Find the area under one arch of the trochoid x=10θ−1sin(θ),y=10−1cos(θ).?

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  • Pope
    Lv 7
    4 weeks ago

    x = 10θ - sin(θ)

    dx = [10 - cos(θ)] dθ

    dx = y dθ

    area

    = ∫[0,20π] y dx

    = ∫[0,2π] y² dθ ... (change of limits with change of variable of integration)

    = ∫[0,2π] [10 - cos(θ)]² dθ

    = ∫[0,2π] [100 - 20cos(θ) + cos²(θ)] dθ

    = ∫[0,2π] {201/2 - 20cos(θ) + [2cos²(θ) - 1]/2} dθ

    = ∫[0,2π] [201/2 - 20cos(θ) + cos(2θ)/2] dθ

    = [201θ/2 - 20sin(θ) + sin(2θ)/4] | [0,2π]

    = 201π

    Since I see a question about this, yes, the given equations define a trochoid. Using the geometric model of a rolling circle, the circle has radius 10 and rolls on the top of the x-axis. The trace point is offset from center by one unit.

  • rotchm
    Lv 7
    4 weeks ago

    Your question doesn't make sense since the eqs you give is not quite a trochoid.

    **Correction. Yes it is a trochoid. Just visually, its quite elongated hence my too quick conclusion.

    One arch= cycle is trivially found to be 2π. So, evaluate

    ∫y dθ where y = your y(θ).

    0

    Done.

    • Pope
      Lv 7
      4 weeks agoReport

      The result of your integral would be 20π. Let me suggest that you graph the curve, at least roughly. In the x direction, one cycle has breadth 20π, and y ≥ 9 throughout. That makes the area clearly greater than 180π. Notice also that your calculation is entirely independent of x.

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