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Math Help - Centroid

  1. #1
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    Centroid

    Find the centroid of the region bounded by the following curves  y = x^2, y = x+3

    Centroid : (___, ___)
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  2. #2
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    Quote Originally Posted by viet View Post
    Find the centroid of the region bounded by the following curves  y = x^2, y = x+3

    Centroid : (___, ___)
    I will assume it is a homogenous density.

    In that case fist find the mass.
    In this case simply the area.

    \int_D \int dA
    To express the region we need the intersection points between the line and parabola.
    x+3=x^2
    x^2-x-3=0
    x=.5\pm .5\sqrt{13}
    The upper curve is,
    x+3 and lower is x^2.
    Thus,
    m=\int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}} \int_{x^2}^{x+3} dy\, dx
    Thus,
    \int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}}x+3-x^2dx\approx 5.67.

    Now we find the first moments.

    M_x=\int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}} \int_{x^2}^{x+3}y dy\, dx
    Thus,
    \int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}}\frac{1}{2}(x+3)^2-\frac{1}{2}(x^2)^2 dx\approx 11.84

    M_y=\int_{.5-.5\sqrt{13}}^{.5+.5\sqrt{13}} \int_{x^2}^{x+3} xdy\, dx
    Thus,
    \int_{-.5-.5\sqrt{13}}^{.5+.5\sqrt{13}}x(x+3)-x(x^2)dx=8.14

    Thus,
    \bar{x}= \frac{M_x}{M}=\frac{11.84}{5.67}\approx 2.08

    \bar{y}=\frac{M_y}{M}=\frac{8.14}{5.67}\approx 1.43
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  3. #3
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    the y coordinate y_c is given by
    \frac{1}{2A}\int_a^b(f(x)-g(x))(f(x)+g(x))dx.
    and the x coordinate x_c is given by
    \frac{1}{A}\int_a^bx(f(x)-g(x))dx

    A is the area of the region between the two graphs and a,b are the points where they intersect.
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  4. #4
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    i got it thanks
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