dx dy dz
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This problem is asking for the computation of the outward flux of the vector field Fx y z 4xi 5yj 5zk across the boundary of the given rectangular pri... View the full answer
Related Book For
Calculus Of A Single Variable
ISBN: 9781337275361
11th Edition
Authors: Ron Larson, Bruce H. Edwards
Posted Date:
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