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Compute the outward flux of the vector field F(x, y, z) = 4ci +3yj + 3k across the boundary of the right cylinder with radius 5 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), y₁(r, 0), z₁(r, 0) where 0 ≤0 ≤ 2,0

Compute the outward flux of the vector field F(x, y, z) = 4ci +3yj + 3k across the boundary of the right cylinder with radius 5 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), y₁(r, 0), z₁(r, 0) where 0 ≤0 ≤ 2,0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Compute the outward flux of the vector field F(x, y, z) = 4ci + 3yj + 3k across the boundary of the right cylinder with radius 5 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), z₁(r, 0) where 0 ≤ 0 ≤ 2,0 < r ≤ 5 and r cos (theta) rsin(theta) ₁ (r, 0) = yı(r,0) = z₁(r, 0) = -3 The top face: 0₂ (r, 0) = (x2(r, 0), Y2(r, 0), z₂(r, 0) where 0 ≤ 0 ≤ 2,0 ≤r≤ 5 and #₂ (r, 0) =r cos theta 32 (r, 6) = r sin theta 22(1,0) = 4 The lateral face: 03(0, 2) = (x3(0, 2), y3(0, z), z3(0, 2) where 0 ≤ 0 ≤ 2, -3 ≤ z ≤ 4 and 3 (0) = 5 cos theta ys (0) = 5 sin theta 23 (2) = Z Then (mind the orientation) ασι dr de + dr d0 + F(0₂). F.ndS= F(03). X [[P-nds - ["" ["P(0₂) - (x ) do 1² - ["²" ["Por) (x ² ) + 49 + [²* [*, ³903) · (² × ³ ) de 19 dz de 20 ər 0+0+0 Part 2 - Using the Divergence Theorem [[F.nds = [[[dis -IITO = divF dV dz dr do
Transcribed Image Text:Compute the outward flux of the vector field F(x, y, z) = 4ci + 3yj + 3k across the boundary of the right cylinder with radius 5 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), z₁(r, 0) where 0 ≤ 0 ≤ 2,0 < r ≤ 5 and r cos (theta) rsin(theta) ₁ (r, 0) = yı(r,0) = z₁(r, 0) = -3 The top face: 0₂ (r, 0) = (x2(r, 0), Y2(r, 0), z₂(r, 0) where 0 ≤ 0 ≤ 2,0 ≤r≤ 5 and #₂ (r, 0) =r cos theta 32 (r, 6) = r sin theta 22(1,0) = 4 The lateral face: 03(0, 2) = (x3(0, 2), y3(0, z), z3(0, 2) where 0 ≤ 0 ≤ 2, -3 ≤ z ≤ 4 and 3 (0) = 5 cos theta ys (0) = 5 sin theta 23 (2) = Z Then (mind the orientation) ασι dr de + dr d0 + F(0₂). F.ndS= F(03). X [[P-nds - ["" ["P(0₂) - (x ) do 1² - ["²" ["Por) (x ² ) + 49 + [²* [*, ³903) · (² × ³ ) de 19 dz de 20 ər 0+0+0 Part 2 - Using the Divergence Theorem [[F.nds = [[[dis -IITO = divF dV dz dr do
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