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Finding the centroid of a cardiod curve.

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data

    A cardioid or heart shaped curve is given in plane polar coordinates [itex]\rho[/itex],[itex]\phi[/itex] by the parametrisation [itex]\rho[/itex]=1-cos[itex]\phi[/itex] or equivilantly
    [itex]\rho[/itex]=2sin[itex]^{2} (\frac{ϕ}{2})[/itex] for [itex]0\leq\phi[/itex][itex]\leq[/itex]2[itex]\pi[/itex]

    A shape resembling an apple is generated by revolving the upper (+y) half of the cardioid curve about the x axis. Show using an integral of functions of [itex]\frac{ϕ}{2}[/itex], that the upper half of the cardioid curve lies at [itex]\bar{y}[/itex]=[itex]\frac{4}{5}[/itex]

    2. Relevant equations


    3. The attempt at a solution

    Sorry about my awful latex usage, I have never used this stuff before! Basically I try the above integral. I found the area of the whole shape (above and below x axis) in a previous part as [itex]\frac{3}{2}[/itex][itex]\pi[/itex] by integrating the limits 0[itex]\leq[/itex][itex]\phi[/itex][itex]\leq[/itex]2[itex]\pi[/itex] and [itex]\rho[/itex] between 0 and 2sin[itex]^{2}[/itex][itex]\frac{ϕ}{2}[/itex] which I know is right.

    For the top part of the integral I am trying it with y=[itex]\rho[/itex]sin[itex]\phi[/itex] with [itex]\phi[/itex] between 0 and [itex]\pi[/itex] and [itex]\rho[/itex] between 0 and 2sin[itex]^{2}[/itex][itex]\frac{ϕ}{2}[/itex] but it does not work as the extra [itex]sin\phi[/itex] means it is impossible to have a [itex]\pi[/itex] in the integral solution so the final answer for [itex]\bar{y}[/itex] will always be over [itex]\pi[/itex] . I think I am setting up the ydA integral wrong but am not sure how. Any guidance wouild be much appreciated, cheers.
  2. jcsd
  3. Sep 16, 2011 #2
    For greater clarity I have attached the question ( part b i), I have already done a) and my attempt at a solution. I just think perhaps I am setting up the integral wrong, any hints to get me on the right track would be appreciated, thanks!

    Attached Files:

  4. Sep 16, 2011 #3


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    You're misinterpreting the problem. It's asking for the centroid of the curve, not the surface.
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