Multivariable calculus

Triple Integral of x + y + z over a Box

Q: Why is the answer not the same as the volume?

The volume is \iiint_B 1 \,dV = 6, which is what you get when the integrand is the constant 1. Here the integrand grows from 0 at one corner to 6 at the opposite corner, averaging 3, so the integral is 3 times the volume.

Q: Does swapping the box edges change the value?

No, and the reason is worth knowing. For a box with edges a, b and c measured from the origin, the integral is the volume times \frac{a + b + c}{2}, and that expression does not care which edge is which. Running x from 0 to 3 and z from 0 to 1 still gives 18. Changing an edge length is what moves the answer: stretching the z edge to 6 gives 12 \cdot \frac{1 + 2 + 6}{2} = 54.

The triple integral of x + y + z over the box where x runs 0 to 1, y runs 0 to 2 and z runs 0 to 3 equals 18. Linearity splits the sum into three integrals contributing 3, 6 and 9. The box has volume 6, so the average value of x + y + z on it is 3.

f(x,y,z) = x + y + z

\int_0^1 \int_0^2 \int_0^3 (x + y + z) \,dz\,dy\,dx = 18

Numerically 18.000000, confirmed by quadrature on every build.

Split the sum, then restore the missing lengths

An integral of a sum is the sum of the integrals, so handle $x$ , $y$ and $z$ one term at a time. The point that decides the answer is that each term still gets integrated over all three variables, so the two idle directions contribute their lengths as plain factors.

\iiint_B x \,dV = \left(\int_0^1 x \,dx\right)(2)(3) = \frac{1}{2}\cdot 6 = 3

\iiint_B y \,dV = (1)\left(\int_0^2 y \,dy\right)(3) = 2 \cdot 3 = 6, \qquad \iiint_B z \,dV = (1)(2)\left(\int_0^3 z \,dz\right) = 2 \cdot \frac{9}{2} = 9

Adding the three gives $3 + 6 + 9 = 18$ .

The centre point check

The integrand is linear, so its average over the box is exactly its value at the centre. The centre of this box is $\left(\tfrac{1}{2}, 1, \tfrac{3}{2}\right)$ , where $x + y + z = 3$ .

\iiint_B (x + y + z) \,dV = 3 \cdot \text{volume} = 3 \cdot (1)(2)(3) = 18

This check only works for integrands that are linear in every variable. A term like $x^2$ is curved, so its average over an interval sits above the value at the midpoint and the shortcut fails.

The mistake: forgetting the other two directions

The common wrong answer is $\frac{1}{2} + 2 + \frac{9}{2} = 7$ , from computing $\int_0^1 x \,dx$ , $\int_0^2 y \,dy$ and $\int_0^3 z \,dz$ and adding them. That treats each term as a single integral instead of a triple one.

Integrating $x$ over the box means summing $x$ over every point of a solid, not over a segment. The $y$ and $z$ integrations of a term with no $y$ or $z$ in it are not free: each contributes the length of that edge.

$\iiint_B x \,dV$ picks up the factor $(2)(3) = 6$ from the $y$ and $z$ edges.
$\iiint_B y \,dV$ picks up $(1)(3) = 3$ from the $x$ and $z$ edges.
$\iiint_B z \,dV$ picks up $(1)(2) = 2$ from the $x$ and $y$ edges.

Frequently asked questions

Why is the answer not the same as the volume?

The volume is $\iiint_B 1 \,dV = 6$ , which is what you get when the integrand is the constant $1$ . Here the integrand grows from $0$ at one corner to $6$ at the opposite corner, averaging $3$ , so the integral is $3$ times the volume.

Does swapping the box edges change the value?

No, and the reason is worth knowing. For a box with edges $a$ , $b$ and $c$ measured from the origin, the integral is the volume times $\frac{a + b + c}{2}$ , and that expression does not care which edge is which. Running $x$ from $0$ to $3$ and $z$ from $0$ to $1$ still gives $18$ . Changing an edge length is what moves the answer: stretching the $z$ edge to $6$ gives $12 \cdot \frac{1 + 2 + 6}{2} = 54$ .