# How do you find the area of r=1+cos(theta)?

Jul 23, 2016

$\frac{3 \pi}{2}$ areal units.

#### Explanation:

If the pole r = 0 is not outside the region, the area is given by

$\left(\frac{1}{2}\right) \int {r}^{2} d \theta$, with appropriate limits.

The given curve is a closed curve called cardioid.

It passes through the pole r = 0 and is symmetrical about the initial

line $\theta = 0$.

As $r = f \left(\cos \theta\right)$, r is periodic with period $2 \pi$.

And so the area enclosed by the cardioid is

$\left(\frac{1}{2}\right) \int {r}^{2} d \theta$, over $\theta \in \left[0 , 2 \pi\right]$.

$\left(\frac{1}{2}\right) \left(2\right) \int {\left(1 + \cos \theta\right)}^{2} d \theta , \theta \in \left[0 , \pi\right]$, using symmetry about $\theta = 0$

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

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

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

=[3/2theta+2sin theta]+(1/2)(1/2)sin 2theta],

between limits $0 \mathmr{and} \pi$

$= \frac{3 \pi}{2} + 0 + 0$