Triple integrals add up a quantity over a solid region $E$ in space: mass from density, total charge, or simply volume. They are computed like double integrals over general regions, one variable at a time, now with three nested integrals.
Key idea
$\iiint_E f\,dV$ is the limit of sums $\sum f(x^*, y^*, z^*)\,\Delta V$ over small boxes. For $$E = \{(x, y) \in D,\ u_1(x, y) \le z \le u_2(x, y)\}\text{:}$$
$$\iiint_E f\,dV = \iint_D \left(\int_{u_1(x, y)}^{u_2(x, y)} f(x, y, z)\,dz\right) dA.$$
The volume of $E$ is $\iiint_E 1\,dV$.
Why it works
Think of the solid as a bundle of vertical sticks. Above each point $(x, y)$ of the shadow $D$, a stick runs from the bottom surface $z = u_1$ to the top surface $z = u_2$; the inner integral adds $f$ along that stick. Then the double integral adds the sticks over the shadow, which you already know how to set up. The outermost limits must be numbers; the innermost may depend on both outer variables.
Worked examples
Worked example: Over a box
Evaluate $\iiint_B xyz\,dV$ over $B = [0, 1] \times [0, 2] \times [0, 3]$.
- All limits are constants and the integrand is a product, so it splits:
- $$\left(\int_0^1 x\,dx\right)\left(\int_0^2 y\,dy\right)\left(\int_0^3 z\,dz\right) = \frac{1}{2} \cdot 2 \cdot \frac{9}{2} = \frac{9}{2}.$$
Answer: $\frac{9}{2}$
Worked example: Volume of a tetrahedron
Find the volume of the solid bounded by the coordinate planes and ${}3x + 2y + z = 6$.
- Top: $z = 6 - 3x - 2y$. It meets the floor along ${}3x + 2y = 6$, so the shadow is ${}0 \le x \le 2$, ${}0 \le y \le 3 - \frac{3x}{2}$.
- Inner: $\int_0^{6 - 3x - 2y} dz = 6 - 3x - 2y$. Middle: $\int_0^{3 - 3x/2}(6 - 3x - 2y)\,dy = \left(3 - \frac{3x}{2}\right)^2$.
- Outer: $\int_0^2 \left(3 - \frac{3x}{2}\right)^2 dx = 6$. Check: $\frac{1}{6}$ of the box ${}2 \times 3 \times 6$ is ${}6$.
Answer: ${}6$
Worked example: An integral over it
For the same tetrahedron $E$, find $\iiint_E z\,dV$ and the height of its centroid.
- Inner: $\int_0^{6 - 3x - 2y} z\,dz = \frac{1}{2}(6 - 3x - 2y)^2$.
- Middle (with $w = 6 - 3x - 2y$): $\frac{1}{12}(6 - 3x)^3$. Outer: $\frac{1}{12}\int_0^2 (6 - 3x)^3\,dx = 9$.
- Centroid height $\bar{z} = \frac{9}{6} = \frac{3}{2}$, a quarter of the way up, as for every tetrahedron.
Answer: ${}9$; $\bar{z} = \frac{3}{2}$
Common mistake
Limits that do not nest. The innermost limits may involve the two outer variables; the middle limits only the outermost one; the outermost are numbers. Writing $\int_0^2\int_0^3\int_0^6 dz\,dy\,dx$ for the tetrahedron integrates over the whole box (volume 36). Check volumes against simple shapes, like $\frac{1}{6}abc$ for a corner tetrahedron.