Question

find the area outside circle r = 1/2 and inside circle r = cosθ


Answer 1

The two curves intersect where $1/2=\cos \theta.$ So $\theta=-\pi/3$ or $\pi/3.$

The area we want is then

$$\dfrac{1}{2}\displaystyle\int_{-\pi/3}^{\pi/3}((\cos \theta)^2-(1/2)^2)d\theta$$

$$=\dfrac{1}{8}\displaystyle\int_{-\pi/3}^{\pi/3}(2(1+\cos(2\theta))-1)d\theta$$

$$=\dfrac{1}{8}\displaystyle\int_{-\pi/3}^{\pi/3}(1+2\cos(2\theta))d\theta$$

$$=\dfrac{1}{8}[\theta+\sin(2\theta)]\begin{matrix} \pi/3 \\ -\pi/3 \end{matrix}$$

$$=\dfrac{1}{8}(\dfrac{\pi}{3}+\sin(\dfrac{2\pi}{3})-(-\dfrac{\pi}{3})-\sin(\dfrac{-2\pi}{3}))$$

$$=(\dfrac{\pi}{12}+\dfrac{\sqrt{3}}{8})({units}^2)$$

The area outside circle $r = 1/2$ and inside circle $r = \cosθ$ is $(\dfrac{\pi}{12}+\dfrac{\sqrt{3}}{8})$ square units.