# Finding the centroid of a cardiod curve.

1. Sep 13, 2011

### PhyStan7

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

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$$\leq$2$\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 $\bar{y}$=$\frac{4}{5}$

2. Relevant equations

$\bar{y}$=$\int$ydA$\div$$\int$dA

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

For the top part of the integral I am trying it with y=$\rho$sin$\phi$ 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.

2. Sep 16, 2011

### 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!

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3. Sep 16, 2011

### vela

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

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