Compute the outward flux of the vector field F(x, y, z) = 5xi + 3yj +5ek...

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Compute the outward flux of the vector field F(x, y, z) = 5xi + 3yj +5ek across the boundary of the right cylinder with radius 7 with bottom edge at height z = -3 and upper edge at z = 4. Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the cylinder to be positive. Part 1 - Using a Surface Integral First we parameterize the three faces of the cylinder. Note: in the following type as theta. The bottom face: 01(r, 0) = (x₁(r, 0), yı(r,0), zı(r, 0) where 0 ≤ 0 ≤ 2T, 0 ≤ r ≤ 7 and x1 (r, 0) = y₁ (r, 0) = z1(r, 0) = y2 (r, 0) The top face: 0₂ (r, 0) = (x2(r, 0), 2(r, 0), 2(r, 0) where 0≤0 2T, 0≤r≤7 and x₂ (r, 0) = 22 (r, 0) == -3 = = 4 The lateral face: 03(0, 2) = (x3(0, 2), y3(0, 2), 23(0, 2) where 0≤0≤ 2π, -3 ≤ z ≤ 4 and x3(0) = Y3 (0) = 23(z) = Then (mind the orientation) [[F.nds - 1² [² F (01₁) · (801 -√²² J² P(0₂). F(02) - (307 2πT 202 202 dr de + Ər 20 0 203 20 2π + - [²* *, F(03). (² Əz 0 -3 = + = доз X dz de + Part 2 - Using the Divergence Theorem [[F.nas [//di = 1 div F dV 201 Ər dr de dz dr de Compute the outward flux of the vector field F(x, y, z) = 5xi + 3yj +5ek across the boundary of the right cylinder with radius 7 with bottom edge at height z = -3 and upper edge at z = 4. Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the cylinder to be positive. Part 1 - Using a Surface Integral First we parameterize the three faces of the cylinder. Note: in the following type as theta. The bottom face: 01(r, 0) = (x₁(r, 0), yı(r,0), zı(r, 0) where 0 ≤ 0 ≤ 2T, 0 ≤ r ≤ 7 and x1 (r, 0) = y₁ (r, 0) = z1(r, 0) = y2 (r, 0) The top face: 0₂ (r, 0) = (x2(r, 0), 2(r, 0), 2(r, 0) where 0≤0 2T, 0≤r≤7 and x₂ (r, 0) = 22 (r, 0) == -3 = = 4 The lateral face: 03(0, 2) = (x3(0, 2), y3(0, 2), 23(0, 2) where 0≤0≤ 2π, -3 ≤ z ≤ 4 and x3(0) = Y3 (0) = 23(z) = Then (mind the orientation) [[F.nds - 1² [² F (01₁) · (801 -√²² J² P(0₂). F(02) - (307 2πT 202 202 dr de + Ər 20 0 203 20 2π + - [²* *, F(03). (² Əz 0 -3 = + = доз X dz de + Part 2 - Using the Divergence Theorem [[F.nas [//di = 1 div F dV 201 Ər dr de dz dr de

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To compute the outward flux of the vector field Fx y z 5xi 3yj 5zk across the boundary of the right cylinder with radius 7 and bottom edge at height z 3 and upper edge at z 4 we can use both a surface ... View the full answer