# 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.

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Why do you say it's incorrect? – user61527 Dec 28 '13 at 6:33
Because this is a question to a problem online (it's a 'brilliant' question) and the answer is supposed to be an integer and so when i enter 27/4 the answer is not accepted. May be the system has it wrong then i suppose =/ – genius12 Dec 28 '13 at 6:37

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$\overbrace{\int_{0}^{1}\dd x\,x}^{\ds{1 \over 2}}\ \overbrace{\int_{-1}^{2}\dd y\,y}^{\ds{3 \over 2}}\ \overbrace{\int_{0}^{3}\dd z\,z^{2}}^{\ds{9}} = {27 \over 4}$$

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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*}

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are u sure this is right? i mean this is a "brilliant" question and so the answer should be an integer but its not which is why i have doubts... – genius12 Dec 28 '13 at 6:44
@genius12: The answer is right. This is strange to me that you said the result should be integer! Who does think like this?! :-) – Babak S. Dec 28 '13 at 7:34

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 $\frac{1}{2}\cdot \frac{3}{2} \cdot 9=\frac{27}{4}$.

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@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

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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...]

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