## Thomas' Calculus 13th Edition

$\frac{sin^2 4}{2}$
$\int^2_0 \int^{4-x^2}_0 \int^x_0 \frac{sin2z}{4-z}$ dy dz dx =$\int^2_0 \int^{4-x^2}_0 \frac{x \sin 2z}{4z}$ dz dx =$\int^4_0 \int^{\sqrt{4-z}}_0 (\frac{sin 2z}{4-z})dx$ dz =$\int^4_0 (\frac{sin2z}{4-z})\frac{1}{2} (4-z)dz$ =$[-\frac{1}{4}cos2z]^4_0$ =$[-\frac{1}{4}+\frac{1}{2}sin^2z]^4_0$ =$\frac{sin^2 4}{2}$