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Math Help - Area and centroid

  1. #1
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    Post Area and centroid

    Hi All,

    a) Find the area bounded by the curve y=x^2 and the line y=4. Determine the co-ordinates of the centroid of the area.

    (I know about the centroid equations but do not know how to apply to this question. Also I think I know the area bounded equation \int\pi{y^2}dx but not to sure if you have to apply it to this question)


    b) A uniform wire is bent in the form of a three quarter circle (picture below). Determine the co-ordinates of the mass centre.

    (Also what kind of topic is this?)

    I'm sure you guys can help.

    Thanks
    Attached Thumbnails Attached Thumbnails Area and centroid-masscentre.gif  
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  2. #2
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    Quote Originally Posted by dadon
    Hi All,

    a) Find the area bounded by the curve y=x^2 and the line y=4. Determine the co-ordinates of the centroid of the area.

    (I know about the centroid equations but do not know how to apply to this question. Also I think I know the area bounded equation \int\pi{y^2}dx but not to sure if you have to apply it to this question)

    b) A uniform wire is bent in the form of a three quarter circle (picture below). Determine the co-ordinates of the mass centre....
    Hello,

    to a)

    The coordinates of a centroid which is bounded "at the top" by the curve f and is bounded "at the bottom" by the curve g can be calculated like this:

    x_C=\frac{\int^{a}_{b}{(x\cdot (f(x)-g(x) )) dx}}{\int^{a}_{b}{(f(x)-g(x))dx}} and

    y_C=\frac{\int^{a}_{b}{(f(x))^2-(g(x))^2 ) dx}}{2 \cdot \int^{a}_{b}{(f(x)-g(x))dx}}

    With your problem
    f(x) = 4
    g(x) = x^2
    b = -2
    a = 2

    Plug in all values you got, calculate the integrals and you'll get C \left(0 , \frac{12}{5} \right)

    I've attached a diagram to show the situation.

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails Area and centroid-centr_parab.gif  
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  3. #3
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    Post re:

    HEY! thanks for the reply!

    for yc I keep getting 6/5 rather then 12/5

    heres what I do:

    TOP

    y_c =\left( \frac{x^5}{5} \right)

    Limit -2 and 2

    Therefore:-

    \frac{(2)^5}{5} - \frac{(-2)^5}{5} = 6.4

    BOTTOM

    y_c =\left( \frac{x^3}{3} \right)

    Limit -2 and 2

    Therefore:-

     \frac{(2)^3}{3} - \frac{(-2)^3}{3} = \frac{16}{3}

    6.4 / 16/3 = 6/5 ?

    lol I don't know where I am going wrong!
    Last edited by dadon; May 2nd 2006 at 02:03 PM.
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  4. #4
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    Quote Originally Posted by dadon
    HEY! thanks for the reply!

    lol I don't know where I am going wrong!

    Hello,

    TOP

    y_c =\left( \frac{x^5}{5} \right)

    Limit -2 and 2

    Therefore:-

    \frac{(2)^5}{5} - \frac{(-2)^5}{5} = 6.4  \ \ \longleftarrow \mbox{this here looks funny to me}

    Greetings

    EB
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  5. #5
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    Post re:

    Thanks for that earboth!

    Sorry my mistake should have been 12.8 not 6.4 and hence gives the answer 12/5.

    Kind Regards,

    dadon
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  6. #6
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    Post re:

    oh sorry I forgot to ask where it says area bounded to the curve do I just integrate and plug in the limits (2 & -2) to get the area.

    Thanks

    If that is the case I get 16/3
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  7. #7
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    Quote Originally Posted by dadon
    oh sorry I forgot to ask where it says area bounded to the curve do I just integrate and plug in the limits (2 & -2) to get the area.

    Thanks

    If that is the case I get 16/3
    This is case of top curve minus bottom curve.
    The top curve is y=4
    The bottom curve is y=x^2
    From -2\leq x\leq 2 because they intersect at -2,2.
    Thus,
    \int^2_{-2} 4-x^2 dx=\left 4x-\frac{x^3}{3} \right|^2_{-2}=4(2)-\frac{8}{3}-4(-2)+\frac{-8}{3}= 16-\frac{16}{3}=\frac{32}{3}
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  8. #8
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    Thumbs up

    Thanks guys for that!

    And for correcting my mistakes. Hopefully I will learn from them.

    Kind Regards,

    dadon
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