Answer
$$ \int_{0}^{3} \int_{0}^{2} \int_{0}^{1}(x+y+z) d x d z d y=18$$
Work Step by Step
Given $$ \int_{0}^{3} \int_{0}^{2} \int_{0}^{1}(x+y+z) d x d z d y$$ So we get \begin{aligned} I&= \int_{0}^{3} \int_{0}^{2} \int_{0}^{1}(x+y+z) d x d z d y \\&=\int_{0}^{3} \int_{0}^{2}\left[\frac{x^{2}}{2}+x y+x z\right]_{0}^{1} d z d y\\ &=\int_{0}^{3} \int_{0}^{2}\left(\frac{1}{2}+y+z\right) d z d y\\ &=\int_{0}^{3}\left[\frac{1}{2} z+y z+\frac{z^{2}}{2}\right]_{0}^{2} d y \\ &=\int_{0}^{3}(1+2 y+2) d y \\ &=\int_{0}^{3}(3+2 y) d y\\ &=\left[3 y+y^{2}\right]_{0}^{3}\\ &=9+9-0\\ &=18 \end{aligned}
