# Area of a cardioid and a circle

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 $$\frac{9\pi}{4}$$.

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

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 $$\pi.$$