<s>[INST] Help with Evaluating triple integral Evaluate $\int \int \int_B xyz^2 dV$, where B is a cuboid bounded by the regions $ 0 \le x \le 1 $, $ -1 \le y \le 2 $, $ 0 \le z \le 3 $. 
I keep getting $ \frac{27}{4}$ as my answer but apparently it's incorrect...Any help would be appreciated.
 [/INST] Just as another check on this, we can use the "odd symmetry" about $ \ y = 0 \ $ of the integrand function $ \ xyz^2 \ $ to cancel the contributions from the portions of the volume over $ \ -1 \ \le \ y \ \le \ 0 \ $ and $ \ 0 \ \le \ y \ \le \  1 \ , $ leaving integration only over $ \ 1 \ \le \ y \ \le \  2 \ . $  I won't even bother "separating" the variables; we still find
$$ \int_0^3 \int_1^2 \int_0^1 \ xyz^2 \ \ dx \ dy \ dz \ \ = \ \ \int_0^3 \int_1^2  \ \left(\frac{1}{2}x^2yz^2 \right) \vert_{x=0}^{x=1} \ \  dy \ dz $$
$$ = \ \ \frac{1}{2} \ \int_0^3 \int_1^2  \ yz^2 \ \  dy \ dz \ \ = \ \ \frac{1}{2} \ \int_0^3  \ \left(\frac{1}{2}y^2z^2 \right) \vert_{y=1}^{y=2} \ \  dz  $$
$$ = \ \ \frac{1}{2} \cdot \frac{3}{2} \ \int_0^3  \ z^2   \ \  dz \ \ = \ \ \frac{1}{2} \cdot \frac{3}{2} \ \left(\frac{1}{3}z^3 \right) \vert_{z=0}^{z=3} \ \ = \ \ \frac{1}{2} \cdot \frac{3}{2} \cdot 9 \ \ . $$
[So, 'nother county heard from...]
</s>