Use horizontal strips for an inverse-sine region

$x=2\sin y$, so $A_1=\int_{\pi/12}^{\pi/6}2\sin y\,dy=2[\cos(\pi/12)-\cos(\pi/6)]\approx0.19980=0.200$ square units (3 s.f.).

Let $c=2\sin(\pi/12)$. The same area is $\pi/6-(\pi/12)c-\int_c^1\sin^{-1}(x/2)\,dx$, giving $0.200$ square units. The first term is the enclosing rectangle of width $1$ and height $\pi/6$.

For $0\le x\le c$, the radii are $\pi/6$ and $\pi/12$; for $c\le x\le1$, the inner radius is $\sin^{-1}(x/2)$. Equivalently $V=\pi(\pi/6)^2-\pi(\pi/12)^2c-\pi\int_c^1[\sin^{-1}(x/2)]^2dx\approx0.510$ cubic units (3 s.f.). Horizontal shells verify $V=4\pi\int_{\pi/12}^{\pi/6}y\sin y\,dy$.

The upper-right and lower-left rectangle terms give $A_2=\sqrt3(\pi/3)-\pi/6-\int_{\pi/6}^{\pi/3}2\sin y\,dy=\tfrac\pi6(2\sqrt3-1)+1-\sqrt3$.

$\int\sin^{-1}(x/2)\,dx=x\sin^{-1}(x/2)+\sqrt{4-x^2}+C$. Evaluating from $1$ to $\sqrt3$ gives $\sqrt3(\pi/3)+1-\pi/6-\sqrt3=\tfrac\pi6(2\sqrt3-1)+1-\sqrt3$, as above.