Triple Integral of xyz + x^2 over a Cube

Use symmetry before you use calculus

The region is symmetric about every coordinate plane, so check each term for odd symmetry first. A term that changes sign when one variable flips, over an interval symmetric about zero, integrates to zero.

Replacing $x$ by $-x$ turns $xyz$ into $-xyz$, so that term is odd in $x$ and contributes nothing. It is also odd in $y$ and in $z$, so it dies three times over.

The surviving term is even in $x$ and free of $y$ and $z$, so it separates into one real integral times two edge lengths of $2$.

The mistake: assuming symmetry kills everything

Having watched $xyz$ vanish, students often declare the whole integral zero. Symmetry is a term by term test, not a verdict on the integrand as a whole, and $x^2$ is even, so it survives with everything it has.

The opposite error is refusing to use symmetry at all and grinding through the iterated integral. That works, and the $xyz$ term actually dies at the first stage, since $\int_{-1}^{1} xyz \,dz = xy\left[\frac{z^2}{2}\right]_{-1}^{1} = 0$. Symmetry tells you that before you write anything down, and it keeps working on regions like a ball where the antiderivatives are far less pleasant.

• Odd in at least one variable, over a symmetric interval in that variable: the term integrates to zero.
• Even in every variable, or free of some variables: the term survives and must be computed.
• The test needs the interval to be symmetric about zero. Over $[0, 1]$ the term $xyz$ contributes $\frac{1}{8}$, not $0$.

Reading the answer

The cube has volume $8$, so the average value of $xyz + x^2$ on it is $\frac{8/3}{8} = \frac{1}{3}$. That matches the average of $x^2$ alone on $[-1, 1]$, which is $\frac{1}{3}$, confirming that the $xyz$ term contributed nothing to the mean.

The integrand itself does take negative values, on the octants where $xyz < 0$ and $x^2$ is small. A positive total does not mean a positive integrand, only that the positive regions win.