Question

How do you find the area inside of the Cardioid `r=3+2cos(theta)` for `0 <= (theta) <= 2pi`?

Answer 1

`11pi`

Explanation:

`A=1/2int r^2d theta = 1/2 int_o^(2pi) (3+2costheta)^2d theta`

`A=1/2 int_o^(2pi) (9+12costheta+4cos^2theta) d theta`

`A=1/2 (9theta +12sintheta+4int_0^(2pi) (1+cos2theta)/2 d theta)`

`A=1/2 (9theta +12sintheta+4(theta/2+1/4sin2theta))|_0^(2pi)`

`A=1/2(18pi+4pi)=11pi`