# Find the area of the region enclosed by the inner loop of the curve:

$r = 1 + 2sin \theta$

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.

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.

Figure 1

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.