Finding the centroid of a cardiod curve.

[PhyStan7]

Homework Statement

A cardioid or heart shaped curve is given in plane polar coordinates \rho,\phi by the parametrisation \rho=1-cos\phi or equivilantly
\rho=2sin^{2} (\frac{ϕ}{2}) for 0\leq\phi\leq2\pi

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 \frac{ϕ}{2}, that the upper half of the cardioid curve lies at \bar{y}=\frac{4}{5}

Homework Equations

\bar{y}=\intydA\div\intdA

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 \frac{3}{2}\pi by integrating the limits 0\leq\phi\leq2\pi and \rho between 0 and 2sin^{2}\frac{ϕ}{2} which I know is right.

For the top part of the integral I am trying it with y=\rhosin\phi with \phi between 0 and \pi and \rho between 0 and 2sin^{2}\frac{ϕ}{2} but it does not work as the extra sin\phi means it is impossible to have a \pi in the integral solution so the final answer for \bar{y} will always be over \pi . I think I am setting up the ydA integral wrong but am not sure how. Any guidance wouild be much appreciated, cheers.

 

[PhyStan7]

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!

 

[vela]

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