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Math Help - Question about a center of mass/integrals problem

  1. #1
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    Question about a center of mass/integrals problem

    Set up the integrals to find the center of mass of an equilateraly triangle with density ρ(x,y)=K√ (x+y), with sides of length two and the midpoint of one side is the origin.

    Ok first question..I know that the limits on the x integral are from -1 to 1, but I am not sure what to put for y's limits. I am not sure if it would be 0 to 2-2x, 0 to 2+2x, or 2+2x to 2-2x, so I need some help with this please.

    Also, I know that I need to set up the integral for mass first. But after this, in my book, sometimes integrals are set up for x bar and y bar, and then in other cases integrals are set up for M_y and M_x. So my question is, which two should I do in the case of a triangle? Does it even make a difference?

    Thanks so much for your help!
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  2. #2
    MHF Contributor chisigma's Avatar
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    The triangle whose coordinates of the center of mass You have to compute is illustrated in the figure...



    The computation is performed in two steps...

    a) first computing the mass m of the triangle...

    \displaystyle m= \int \int _{A} \rho (x,y)\ dx\ dy (1)

    b) then computing the coordinates of the center of mass...

    \displaystyle x_{m} = \frac{1}{m}\ \int \int _{A} x\ \rho (x,y)\ dx\ dy

    \displaystyle y_{m} = \frac{1}{m}\ \int \int _{A} y\ \rho (x,y)\ dx\ dy (2)

    For the function \rho (x,y) we can set k=1 so that is \rho = \sqrt {x^{2} + y^{2}} and the integral (1) becomes...

    \displaystyle m= \int_{0}^{\sqrt{3}} \int_{-1 + \frac{x}{\sqrt{3}}}^{1 - \frac{x}{\sqrt{3}}} \sqrt{x^{2} + y^{2}}\ dx\ dy (3)

    ... and integrals (2)...

    \displaystyle x_{m}= \frac{1}{m}\ \int_{0}^{\sqrt{3}} \int_{-1 + \frac{x}{\sqrt{3}}}^{1 - \frac{x}{\sqrt{3}}}  x\ \sqrt{x^{2} + y^{2}}\ dx\ dy

    \displaystyle y_{m}= \frac{1}{m}\ \int_{0}^{\sqrt{3}} \int_{-1 + \frac{x}{\sqrt{3}}}^{1 - \frac{x}{\sqrt{3}}} y\ \sqrt{x^{2} + y^{2}}\ dx\ dy (4)

    Kind regards

    \chi \sigma
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