Triple Integral of xyz over the Unit Cube

Work from the inside out

A box has constant limits in every variable, so each stage of the iterated integral is an ordinary single variable integral. Start with the innermost one, in $z$, and treat $x$ and $y$ as fixed numbers.

That result no longer mentions $z$. Feed it to the middle integral in $y$, holding $x$ fixed, then to the outer integral in $x$.

Every stage must eliminate its variable completely. If a $z$ survives into the $y$ integral, a limit was substituted wrongly.

The shortcut when the integrand factors

Because $xyz$ splits into a function of $x$, a function of $y$ and a function of $z$, and because every limit is a constant, the triple integral splits into a product of three one variable integrals.

Two conditions have to hold together, and both are easy to check before you start.

• The integrand is a product with each factor using only one variable.
• The region is a box, so no limit mentions another variable.
• Fail either one and the product form is wrong, even if the integrand looks separable.

The mistake: splitting a region that is not a box

The product trick gets used on regions where it does not apply. If $z$ runs from $0$ to $xy$ instead of $0$ to $1$, the inner integral gives $\frac{x^3y^3}{2}$, which still carries $x$ and $y$, so there is no clean factor of $\int z \,dz$ to pull out.

The other slip is dropping a factor. Students compute $\int_0^1 z \,dz = \frac{1}{2}$ once, then write $\frac{1}{2}$ as the answer instead of cubing it. Three integrations means three factors, and each one has to appear.

A quick sanity check catches both. On the unit cube the volume is $1$, and $xyz$ stays between $0$ and $1$ inside, so the answer has to land between $0$ and $1$. It does, at $0.125$.