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Given the Cardioid by $f(\varphi)=3-3\cos(\varphi)$ and the circle given by $g(\varphi)=-6\cos(\varphi)$. I have 2 questions regarding its areas:

  1. Why $\frac{1}{2}\int_{0}^{\pi}g(\varphi)^2d\varphi=9\pi$ when it should be $\frac{9\pi}{2}$?
  2. Why $\frac{1}{2}\int_{0}^{\pi}g(\varphi)^2 - f(\varphi)^2d\varphi=\frac{-9\pi}{4}$ when it should be $\frac{9\pi}{4}$?

The first question regards the area of a circle of diameter $6$ and the second questions is about how much more area has the Cardioid than the circle. About how I know that the results should be those: The circle has diameter $6$ so it's area is $9\pi$ and hence the integral of the first halve of the circle in polar coordinates should be $\frac{9\pi}{2}$. About the other result, if I calculate $\frac{1}{2}\int_{0}^{\pi}g(\varphi)^2$ and substract the half area of the circle it gives $\frac{9\pi}{4}$.

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  • $\begingroup$ If you plot the two curves, you will easily see that the cardioid encircles a bigger area than the circle $\endgroup$ Commented Jun 2 at 17:58
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    $\begingroup$ Please provide an explicit calculation that leads you to claim "it should be $\frac{9\pi}2$". $\endgroup$
    – Lieven
    Commented Jun 2 at 17:58
  • $\begingroup$ @Lieven Hope that helps $\endgroup$
    – MiguelCG
    Commented Jun 2 at 18:08

1 Answer 1

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It's not the first half of the circle but the full circle. The origin lies on the circumference, not at the centre. All of the circle lies between the polar arguments $0$ and $\pi.$

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