S is parametrized by X(s, t) = (s cos(t), s sin(t), t), 0 ≤ s ≤ 4, 0 ≤ t ≤ π/2 F = zi + 2xj + yk?

S is parametrized by X(s, t) = (s cos(t), s sin(t), t), 0 ≤ s ≤ 4, 0 ≤ t ≤ π/2

F = zi + 2xj + yk

Verify Stoke's theorem.

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  • 1 decade ago
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    The boundary of S can be decomposed into 4 parts; a segment from (0,0,0) to (4,0,0), a helix from (4,0,0) to (0,4,π/2), a segment from (0,4,π/2) to (0,0,π/2) and another segment from (0,0,π/2) to (0,0,0):

    δ(S) = L1 U L2 U L3 U L4

    L1(s) = (s, 0, 0), 0 ≤ s ≤ 4; dL1 = (1,0,0) ds

    L2(t) = (4 cos(t), 4 sin(t), t), 0 ≤ t ≤ π/2; dL2 = (-4 sin(t), 4 cos(t), 1) dt

    L3(s) = (0, 4-s, π/2), 0 ≤ s ≤ 4; dL3 = (0, -1, 0) ds

    L4(t) = (0, 0, π/2-t), 0 ≤ t ≤ π/2; dL4 = (0,0,-1) dt

    ∫(over δ(S)) F . dL =

    = ∫(from s=0 to 4) (0, 2s, 0).(1, 0, 0) ds +

    + ∫(from t=0 to π/2) (t, 8 cos(t), 4 sin(t)).(-4 sin(t), 4 cos(t), 1) dt +

    + ∫(from s=0 to 4) (π/2, 0, 4-s).(0, -1, 0) ds +

    + ∫(from t=0 to π/2) (π/2-t, 0, 0).(0, 0, -1) dt =

    = 0 +

    + ∫(from t=0 to π/2) [-4 t sin(t) + 32 cos²(t) + 4 sin(t)] dt +

    + 0 +

    + 0 =

    = -4 + 8 π + 4 = 8 π

    ====

    curl(F) = i + j + 2k

    dS = ∂S/∂s X ∂S/∂t ds dt = (cos(t), sin(t), 0) X (-s sin(t), s cos(t), 1) ds dt=

    = (sin(t), -cos(t), s) ds dt

    ∫(over S) curl(F) . dS =

    = ∫(from s=0 to 4) ∫(from t=0 to π/2) (1,1,2).(sin(t), -cos(t), s) ds dt =

    = ∫(from s=0 to 4) ∫(from t=0 to π/2) [sin(t) - cos(t) + 2s] ds dt

    = ∫(from s=0 to 4) (1 - 1 + s π) ds =

    = (4²-0²)π/2 = 8 π

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