Answer
$21$
Work Step by Step
Let us consider that $I=\int \int \int (xy+z^{2}) dV$
Here, we have $E = (x,y,z) | 0 \leq x \leq 2, 0 \leq y \leq\ 1, 0 \leq z \leq 3$
Now, integrating the integral $\int \int \int (xy+z^{2}) dV$ initially with respect to $x$, then $y$, and then $z$
$I = \int^{3}_{0} \int^{1}_{0} \int^{2}_{0} (xy+z^{2}) dx dy dz=\int^{3}_{0} \int^{1}_{0} [\dfrac{x^{2}y}{2}+z^{2}x]^{x=2}_{x=0}dy dz$
This implies that
$I=\int^{3}_{0} \int^{1}_{0} 2y+2z^{2}dydz=\int^{3}_{0} [y^{2} + 2z^{2}y]^{1}_{0}dz$
Hence, we have $I=\int^{3}_{0} 1+2z^{2}dz=|z+(2/3)z^{3}|^{3}_{0} =(3-0)+[\dfrac{2}{3} (27)-0]=21$
