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Centroid Question

  1. Apr 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Just studying statics in my first year of university and i do not understand the centroid topic at all. The question goes like this, given a equation y=x^2 , and the x-axis is 0 - 4 , with the lower portion of the graph is highlighted. It is given by diagram but sorry I don't have scanner so can't provide the diagram, but i think the info i gave is enough to illustrate that simple graph. The question ask to find the centroid of it.

    2. Relevant equations
    The equations i know that i have to use is X-dash and y-dash formula. x-dash formula is (let assume | is integration) "|xdA within 0-4 divided by |dA, (A=|ydx)" and
    "|ydA within 0-16 divided by |dA, (A=|xdy)".

    3. The attempt at a solution

    I go through all the equations and I found X-dash is 3 which I think is correct, but my y-dash is 9.6, which is doubled of the answer provided to me, 4.8... I don't understand what's wrong with my calculation ?

    p/s : I asked my lecturer and he said the |ydA shud be |(y/2)dA , but why ? Please do help me because my exam is coming and I still dun understand this small parts of statics. Thanks for any help.
    Last edited: Apr 19, 2007
  2. jcsd
  3. Apr 19, 2007 #2


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    I've no idea what your formulae mean, but you get the centroid x coord by integrating x*y*dx from 0-4 and dividing that by the integral of y*dx from 0-4.

    The rotate the axes and repeat for y.
  4. Apr 20, 2007 #3
    For the Y centroid

    Hey Mentz114, thanks for the help, the formulae u provided helped me to find the x-centroid, which is 3, but i still find the y centroid to be 9.6. Isn't that the formula to find y centroid is integrating y*x*dy from 0-16 and dividing that by the integral of x*dy from 0-16 ?
  5. Apr 20, 2007 #4


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    I got the x centroid to be 8/3, but I could be wrong.

    To do y, keep the same vars, but invert the function so

    y = sqrt(x)
  6. Apr 20, 2007 #5


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    Should you be taking the square root of 9.6 ?
  7. Apr 20, 2007 #6
    I found the X centroid to be 3 if using my own formula and yours.
    You said "integrating x*y*dx from 0-4 and dividing that by the integral of y*dx from 0-4" to find x-centroid right ?

    So |x*y*dx will turned into |x^3dx since y=x^2 , hence i got 64 after integration.

    Then |y*dx will turned into |x^2dx since y=x^2 as well, hence I got 21.333 after integration.

    So |x*y*dx divide by |y*dx = 64/21.3333 = 3 ( X-Centroid)
  8. Apr 20, 2007 #7
    You said to keep the same variants, means use 0-4 ? But the function is inverted, shouldn't be using 0-16 ? since y=x^2 , and shouldn't be x=sqrt(y) ?
  9. Apr 20, 2007 #8


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    The integral of x^3dx is (x^4)/4 and the integral of x^2dx is (x^3)/3.

    As I said, my answer could be wrong.

    I think you've got the hang of it so I'll leave you to it.

    Try to understand why the formulae are correct.
  10. Apr 20, 2007 #9
    Hmm , I can understand for X centroid , but how about Y centroid ? Could you may be write the formula for me to find Y centroid ? As I still haven't found the Y centroid yet .... not the correct one at least ...
  11. Apr 20, 2007 #10


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    I get 9.6 !

    [tex]\int_0^{16} x^{\frac{3}{2}} dx[/tex]

    divided by

    [tex]\int_0^{16} x^{\frac{1}{2}} dx[/tex]
  12. Apr 20, 2007 #11
    I got 9.6 as well, but my lecturer's answer is 4.8 which is half of it ... that's make me confusing ... :confused:
  13. Apr 20, 2007 #12


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    Metz114 is finding the Y centroid of the wrong area.

    Draw a picture. The "width" of the area is that you are integrating over is [tex]4-y^{1/2}[/tex], not [tex]y^{1/2}[/tex].

    So the integral you need to evaluate is [tex]\int_0^{16} y(4-y^{1/2})dy[/tex]

    The fact that you get twice the right answer is a conincidence.
  14. Apr 20, 2007 #13


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    There we are - corrected.
    I imagined the shape wrongly. Thank you, AlephZero
    Last edited: Apr 20, 2007
  15. Apr 20, 2007 #14
    Ok, finally got the answer, but still forgive for my noob-ness, why when find the y-centroid, the integrating have to be [​IMG] ? Why the x had become 4-y^(1/2) ? Sorry but anyway, thanks for all the helps.
  16. Apr 21, 2007 #15


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    Here's a picture to show what's going on when you evaluate the two integrals.

    Attached Files:

  17. Apr 21, 2007 #16
    huh ? It seems like the atcachment is waiting for approval ... Why need to have approval to show attachment ? Or may be Alephzero is kind enough to send me the link of that gif file ? Thanks ...
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