Answer
$ \dfrac{27}{4}$
Work Step by Step
Here, we have $B = (x,y,z) | 0 \leq x \leq 1, -1 \leq y \leq\ 2, 0 \leq z \leq 3$
Consider $I=\int \int \int xyz^{2} dV $
We need to integrate the integral first with respect to $y$, then $z$, and then $x$
$I= \int^{1}_{0} \int^{3}_{0} \int^{2}_{-1} xyz^{2} dy dz dx= \int^{1}_{0} \int^{3}_{0} (1/2) [xy^{2}z^{2}]^{2}_{-1}dz dx $
$\int^{1}_{0} \int^{3}_{0} (\dfrac{3}{2}) xz^{2}dzdx=\int^{1}_{0} (\dfrac{1}{2}) [xz^{3}]^{3}_{0}dx$
Therefore,
$\int \int \int xyz^{2} dV =(\dfrac{27}{4}) [x^{2}]^{1}_{0} = \dfrac{27}{4}$
