# How do I find the area inside the cardioid r=1+costheta?

Aug 8, 2018

$4 \pi$

#### Explanation:

We will integrate the area differential for a polar function:
$\mathrm{dA} = d \left(\frac{1}{2} {r}^{2} \theta\right) = r \theta d r + \frac{1}{2} {r}^{2} d \theta$

We can integrate this just $d \theta$:
$\mathrm{dA} = \left(r \theta \frac{\mathrm{dr}}{d \theta} + \frac{1}{2} {r}^{2}\right) d \theta$

This yields
$A = \int \setminus \setminus \mathrm{dA} = {\int}_{0}^{2 \pi} \left(\left[1 + \cos \theta\right] \cdot \theta \cdot \left(- \sin \theta\right) + \frac{1}{2} {\left(1 + \cos \theta\right)}^{2}\right) d \theta$
$A = {\int}_{0}^{2 \pi} \left[\frac{3}{4} + \frac{1}{4} \cos 2 \theta + \cos \theta - \theta \sin \theta - \frac{1}{2} \theta \sin 2 \theta\right] d \theta$

This integral is pretty simple, with a few integrations by parts.

The ultimate answer turns up to be $A = 4 \pi$.