Answer
$\dfrac{16}{15}$
Work Step by Step
Integrate the integral first with respect to $x$, then $y$, and then $z$.
$ \int^{2}_{0} \int_{0}^{z^2} \int_{0}^{y-z} (2x-y) dx dy dz= \int^{2}_{0} \int_{0}^{z^2} (-yz+z^2)dy dz$
This implies that
$\int^{2}_{0}[\dfrac{-y^2z}{2}+yz^2]_0^{z^{2}} dz=[\dfrac{(-z)^{6}}{12}+\dfrac{(z)^5}{5}]^{2}_{0}$
Thus, $ \int^{2}_{0} \int_{0}^{z^2} \int_{0}^{y-z} (2x-y) dx dy dz=[\dfrac{-(2)^{6}}{12}+\dfrac{(2)^5}{5}]^{2}_{0}=\dfrac{16}{15}$
