## Calculus 8th Edition

$\dfrac{\pi^2}{2}-2$
Let us consider that $I=\iiint_E \sin y dV$ $I=\int_{0}^{\pi} \int_{0}^{\pi-x} \int_{0}^{x}\sin y dzdy dx$ Further, we have $\int_{0}^{\pi} [-x (\cos y)]_{0}^{\pi-x} dx=\int_{0}^{\pi} [x (\cos x)+x] dx$ and $[\dfrac{x^2}{2}]_{0}^{\pi} \int_0^{\pi} x \cos x dx=\dfrac{\pi^2}{2}+[x \sin x+\cos x]_{0}^{\pi}$ or, $\dfrac{\pi^2}{2}+\pi \sin \pi-0+\cos \pi-\cos 0=\dfrac{\pi^2}{2}-2$