How do I find the area inside a limacon?

Hint: In geometry, a limaçon, is defined as a roulette formed by the path of a point fixed to a circle when that circle rolls around the outside of a circle of equal radius. Limaçon curves look like circles. They have various types depending on the values in their equations. The polar equation of the limacon is \[r=b+a\cos \theta \]. We will assume that the limacon does not cross itself, for this condition to be true \[\left| b \right|\ge \left| a \right|\]. The infinitesimal segment of limacon has an area \[\dfrac{1}{2}{{r}^{2}}d\theta \]. To find the area of limacon, we have to integrate this over the range \[0\] to \[2\pi \].

Complete step by step solution:
We are asked to find the area inside a limacon, we know that the polar equation of a limacon is \[r=b+a\cos \theta \]. We know that the infinitesimal segment of limacon has an area \[\dfrac{1}{2}{{r}^{2}}d\theta \]. To find the area of limacon, we have to integrate this over the range \[0\] to \[2\pi \].
We can do this as follows,
\[\int\limits_{0}^{2\pi }{\dfrac{1}{2}{{r}^{2}}d\theta }=\int\limits_{0}^{2\pi }{\dfrac{1}{2}{{\left( b+a\cos \theta \right)}^{2}}d\theta }\]
Simplifying the above expression, we get
\[\int\limits_{0}^{2\pi }{\dfrac{1}{2}{{r}^{2}}d\theta }=\int\limits_{0}^{2\pi }{\dfrac{1}{2}\left( {{b}^{2}}+{{a}^{2}}{{\cos }^{2}}\theta +2ab\cos \theta \right)d\theta }\]
We can separate the integration over the addition of functions, thus we can simplify the above expression as
\[\Rightarrow \dfrac{1}{2}\left( \int\limits_{0}^{2\pi }{{{b}^{2}}d\theta }+\int\limits_{0}^{2\pi }{{{a}^{2}}{{\cos }^{2}}\theta d\theta }+\int\limits_{0}^{2\pi }{2ab\cos \theta d\theta } \right)\]
Integrating the above expression, we get
\[\Rightarrow \dfrac{1}{2}\left( 2\pi {{b}^{2}}+\pi {{a}^{2}} \right)\]
We can simplify the above expression to express it as
\[\Rightarrow \pi \left( {{b}^{2}}+\dfrac{1}{2}{{a}^{2}} \right)\]

Note: As we already said that limaçon curves look like circles. They have various types depending on the values in their equations. If the value of a in the polar equation of limacon is 0. Then, it becomes a special case that represents the circle. The radius of the circle is b. The area equation is also simplified as \[\pi {{b}^{2}}\].