Question

How do I find the area inside the cardioid `r=1+costheta`?

Answer 1

`4pi`

Explanation:

We will integrate the area differential for a polar function:
`dA = d(1/2 r^2 theta) = r theta d r + 1/2 r^2 d theta`

We can integrate this just `d theta`:
`dA = (r theta (dr)/(d theta) + 1/2 r^2) d theta`

This yields
`A = int\ \ dA = int_0^(2pi) ([1 + cos theta] * theta * (- sin theta) + 1/2 (1 + cos theta)^2) d theta`
`A = int_0^(2pi) [3/4 + 1/4 cos2theta + cos theta - theta sin theta - 1/2 theta sin 2theta] d theta`

This integral is pretty simple, with a few integrations by parts.

The ultimate answer turns up to be `A = 4pi`.