Reuse a substitution result in a circle-area problem

Put $x=\sin\theta$, so $dx=\cos\theta\,d\theta$. The limits become $0$ and $\pi/6$. On this interval $\cos\theta\ge0$, so $\sqrt{1-\sin^2\theta}=\cos\theta$ and the integral is $\int_0^{\pi/6}\cos^2\theta\,d\theta$.

Use $\cos^2\theta=(1+\cos2\theta)/2$. Thus $\frac12[\theta+\frac12\sin2\theta]_0^{\pi/6}=\pi/12+\sqrt3/8$, as required.

From $(y-1)^2=1-x^2$ obtain $y=1\pm\sqrt{1-x^2}$. The shaded boundary has $y<1$, hence choose $y=1-\sqrt{1-x^2}$. Split the area at $x=1/2$: the left piece $A_1$ is under this circle branch; the right piece $A_2$ is a triangle ending at the line intercept $x=1/\sqrt3$.

$A_1=\int_0^{1/2}(1-\sqrt{1-x^2})\,dx=\frac12-(\frac\pi{12}+\frac{\sqrt3}8)$. The Teacher circles and labels the triangle contribution: $A_2=\frac12(\frac1{\sqrt3}-\frac12)(1-\frac{\sqrt3}2)$. Adding gives $A_1+A_2=\frac1{2\sqrt3}-\frac\pi{12}$ square units, retaining the required exact answer.