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.

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.