How have I incorrectly calculated the area A inside the curve $r =1$ and outside the curve $r = 2\cos(\theta)$ I hope this isn't against guidelines but I want to show how I solved this on white paper, it might be clear where my logic went wrong in solving.
Find the are A of the region inside the curve $r = 1$ and outside the curve $r = 2cos(\theta)$

Thank you
 A: You cannot integrate both over the interval $\left[\frac{\pi}{3},\pi\right]$.
The largest part is the quarter circle in the second quadrant which we know must have area equal to $\dfrac{\pi}{4}$ but it can be easily verified:
\begin{equation}
\frac{1}{2}\int_{\pi/2}^\pi 1\,d\theta=\left[\frac{\theta}{2}\right]_{\pi/2}^\pi=\frac{\pi}{4}
\end{equation}
It is only the part in quadrant I that is bounded by both functions:
\begin{eqnarray}
\frac{1}{2}\int_{\pi/3}^{\pi/2}1-4\cos^2\theta\,d\theta&=& \text{ continue from here}\\
\end{eqnarray}
A: 
The area of the blue and yellow pieces together is,
$$\int_{\frac{\pi}{3}}^{\pi} \frac{1}{2}(1)^2 d\theta=\frac{\pi}{3}$$
The area of the blue piece by itself is,
$$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{2}(2\cos (\theta))^2 d\theta$$
$$=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 2\cos^2(\theta) d\theta$$
$$=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 2\left(\frac{1+\cos(2\theta)}{2} \right) d\theta$$
$$=\frac{\pi}{6}-\frac{\sqrt{3}}{4}$$
Thus the area of the yellow piece is,
$$\frac{\pi}{3}-(\frac{\pi}{6}-\frac{\sqrt{3}}{4})=\frac{\pi}{6}+\frac{\sqrt{3}}{4}$$
The area you are interested in is twice this:
$$A=\frac{\pi}{3}+\frac{\sqrt{3}}{2}$$
