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Math Help - Mass of enclosed area

  1. #1
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    Mass of enclosed area

    I really do not understand this and of course I need to by a few hours from now. If any one can help I would really appreciate it!


    I have an area enclosed by y = 3x and y = x^2 with x greater than or equal to 1 and less than or equal to 2


    I need to find the mass of the region and then find the moment about the x axis and the moment about the y axis.

    Thanks for all the help I have received this term!!!!!
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  2. #2
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    Quote Originally Posted by Frostking View Post
    I really do not understand this and of course I need to by a few hours from now. If any one can help I would really appreciate it!


    I have an area enclosed by y = 3x and y = x^2 with x greater than or equal to 1 and less than or equal to 2


    I need to find the mass of the region and then find the moment about the x axis and the moment about the y axis.

    Thanks for all the help I have received this term!!!!!
    The mass is just the integral (the area) if density is constant. the line is above the parabola for 1 < x < 2

    so we get

    m=\int_{1}^{2}(3x-x^2)dx=\frac{3}{2}x^2-\frac{1}{3}x^3 \bigg|_{1}^{2}=6-\frac{8}{3}-(\frac{3}{2}-\frac{1}{3})=\frac{13}{6}

    The moment about the y axis is

    M_y=\rho\int_{a}^{b}x[f(x)-g(x)]dx where rho is the density

    M_y=\int_{1}^{2}x(3x-x^2)dx=\frac{13}{4}

    The moment about the x axis is

    M_x=\rho\int_{a}^{b}\frac{1}{2}\{ [f(x)]^2-[g(x)]^2\}dx

    M_x=\frac{1}{2}\int_{1}^{2} \{ [3x]^2-[x^2]^2 \} dx =\frac{74}{5}

    Note that the center of mass is at the point \left(\frac{M_x}{m},\frac{M_y}{m}\right)

    Good luck.
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