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Find centroid two-dimensional region

  1. Apr 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the centroid of the region bounded by the given curves.

    [tex]y = x^3[/tex][tex]x + y = 10[/tex][tex]y = 0[/tex]

    2. Relevant equations

    [tex]\bar x = \frac{1}{A}\int^b_a xf(x) dx[/tex][tex]\bar y = \frac{1}{A}\int^b_a \frac{1}{2}(f(x))^2 dx[/tex]
    Where A is the area of the region containing the centroid.

    3. The attempt at a solution

    First thing I did was find the intersection of the curves [tex]x^3 = 10-x \Rightarrow x = 2 \Rightarrow P(2,8)[/tex]
    Then I put things in terms of y to integrate to find the area A of the region.[tex]y = x^3 \Rightarrow x = \sqrt[3] y[/tex][tex]x + y = 10 \Rightarrow x = 10 - y[/tex][tex]A = \int^8_0 ((10-y)-(\sqrt[3] y)) dy = 36[/tex]
    So [tex]\bar x = \frac{1}{A}\int^b_a yf(y) dy = \frac{1}{36}\int^8_0 y((10-y)-(\sqrt[3] y))dy = \frac{496}{189} \approx 2.63[/tex][tex]\bar y = \frac{1}{A}\int^b_a \frac{1}{2}(f(y))^2 dy = \frac{1}{36}\int^8_0 \frac{1}{2}((10-y)^2-(\sqrt[3] y)^2)dy = \frac{584}{135} \approx 4.33[/tex]
    So the centroid is at [tex]\left(\frac{496}{189},\frac{584}{135}\right)[/tex]
    What I'm getting here isn't correct. Thanks for commenting!
     
  2. jcsd
  3. Apr 1, 2012 #2

    HallsofIvy

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    This is completely wrong. If you look at a graph of x+y= 10 and [itex]y= x^3[/itex], then you will see that, for each y, x ranges from 0 on the left to
    1)[itex] y^{1/3}[/itex] for y between 0 and 8 and
    2) [itex]10 - y[/itex] for y between 8 and 10.
    The area is given by
    [tex]\int_{y= 0}^8 y^{1/3}dy+ \int_{y= 8}^{10} (10- y) dy[/tex].

    Much simpler, and what most people would do, is to integrate with respect to x. For each x, between 0 and 2, y ranges from [itex]x^3[/itex] to [itex]10- x[/itex] so the area would be
    [tex]\int_{x=0}^2(10- x- x^3) dx[/tex]

     
  4. Apr 1, 2012 #3
    Thank you for responding, HallsofIvy!

    I think I'm still confused. I thought this was the region I needed to find the centroid of:

    wh9tq.png

    Which has an area [tex]A = \int^2_0 x^3 dx + \int^{10}_2 (10-x) dx = 36 [/tex]
     
  5. Apr 1, 2012 #4

    HallsofIvy

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    Oh, blast! I misead "y= 0" as "x= 0".
     
  6. Apr 1, 2012 #5
    I had another look and figured it out. I guess when I did this in terms of y my x and y coordinates got flipped. The values of the integral are correct, but the coordinates are backwards, so the centroid is at[tex]\left(\frac{584}{135},\frac{496}{189}\right)[/tex]
     
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