# How do I find the area inside a cardioid?

Oct 25, 2014

The polar equation of a cardioid is

$r = 2 a \left(1 + \cos \theta\right)$,

which looks like this with a=1:

So, the area inside a cardioid can be found by

$A = {\int}_{0}^{2 \pi} {\int}_{0}^{2 a \left(1 + \cos \theta\right)} r \mathrm{dr} d \theta$

$= {\int}_{0}^{2 \pi} {\left[{r}^{2} / 2\right]}_{0}^{2 a \left(1 + \cos \theta\right)} d \theta$

$= {\int}_{0}^{2 \pi} \frac{{\left[2 a \left(1 + \cos \theta\right)\right]}^{2}}{2} d \theta$

$= 2 {a}^{2} {\int}_{0}^{2 \pi} \left(1 + 2 \cos \theta + {\cos}^{2} \theta\right) d \theta$

$= 2 {a}^{2} {\int}_{0}^{2 \pi} \left[1 + 2 \cos \theta + \frac{1}{2} \left(1 + \cos 2 \theta\right)\right] d \theta$

$= 2 {a}^{2} {\left[\theta + 2 \sin \theta + \frac{1}{2} \left(\theta + \frac{\sin 2 \theta}{2}\right)\right]}_{0}^{2 \pi}$

$= 2 {a}^{2} \left[2 \pi + \frac{1}{2} \left(2 \pi\right)\right] = 6 \pi {a}^{2}$

I hope that this was helpful.