Question

How to find the area of the region inside the cardioid?

r = 2 + 2cosθ and outside the circle r = 3

Answer 1

`(27sqrt(3))/2 - 3pi approx 13.96`

Explanation:

First, we need to figure out what angles the two cross at. This is quite simple if we equate the two radii:
`3 = 2 + 2 costheta implies costheta = 1/2 implies theta = pm pi/3`

Now we know our bounds of integration. From there, we only need to figure our infinitesimal area.

We know that the sector of a circle has area
`A = 1/2 r^2 theta`
Imagining an infinitesimal version of this:
`dA = d(1/2 r^2 theta) = 1/2 r^2 d theta + r theta\ dr`
Since we know `r(theta)`, we can rewrite this as
`dA = (dA)/(d theta)d theta = (1/2 r^2 + r\ theta (dr)/(d theta))d theta`

This brings us to our integration:
`A_(card) = int_(-pi/3)^(pi/3) [1/2 (2costheta +2)^2 + (2costheta + 2) (2sintheta) theta ]d theta`
`= 2int(cos^2theta + 2cos theta + 1)d theta + 4int (theta cos theta sin theta + theta sin theta) d theta`
The first can be done analytically moderately easily with either half angle identities or integration by parts. The second is not impossible via double angle identities and integration by parts again. I will simply give the answer from WolframAlpha:
`A_(card) = (27sqrt(3))/2`

Similarly,
`A_(cir\c) = 1/2 r^2 theta = 1/2 * 9 * (2pi)/3 = 3pi`

Therefore, the total area outside the circle but inside the cardioid is
`A_(t\ot) = A_(card) - A_(circ) = (27sqrt(3))/2 - 3pi approx 13.96`