Triple Integral of x^2 + y^2 + z^2 over a Box

Three separable pieces, each with its own edge factors

The integrand is $\lvert \mathbf{r} \rvert^2$, the squared distance from the origin. Split it by linearity. Each term is separable, so it becomes one real integral times the two edge lengths belonging to the other variables.

The total is $\frac{2}{3} + \frac{8}{3} + \frac{2}{3} = \frac{12}{3} = 4$. The $y$ term dominates because that edge is twice as long, and cubing $2$ in $\int_0^2 y^2 \,dy = \frac{8}{3}$ magnifies the difference.

Why this integral keeps appearing

Moment of inertia about an axis uses the squared distance to that axis, so the $z$ axis version integrates $x^2 + y^2$ and the polar version integrates all three terms. Once you can evaluate one term over a box you can evaluate any of them.

• About the $z$ axis: $\iiint_B \left(x^2 + y^2\right) \,dV = \frac{2}{3} + \frac{8}{3} = \frac{10}{3}$.
• About the $x$ axis: $\iiint_B \left(y^2 + z^2\right) \,dV = \frac{8}{3} + \frac{2}{3} = \frac{10}{3}$.
• About the origin, all three terms: $4$.

With a variable density the terms stop being separable and the full iterated integral has to be worked, but the splitting by linearity still holds.

The mistake: attaching the wrong edges to a term

The frequent error is pairing $\int_0^2 y^2 \,dy$ with the $x$ term, or reusing the same pair of edge lengths for all three pieces. Each term keeps its own variable live and borrows the lengths of the other two.

A second error is claiming $x^2 + y^2 + z^2$ is separable as a product. It is a sum, not a product, so the whole integral is not $\left(\int x^2\right)\left(\int y^2\right)\left(\int z^2\right)$. That product would give $\frac{1}{3}\cdot\frac{8}{3}\cdot\frac{1}{3} = \frac{8}{27}$, nowhere near $4$.

Bound the answer to be safe. On this box the largest value of $x^2 + y^2 + z^2$ is $1 + 4 + 1 = 6$ at the far corner and the smallest is $0$, so with volume $2$ the integral has to sit between $0$ and $12$.