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brainshot3

  • 4 years ago

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the portion of the elliptic cylinder r(u,v) = (2cos v) i + (sin v) j + (u) k with 0 ≤ u ≤ 5, 0 ≤ v ≤ 2pi.

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  1. kumar2006
    • 4 years ago
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    ∫∫s F dS ... F = e^(-y) i - y j + x sin(z) k s: the elliptic cylinder r(u,v) = 2 cos(v) i + sin(v) j + u k ... 0 ≤ u ≤ 5, 0 ≤ v ≤ 2π ∫∫s F dS = ∫∫∫v ∇·F dV ∇·F = x cos(z) - 1 = ∫∫∫v x cos(z) - 1 dx dy dz v: x² + 4y² ≤ 4 ; 0 ≤ z ≤ 5 let u = x , v = 2y , z = z ∂(u,v)/∂(x,y) = 2 = 1/2 ∫∫∫v u cos(z) - 1 du dv dz v: u² + v² ≤ 4 ; 0 ≤ z ≤ 5 let u = r cos(θ) ; v = r sin(θ) ; z = z ∂(u,v)/∂(r,θ) = r = 1/2 ∫∫∫ (r cos(θ) cos(z) - 1) r dr dθ dz {(r,θ,z) | 0 ≤ r ≤ 2 ; 0 ≤ θ ≤ 2π ; 0 ≤ z ≤ 5} = 1/3 ∫∫ (4 cos(θ) cos(z) - 3) dθ dz {(θ,z) | 0 ≤ θ ≤ 2π ; 0 ≤ z ≤ 5} = -2π ∫ dz {(z) | 0 ≤ z ≤ 5} = -10π

  2. kumar2006
    • 4 years ago
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    ∫c F·dr .... F = (x + y) i + (xy) j - (z^2) k ; c: from (0,0,0) to (1,3,1) to (2,-1,4) → (0,0,0) to (1,3,1) x = t ; y = 3t ; z = t ; 0 ≤ t ≤ 1 dx = 1 dt; dy = 3 dt; dz = 1 dt ∫c F·dr = ∫ (x + y) dx + xy dy - z^2 dz = ∫ (t + 3t) + 9t² - t² dt [0,1] = 14/3 → (1,3,1) to (2,-1,4) x = t + 1 ; y = 3 - 4t ; z = 1 + 3t dx = 1 dt; dy = -4 dt; dz = 3 dt ∫c F·dr = ∫ (x + y) dx + xy dy - z^2 dz = ∫ (t + 1 + 3 - 4t) - 4(t + 1)(3 - 4t) - 3(1 + 3t)^2 dt [0,1] = ∫ -11t^2 - 17t - 11 dt [0,1] = -139/6 14/3 - 139/6 = -37/2 Answer (2): -37/2

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