Area of region of a limaçon

Question

I have to generalize the concept of finding the region $D$ inside of limaçons without a loop. In particular $r=a+b\cos(\theta)$, where $0\leq\theta\leq2\pi$, and to prevent the loop $a>b>0$.

My naive approach is to find: $$\int\int_{D}1dxdy = \int_{a-b}^{a+b}\int_{0}^{2\pi}rd\theta dr$$ But then I loose information that $r$ is $a+b\cos\theta$.

I checked how the limaçons with different $a$ and $b$ look like via desmos, but this does not give me any intuition.

Answer 1

By analogy to the rectangular integral to find the area of $\int y\; dx$ the usual way to find the area in polar coordinates is $\int_0^{2 \pi} \frac 12r^2\; d\theta$. The area element is approximately a triangle with height $r$ and base $r\; d\theta$, so area $\frac 12 r^2 d\theta$. Now you substitute in your formula for $r$ to get $\frac 12\int_0^{2\pi} (a+b\cos \theta)^2\; d\theta$. A good picture is here.

Answer 2

An example, symmetric to x-axis.

$$r= 3+2 \cos \theta, Area= 2\int _0^\pi \frac12 r^2 d \theta $$

Answer 3

I approached this from the complex plane, where the area is given by

$$A=\frac{1}{2}\int \Im\{z^*\dot z\}\ d\theta$$

Then,

$$ z=re^{i\theta}\\ z*=re{-i\theta}\\ \dot z=(ir+\dot r)e^{i\theta}\\ \Im\{z^*\dot z\}=r^2 $$

And finally

$$A=\frac{1}{2}\int_{-\pi}^{\pi}(a+b\cos\theta)^2\ d\theta=\frac{1}{2}\pi(2a^2+b^2) $$

According to A Catalog of Special Plane Curves by J. Denus Lawrence, Dover Press. 1972, this is the total area for a single or double loop limaçon.