\[ r = 1 + 2sin \theta \]

This problem aims to find the area of the region enclosed by a **limacon curve** whose equation is $ r = 1 + 2sin\theta$, where $r$ is the radius of the curve. This problem requires knowledge of **coordinate systems**, the formation of a limacon curve, and the formula to find the area of the inner and the outer loop of a limacon curve.

A **coordinate system** is utilized to determine the area of a point in space. Most of the time, we use the** rectangular **or **Cartesian coordinate system** in our mathematical problems. A **rectangular grid system** is used to determine the location of a point in space. We can also determine the location of that exact point by describing its location and distance from a fixed point as a reference.

## Expert Answer

A limacon is an **anallagmatic** **curve** that looks like a circle but instead has a small indent on one side of it. Equations of the form $ r = a + bsin\theta $, $ r = a – bsin\theta $, $ r = a + bcos\theta $, and $ r = a – bcos\theta $ will produce **limacons**.

If the value of $a$ is slightly less than the value of $b$, then the graph would form a **limacon** with an internal loop as seen in the figure below.

So as the first step, we are going to find the interval on which the **internal loop** exits.

Given the equation $ r = 1 + 2sin\theta $, we will be taking $r=0$

\[ 1 + 2sin\theta = 0 \]

\[ sin \theta = \dfrac{-1}{2} \]

\[ \theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6} \]

We can find the area under the inner loop of the limacon curve by accomplishing a **definite integral** between the two solid points. To locate the **area** under the **curve** $r$ between $x = \theta_1$ & $x = \theta_2$, we will integrate $r$ between the limits of $\theta_1$ & $\theta_2$.

Modifying the **integral** as per the required variables:

\[ Area = \int_{\theta 1}^ {\theta2} \dfrac{1}{2}r^ 2 d\theta \]

Putting the values in the Formula:

\[ Area = \int_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}} \dfrac{1}{2}(1+2sin\theta)^ 2 d\theta \]

\[ = \int_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}} \dfrac{1}{2}(1+2sin\theta)^ 2 d\theta \]

\[ = \int_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}} \dfrac{1}{2}+2sin\theta + 2sin^ 2\theta d\theta \]

\[ = \int_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}} \dfrac{3}{2}+2sin\theta – cos2\theta d\theta \]

\[ = \left[ \dfrac{3\theta}{2}-2cos\theta – \dfrac{1}{2} sin2\theta \right]_{\dfrac{7\pi}{6}}^ {\dfrac{11\pi}{6}} \]

\[ = \dfrac{11\pi}{4} – 2 \times \dfrac{\sqrt{3}}{2} – \dfrac{1}{2} \left( – \dfrac{\sqrt{3}}{2}\right) – \left(\dfrac{-7\pi}{4} -2\left(-\dfrac{\sqrt{3}}{2} \right) – \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}\right) \]

\[ = \dfrac{11\pi}{4} – \dfrac{7\pi}{4} -\sqrt{3} + \dfrac{\sqrt{3}}{4} -\sqrt{3} + \dfrac{\sqrt{3}}{4} \]

## Numerical Result

\[Area = \pi – \dfrac{3\sqrt{3}}{2}\]

## Example

Find the **area** of the **region** enclosed by the inner loop of the **polar curve**:

\[ r = 2+4cos\theta \]

\[ cos \theta = \dfrac{-1}{2} \]

\[ \theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}\]

Putting the values in the **Formula**:

\[ Area = \int_{\dfrac{2\pi}{3}}^{\dfrac{4\pi}{3}} \dfrac{1}{2}(2+4cos\theta)^2 d\theta\]

By solving the integrals, the** area under the curve** comes out to be:

\[ A = 2(2\pi – 4\sqrt{3} + \sqrt{3})\]

\[ A = 4\pi – 6\sqrt{3}\]

*Images/mathematical drawings are created with GeoGebra.*