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Math Help - Center of Mass problem

  1. #1
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    Center of Mass problem

    Let y=9-x^2, y=0, x=0

    Find a number h such that y=h would cut the region into two equal areas.

    Now, I found out that the center of mass is (1,3.6), but how do I find the line? Do I just find the line that cut across (0,0) and (1,3.6)? The problem is the hint says that I need to integrate the area with respect to y, but I don't know how that is going to help.

    Thanks!
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  2. #2
    Senior Member vincisonfire's Avatar
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    What you really seach for is the mean of y.
    The definition of the mean is
     \frac{\int yf(y)dy}{\int f(y)dy} *
    In your case  h = \frac{\int_0^9 y\sqrt{9-y}dy}{\int_0^9 \sqrt{9-y}dy}=\frac{324}{5}\frac{1}{18}=\frac{18}{5}
    You must do the top part using part integration.
     \int y\sqrt{9-y}dy = -\frac{12}{5}(9-y)^{3/2}-\frac{2}{5}y(9-y)^{3/2}
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by tttcomrader View Post
    Let y=9-x^2, y=0, x=0

    Find a number h such that y=h would cut the region into two equal areas.

    Now, I found out that the center of mass is (1,3.6), but how do I find the line? Do I just find the line that cut across (0,0) and (1,3.6)? The problem is the hint says that I need to integrate the area with respect to y, but I don't know how that is going to help.

    Thanks!
    The area between y=0 and y=h \le 9 is:

    A(h)=\int_{y=0}^h \sqrt{9-y}\ dy=\frac{2}{3}[(9-h)^{3/2}-27]

    So the total area is A(9)=18 , so we look for h such that A(h)=9

    CB
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by vincisonfire View Post
    What you really seach for is the mean of y.
    The definition of the mean is
     \frac{\int yf(y)dy}{\int f(y)dy} *
    In your case  h = \frac{\int_0^9 y\sqrt{9-y}dy}{\int_0^9 \sqrt{9-y}dy}=\frac{324}{5}\frac{1}{18}=\frac{18}{5}
    You must do the top part using part integration.
     \int y\sqrt{9-y}dy = -\frac{12}{5}(9-y)^{3/2}-\frac{2}{5}y(9-y)^{3/2}
    For this to be correct would require that median = mean (that is we seek the median not the mean and these are not in generaly equal)

    CB
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  5. #5
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    I used to do these problems by simply finding the bounds of the area of integration - in this case you need to know the y min and max values of the region. Then you just say \int_{a}^{h} f(x)dx = \int_{h}^{b} f(x)dx, where a and b are your min and max.

    This is essentially the same thing that CB suggested, but to me it is easier this way.
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