Evaluate the integral:int int int_E(xy+z^2)dV, where E={(x,y,z)|0<=x<=2,0<=y<=1,0<=z<3}

Here we have to evalute the integral ${\displaystyle \int \int \int (xy+z^2)dxdydz}$ over the regin bounded by ${\displaystyle 0\le x\le 2,0\le y\le 1,0\le z<3}$. Then
${\displaystyle \int \int \int (xy+z^2)dxdydz={\int }_{x=0}^2{\int }_{y=0}^1{\int }_{z=0}^3(xy+z^2)dxdydz}$
${\displaystyle ={\int }_{x=0}^2{\int }_{y=0}^1(xyz+\left(\frac{z^3}{3}\right){\mid }_{z=0}^{3}dxdy}$
${\displaystyle ={\int }_{x=0}^2{\int }_{y=0}^1(3xy+9)dxdy}$
${\displaystyle ={\int }_{x=0}^2(3x\frac{y^2}{2}+9y)\mid +{(y=0)}^{1}dx}$
${\displaystyle ={\int }_{x=0}^2(\frac{3x}{2}+9)dx}$
${\displaystyle =(\frac{3x^2}{4}+9x){\mid }_{x=0}^2}$
=(3+18)
=21