Answer
$\dfrac{5(2-\sqrt{3})}{4}$
Work Step by Step
Our aim is to integrate the integral as follows:
$\int^{\pi/6}_0 \int^1_0 \int^3_{-2} dx \space dy \space dz =\int^{\pi/6}_0 \int^1_0 5y\sin (z) \space dy \space dz $
or,$ =\dfrac{5}{2}\int^{\pi/6}_0 sin z \space dz $
or, $=\dfrac{5(2-\sqrt{3})}{4}$
