Question

How do I find the area inside a limacon?

Answer 1

The area enclosed by the limaçon `r = b + a cos theta` is `pi(b^2+1/2 a^2)`

Explanation:

Consider a limaçon with polar equation:

`r = b + a cos theta`

Since the question is asked in a simple form, I will make a simplifying assumption that the limaçon does not self cross, so `abs(a) <= abs(b)`.

Dissecting the limaçon into infinitesimal segments about the origin note that each segment has area `1/2 r^2 d theta`

So the total area of the limaçon is:

`int_0^(2pi) 1/2 r^2 d theta = int_0^(2pi) 1/2 (b+acos theta)^2 d theta`

`color(white)(int_0^(2pi) 1/2 r^2 d theta) = int_0^(2pi) 1/2 (b^2+2ab cos theta+a^2cos^2 theta) d theta`

`color(white)(int_0^(2pi) 1/2 r^2 d theta) = int_0^(2pi) ( 1/2b^2+ab cos theta+1/4a^2(1+cos 2 theta)) d theta`

`color(white)(int_0^(2pi) 1/2 r^2 d theta) = [ 1/2b^2 theta+ab sin theta+1/4a^2(theta+1/2sin 2 theta)]_0^(2pi)`

`color(white)(int_0^(2pi) 1/2 r^2 d theta) = pi(b^2+1/2 a^2)`