Find the area inside $r=2+2\sin(\theta)$ but outside $r=4\sin(\theta)$.

Question

Winplot plot:

I tried this:

$$A = 2 \frac 1 2 \int_0^{\pi/2} (2+2\sin(\theta))^2 - (4\sin(\theta))^2 d\theta + 2 \frac 1 2 \int_{\pi}^{3\pi/2} (2+2\sin(\theta))^2 d\theta$$

Is that right?

How about

$$A = 2 \frac 1 2 \int_{-\pi/2}^{\pi/2} (2+2\sin(\theta))^2 - (4\sin(\theta))^2 d\theta$$

?

Or

$$A = \int_{0}^{2\pi} (2+2\sin(\theta))^2 - (4\sin(\theta))^2 d\theta$$

?

Answer 1

How about working out $$A = 2 \times\frac 1 2 \int_{-\pi/2}^{\pi/2} (2+2\sin(\theta))^2 d\theta$$ and then subtract the area of the circle i.e. $4\pi$?