Answer
$\dfrac{16}{15}$
Work Step by Step
Here, we have
$ \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$
$\implies \int^{2}_{0}[(\dfrac{-1}{2})y^2z+yz^2]_0^{z^{2}} dz=\int^{2}_{0}[\dfrac{-1}{2}z^5+z^4] dz$
$\implies [\dfrac{-z^{6}}{12}+\dfrac{z^5}{5}]^{2}_{0}=[\dfrac{-2^{6}}{12}+\dfrac{2^5}{5}]^{2}_{0}$
Thus, $\int^{2}_{0} \int_{0}^{z^2} \int_{0}^{y-z} (2x-y) dx dy dz=\dfrac{16}{15}$
