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Math Help - need help for explaining the following question

  1. #1
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    Question need help for explaining the following question

    Question :
    Find the mass and center of gravity of the lamina with density delta(x,y)=xy, lying in the first quadrant and bounded by the circle x^2 + y^2 = a^2 and the coordinates axes.


    The book said that
    " the center of gravity will be the same for x and y due to the symmetry of the figure. so y bar = x bar. "

    I don't understand it. Could you please explain to me? Thank you very much.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kittycat View Post
    Question :
    Find the mass and center of gravity of the lamina with density delta(x,y)=xy, lying in the first quadrant and bounded by the circle x^2 + y^2 = a^2 and the coordinates axes.


    The book said that
    " the center of gravity will be the same for x and y due to the symmetry of the figure. so y bar = x bar. "

    I don't understand it. Could you please explain to me? Thank you very much.
    i believed i explained this to you before. remember, we look for symmetry about vertical and horizontal lines first, then any other lines that we can fin symmetry about.

    any vertical line that cuts the region into equal halves is x bar

    any horizontal line that cuts the region into halves is y bar

    for a circle centered at the origin, the axis do this. so we have y = x = 0 giving the desired lines


    EDIT: oh sorry, we're not dealing with a full circle, but the essence of my explanation is still true, so it's not specifically x = y = 0 here, what would be the point of the question then?

    the line x = y cuts the quarter-circle into two symmetric halves, that's why they got that answer
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  3. #3
    Eater of Worlds
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    You can use polar.

    x=rcos{\theta}, \;\ y=rsin{\theta}

    Therefore, you have:

    M=\int_{0}^{\frac{\pi}{2}}\int_{0}^{a}r^{3}sin{\th  eta}cos{\theta} \;\ drd{\theta}

    \overline{x}=\overline{y} from the symmetry of the density and the region:

    M_{y}=\frac{1}{M}\int_{0}^{\frac{\pi}{2}}\int_{0}^  {a}r^{4}sin{\theta}cos^{2}{\theta} \;\ drd{\theta}
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  4. #4
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    Hi galactus and Jhevon,

    May I ask you the following question:

    Where is the center of gravity usually located in this kind of integration? Is it located in the Region R ?

    Thank you very much.
    Last edited by kittycat; September 2nd 2007 at 12:19 PM.
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  5. #5
    Eater of Worlds
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    Yes. can you find the coordinates of the center of gravity?. They will be in terms of a.
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  6. #6
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    yes, that is ( (8a)/15 , (8a)/15 ) .
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