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March 26, 2017

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find the centroid of the plane region bounded by the curves y = cos x, y=sinx, x=0,

  • calculus - ,

    Double check my math, because it gets messy...

    Break it into two intervals: [0,pi/4] and [pi/4,pi/2]

    Int(sin x)[0,pi/4] + Int(cos x)[pi/4,pi/2]
    = (-cos x)[0,pi/4] + (sin x)[pi/4,pi/2]
    = 2-√2
    This is the denominator in the formulas for xbar and ybar: D = 2-√2

    Now for the numerators:
    xbarn = Int(x sin x)[0,pi/4] + Int(x cos x)[pi/4,pi/2]
    Recall that using integration by parts,
    Int(x sin x) = -x cos x + sin x)
    Int(x cos x) = x sin x + cos x)

    If my math is right, xbarn = pi/4 * (2-√2)

    So, xbar = xbarn/D = pi/4
    Makes sense, since the area is symmetric about the line x = pi/4.

    Now for ybar

    ybarn = 1/2 Int(sin^2 x)[0,pi/4] + 1/2 Int(cos^2 x)[pi/4,pi/2]

    Recall that
    sin^2 x = (1 - cos(2x))/2
    cos^2 x = (1 + cos(2x))/2

    ybarn = 1/4 Int(1 - cos 2x)[0,pi/4] + 1/4 Int(1 + cos 2x)[pi/4,pi/2]

    = 1/4 (x - 1/2 sin 2x)[0,pi/4] + 1/4 (x + 1/2 sin 2x)[pi/4,pi/2]

    = 1/8 (pi-2)

    So, ybar = ybarn/D = (pi-2)/(8*(2-√2))

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