Answer
$$\int_{0}^{4} \int_{0}^{\pi / 2} \int_{0}^{1-x} x \cos y d z d y d x=-\frac{40}{3}$$
Work Step by Step
Given $$\int_{0}^{4} \int_{0}^{\pi / 2} \int_{0}^{1-x} x \cos y d z d y d x$$
So, we have
\begin{aligned}I&= \int_{0}^{4} \int_{0}^{\pi / 2} \int_{0}^{1-x} x \cos y d z d y d x\\
&=\int_{0}^{4} \int_{0}^{\pi / 2}[(x \cos y) z]_{0}^{1-x} d y d x\\
&=\int_{0}^{4} \int_{0}^{\pi / 2} x(1-x) \cos y \ d y d x \\ &=\int_{0}^{4}x(1-x) [\sin y]_{0}^{\pi / 2} d x\\
&=\int_{0}^{4}x(1-x) [\sin \frac{\pi}{2}-\sin 0] d x\\
&=\int_{0}^{4} x(1-x) d x\\
&=\int_{0}^{4} (x -x^2) d x\\
&=\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{4}\\
&=\frac{4^{2}}{2}-\frac{4^{3}}{3}-0\\
&=8-\frac{64}{3}\\
&=-\frac{40}{3} \end{aligned}
