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