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Math Help - Centroid

  1. #1
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    Centroid

    Find the exact coordinates of the centroid of the lamina bounded by the curve y=1/(1+x^2), the x-axis, the y-axis, and x=1.

    I got the Area to be (Pi/4)

    For X-Bar, I got: (2/Pi)*ln(2)

    I got stuck on the Y-Bar.
    I got;

    (4/Pi)*(1/2)*the integaral (1/(1+x^2)^2) dx.

    This comes to (2/Pi)*the integral (1/(1+x^2)^2) dx.

    I don't know how to integrate that. So can someone tell me if my area and X-Bar are correct and then help me integrate for Y-Bar?
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  2. #2
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    Anybody??? Bueller???
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  3. #3
    Grand Panjandrum
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    Put:

    I=\int \frac{1}{(1+x^2)^2}~dx

    Split using partial fractions:

    2I=\int \frac{1}{1+x^2}~dx+\int \frac{1-x^2}{(1+x^2)^2}~dx

    Now split up the second term:

    2I=\int \frac{1}{1+x^2}~dx+\int \frac{1}{(1+x^2)^2}~dx-\int \frac{x^2}{(1+x^2)^2}~dx

    The middle term is I, so:

    I=\int \frac{1}{1+x^2}~dx-\int \frac{x^2}{(1+x^2)^2}~dx

    The first integral on the right you can do, and the second can be done by integration by parts after spliting into the product of x and \frac{x}{(1+x^2)^2}.

    RonL
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  4. #4
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    Can you walk me through those steps? I am not getting it. I don't understand where the (1-x^2) came from. When I used partial fractions, I didn't get what you did.
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  5. #5
    Super Member wingless's Avatar
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    You asked a centroid question before and I gave a detailed explanation of finding centroids here. See it again, try to apply it to this question and then tell me where you're stuck.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by thegame189 View Post
    Can you walk me through those steps? I am not getting it. I don't understand where the (1-x^2) came from. When I used partial fractions, I didn't get what you did.
    Do you know how to expand something like \frac{1}{(1+u)^2} using partial fractions?

    You are going to write it as:

    \frac{1}{(1+u)^2}=\frac{A}{1+u}+\frac{B+Cu}{(1+u)^  2}

    then find the A,\ B and C that satisfy this.

    Now just replace u with x^2 to get the partial fraction expansion of \frac{1}{(1+x^2)^2}

    RonL
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by thegame189 View Post
    This comes to (2/Pi)*the integral (1/(1+x^2)^2) dx.

    I don't know how to integrate that.
    Of course you could do what people do in the real world and look it up ibn a table of integrals, or ask a symbolic integrator what it is.

    RonL
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  8. #8
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    I'm stuck on that derivative, not the actual centroid.

    And why isn't it (Ax+B)/(1+x^2) + (Cx+D)/((1+x^2)^2) ?

    Quote Originally Posted by CaptainBlack View Post
    Of course you could do what people do in the real world and look it up ibn a table of integrals, or ask a symbolic integrator what it is.

    RonL
    I have to show work or else I would.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by thegame189 View Post
    I'm stuck on that derivative, not the actual centroid.

    And why isn't it (Ax+B)/(1+x^2) + (Cx+D)/((1+x^2)^2) ?


    Try it and see what you get.

    RonL
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  10. #10
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    I got B=1 in yours and D=1 in mine. It's the same thing. I'm just not getting it. I hate days like these. Nothing seems to click.

    Mine: Ax^3+Bx^2+Ax+Cx+B+D = 1, A=0 from the Ax^3=0, B=0 from the Bx^2=0, C=0 from the Ax+Cx=0, and D=1 from B+D=1 and B=0

    Yours: Ax^2+Bx+A+C = 1, A=0 from Ax^2=0, B=0 from Bx=0, and C=1 from A+C=1 and A=0.
    Last edited by thegame189; March 18th 2008 at 12:08 PM.
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