Rhombus and Sectors of Circles.

Let $a$ be a side of rhombus $ABCD$

Let $A_{s}(\theta)$ be the area of sector $BCA \implies A_{s}(\theta) =\dfrac{1}{2}a^2\theta$ and $A_{\triangle{ABC}}(\theta) = \dfrac{1}{2}a^2\sin(\theta) \implies$

$A_{R}(\theta) = 2(A_{s}(\theta) - A_{\triangle{ABC}}(\theta)) = (\theta - \sin(\theta))a^2 \implies$

$\displaystyle\int_{0}^{\dfrac{\pi}{6}} f(\theta) \:\ d\theta = \displaystyle\int_{0}^{\dfrac{\pi}{6}} (\theta - \sin(\theta)) \:\ d\theta = (\dfrac{\theta^2}{2} + \cos(\theta))|_{0}^{\dfrac{\pi}{6}} =$

$\dfrac{\pi^2 + 36\sqrt{3} - 72}{72} \approx \boxed{0.0031032426884575}$ .