Giao điểm của 2 pt đường tròn \(\left(\dfrac{\sqrt{3}}{2};\dfrac{1}{2}\right)\)
\(\left\{{}\begin{matrix}x=r\cos\varphi\\y=r\sin\varphi\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}r\in\left[2\sin\varphi;1\right]\\\varphi\in\left[0;\dfrac{\pi}{6}\right]\\\left|J\right|=r\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^{\dfrac{\pi}{6}}_0d\varphi\int\limits^1_{2\sin\varphi}\sqrt{r^2}.rdr=\int\limits^{\dfrac{\pi}{6}}_0d\varphi\int\limits^1_{2\sin\varphi}r^2dr=\dfrac{1}{3}\int\limits^{\dfrac{\pi}{6}}_0\left(1-8\sin^3\varphi\right)d\varphi=...\)