# How do I find the area inside a limacon?

Mar 30, 2018

The area enclosed by the limaçon $r = b + a \cos \theta$ is $\pi \left({b}^{2} + \frac{1}{2} {a}^{2}\right)$

#### Explanation:

Consider a limaçon with polar equation:

$r = b + a \cos \theta$

Since the question is asked in a simple form, I will make a simplifying assumption that the limaçon does not self cross, so $\left\mid a \right\mid \le \left\mid b \right\mid$.

Dissecting the limaçon into infinitesimal segments about the origin note that each segment has area $\frac{1}{2} {r}^{2} d \theta$

So the total area of the limaçon is:

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

$\textcolor{w h i t e}{{\int}_{0}^{2 \pi} \frac{1}{2} {r}^{2} d \theta} = {\int}_{0}^{2 \pi} \frac{1}{2} \left({b}^{2} + 2 a b \cos \theta + {a}^{2} {\cos}^{2} \theta\right) d \theta$

$\textcolor{w h i t e}{{\int}_{0}^{2 \pi} \frac{1}{2} {r}^{2} d \theta} = {\int}_{0}^{2 \pi} \left(\frac{1}{2} {b}^{2} + a b \cos \theta + \frac{1}{4} {a}^{2} \left(1 + \cos 2 \theta\right)\right) d \theta$

$\textcolor{w h i t e}{{\int}_{0}^{2 \pi} \frac{1}{2} {r}^{2} d \theta} = {\left[\frac{1}{2} {b}^{2} \theta + a b \sin \theta + \frac{1}{4} {a}^{2} \left(\theta + \frac{1}{2} \sin 2 \theta\right)\right]}_{0}^{2 \pi}$

$\textcolor{w h i t e}{{\int}_{0}^{2 \pi} \frac{1}{2} {r}^{2} d \theta} = \pi \left({b}^{2} + \frac{1}{2} {a}^{2}\right)$