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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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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