Answer
$\dfrac{16}{15}$
Work Step by Step
Need to integrate the given 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$
or, $=\int^{2}_{0}[\dfrac{-y^2z}{2}+yz^2]_0^{z^{2}} dz$
or, $=\int^{2}_{0}[\dfrac{-z^5}{2}+z^4] dz$
or, $=[\dfrac{-z^{6}}{12}+\dfrac{z^5}{5}]^{2}_{0}$
or, $=[\dfrac{-2^{6}}{12}+\dfrac{2^5}{5}]^{2}_{0}$
or, $=\dfrac{16}{15}$
