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.
 A: $\newcommand{\+}{^{\dagger}}%
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$$
\overbrace{\int_{0}^{1}\dd x\,x}^{\ds{1 \over 2}}\
\overbrace{\int_{-1}^{2}\dd y\,y}^{\ds{3 \over 2}}\
\overbrace{\int_{0}^{3}\dd z\,z^{2}}^{\ds{9}} = {27 \over 4}
$$
A: This is equivalent to
\begin{align*}
\int_0^1 x dx \int_{-1}^2 y dy \int_0^3 z^2 dz &= \left(\frac{1}{2} x^2 \Big|_0^1\right) \left(\frac 1 2 y^2\Big|_{-1}^2\right) \left(\frac 1 3 z^3 \Big|_0^3 \right) \\
&= \frac{1}{2} \cdot \left(\frac{4}{2} - \frac{1}{2}\right)\cdot \frac{3^3}{3} \\
&= \frac{1}{2} \cdot \frac{3}{2} \cdot 9 \\
&= \frac{27}{4}
\end{align*}
A: Notice given the limits, we can write
$$
\iiint_B xyz^2 \; dV=\int_0^1 x \;dx \int_{-1}^2 y \;dy \int_0^3 z^2\;dz
$$
Now 
$$
\int_0^1 x \;dx=\frac{1}{2}
$$
$$
\int_{-1}^2 y \;dy=\frac{3}{2}
$$
$$
\int_0^3 z^2\;dz=9
$$
So the answer is $\frac{1}{2}\cdot \frac{3}{2} \cdot 9=\frac{27}{4}$.
A: @genius12: the domain of the integral $B$ is of course a parallelepiped whose volume is an integer. But this does not mean that the integral of $f(x,y,z)=xyz^2$ over $B$ is an integer
A: 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...]
