Answer
$\dfrac{\pi^2}{4}-1$
Work Step by Step
$ \int_{0}^{\sqrt{\pi}} \int_{0}^{x} \int_{0}^{xz} x^2 \sin y dy dz dx= \int_{0}^{\sqrt{\pi}} \int_{0}^{x} [ -x^2 \cos y]_{0}^{xz} dz dx$
or, $=\int_{0}^{\sqrt{\pi}} [x^2 (z-(1/x) \sin (xz))]_0^x dx$
or, $=\int_{0}^{\sqrt{\pi}} [x^3 -x \sin x^2] dx$
Plug $a=x^2 \implies 2x dx =da$
or, $= [(1/2) \int_0^{\pi} a-\sin a da$
or, $= (1/2) [\dfrac{1}{2}a^2+\cos a]_0^{\pi} $
or, $=\dfrac{\pi^2}{4}-1$
