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Homework Help: Centroid of parabola

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data

    y = 16 -x^2 find centroid bounded by x axis

    2. Relevant equations

    x = (1/A) ∫ x(f(x)) dx and (1/A) (1/2)(f(x))^2 dx = y

    3. The attempt at a solution

    I just applied it. It is a weird because I would of thought that x would of been at 0. But I didn't get that I x = 9/16 and y = 8/5. Y might be OK but wouldn't x have to be at 0?

    I calculated the area I got 256/3

    So x = (3/256) ∫(x(16-x^2) dx = (3/256)∫ 16x - x^3 dx = (3/256) (8x^2 -(x^4/4)) between [-4,4] = 9/16

    For y = (3/256) ∫ (1/2)(16-x^2)^2 dx = (3/256)∫ 256 - 32x^2 + x^4 dx = (3/512)(256x = (32/2)x^3 - (x^5/5)) between [-4,4] = 8/5

    Like I said it seems that x would be at 0 but I'm not sure I think I did everthing OK.
  2. jcsd
  3. Oct 19, 2013 #2


    Staff: Mentor

    Your intuition is correct, but your work isn't. ##\bar{x}## = 0, which is pretty obvious from the symmetry of the region.
  4. Oct 19, 2013 #3
    So, Even though it is the long way if I did it the way I did it should workout to be 0. Why do I get what I get ? did you look at y ? What did you think of y?
  5. Oct 19, 2013 #4
    I found my mistake for x. Do you think y looks OK I went over it again seems OK to me.
  6. Oct 19, 2013 #5


    Staff: Mentor

    Your integral doesn't look right to me. It should be
    $$\frac 1 A \int_0^{16} y * (x_{right} - x_{left}) dy$$

    The integrand represents the moment of a horizontal strip across the parabola, and the moment in this case is the length of the lever arm (y) times the area of the horizontal strip ΔA = (xright - xleft)Δy.

    Your function is y = 16 - x2, or equivalently, x = ±√(16 - y).

    Part of your problem seems to be treating x and y cavalierly - you can't do that and come up with the right answer.

    With all of these applications of integration, it's extremely helpful to have a good drawing of the situation. Just memorizing some formulas and using them blindly isn't a good strategy.
  7. Oct 19, 2013 #6
    ##\frac 1 A \int_0^{16} (1/2)[f(x)]^2 dx##

    This is what my book gives for calculation of y. I don't understand your formula. Well I do but why would I need to deviate from what the book says for this problem. I don't see why?
    Last edited: Oct 19, 2013
  8. Oct 19, 2013 #7


    Staff: Mentor

    Your formula should be ##\frac 1 A \int_{-4}^{4} (1/2)[f(x)]^2 dx##, with the difference being the limits of integration. You're integrating with respect to x, and x is in the interval [-4, 4].

    With that correction, your formula and mine should agree. The difference between them is that my formula uses horizontal strips that extend across the parabola. For each of these the lever arm is the y value on the strip. Each of my horizontal strips is very narrow, so all the strips have about the same y value. In my formula, the moment is y * <width of strip> * Δy.

    Your formula uses vertical strips of height f(x). The lever arm is at the middle of the strip, so is (1/2)f(x). The moment is f(x) * (1/2)f(x) * Δx.
  9. Oct 19, 2013 #8
    I meant [-4,4] I forgot to change it. Yeah I recalculated y with my formula and I got 32/5.

    In yours what is x left and right? Would it be after subtracting.

    ##\frac 1 A \int_0^{16} y * (2\sqrt(16-y)) dy##
  10. Oct 19, 2013 #9


    User Avatar
    Science Advisor
    Homework Helper

    The books formula is right. It's an equivalent formulation. To show they are the same you use Green's theorem.
  11. Oct 19, 2013 #10


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    Science Advisor
    Homework Helper

    Yes, or you can explain it that way instead of using Green's theorem. Actually somewhat nicer.
  12. Oct 19, 2013 #11
    Pretty out there though for me.
  13. Oct 19, 2013 #12
    What class out greens theorem come up/
  14. Oct 19, 2013 #13
  15. Oct 19, 2013 #14


    User Avatar
    Science Advisor
    Homework Helper

    Vector calculus, but I wouldn't worry about it. Mark44's explanation is fine and much more elementary.
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