Answer
$\dfrac{\pi^2}{2}-2$
Work Step by Step
Consider $I=\iiint_E \sin y dV$
$ I=\int_{0}^{\pi} \int_{0}^{\pi-x} \int_{0}^{x}\sin y dzdy dx=\int_{0}^{\pi} \int_{0}^{\pi-x} x \sin y dy dx$
or, $=\int_{0}^{\pi} [-x \cos y]_{0}^{\pi-x} dx$
or, $= \int_{0}^{\pi} x \cos x+x dx$
or, $= [\dfrac{x^2}{2}]_{0}^{\pi} \int_0^{\pi} x \cos x dx$
or, $=\dfrac{\pi^2}{2}+[x \sin x+\cos x]_{0}^{\pi}$
or, $=\dfrac{\pi^2}{2}-2$
