Finding the centroid of a cardiod curve.

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The discussion focuses on finding the centroid of a cardioid curve defined in polar coordinates. The user has calculated the area of the shape generated by revolving the upper half of the cardioid about the x-axis but struggles with setting up the integral for the y-coordinate of the centroid. They express confusion over the integration limits and the inclusion of sin(φ) in their calculations, which complicates achieving the correct result. A clarification is provided that the problem pertains to the centroid of the curve itself rather than the surface area generated. The user seeks guidance on correctly establishing the integral for the centroid calculation.
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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.
 
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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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You're misinterpreting the problem. It's asking for the centroid of the curve, not the surface.
 
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