A
double integral
adds a function over a flat region. Add one more dimension and you integrate over a
solid E in space. The triple
integral
\iiint_E f(x, y, z)\, dV
sums f over every infinitesimal box
dV = dx\, dy\, dz packed into E. If
f is a density, the result is total mass; if
f \equiv 1, it is the plain volume of the solid:
\operatorname{vol}(E) = \iiint_E 1\, dV.
Iterated integrals: integrate from the inside out
Just as Fubini turned a double integral into two nested single integrals, a triple integral
unwinds into three. Integrate the innermost variable first (the others held
constant), then the middle, then the outer.
Worked example — a box [0,1]\times[0,2]\times[0,3]
Step 1 — set up the iterated integral for
f(x,y,z) = xyz over the box. All limits are constants:
\iiint_E xyz\, dV = \int_0^1 \!\int_0^2 \!\int_0^3 xyz\; dz\, dy\, dx.
Step 2 — innermost integral (over z). Hold
x, y fixed and pull them out:
\int_0^3 xyz\, dz = xy \left[ \frac{z^2}{2} \right]_0^3 = xy \cdot \frac{9}{2} = \frac{9}{2} xy.
Step 3 — middle integral (over y). Now integrate
the result over y, holding x fixed:
\int_0^2 \frac{9}{2} xy\, dy = \frac{9}{2} x \left[ \frac{y^2}{2} \right]_0^2 = \frac{9}{2} x \cdot 2 = 9x.
Step 4 — outer integral (over x).
\int_0^1 9x\, dx = 9 \left[ \frac{x^2}{2} \right]_0^1 = \frac{9}{2}.
So \iiint_E xyz\, dV = \tfrac92. (Because the box has constant
limits and the integrand factors as x \cdot y \cdot z, the whole
thing also equals \big(\int_0^1 x\, dx\big)\big(\int_0^2 y\, dy\big)\big(\int_0^3 z\, dz\big) = \tfrac12 \cdot 2 \cdot \tfrac92 = \tfrac92
— a handy check.)
Coordinates that match the solid's symmetry
Boxes are easy; cylinders and spheres are not, in Cartesian coordinates. As in the
polar
case, the cure is to change coordinates — and the volume element picks up a Jacobian factor.
Cylindrical coordinates: polar in the plane, plus a height
Use polar (r, \theta) for x, y and keep
z as the height: x = r\cos\theta,
y = r\sin\theta, z = z. The polar area
element r\, dr\, d\theta simply gains a dz:
dV = r\, dz\, dr\, d\theta.
Worked example — volume of a cylinder, radius a, height h
Step 1 — set up with f \equiv 1. The solid is
0 \le r \le a, 0 \le \theta \le 2\pi,
0 \le z \le h:
V = \int_0^{2\pi} \!\int_0^a \!\int_0^h 1 \cdot r\, dz\, dr\, d\theta.
Step 2 — innermost (z). Nothing depends on
z, so it contributes a factor h:
\int_0^h r\, dz = r h.
Step 3 — middle (r). Here the Jacobian
r matters:
\int_0^a r h\, dr = h \left[ \frac{r^2}{2} \right]_0^a = \frac{a^2 h}{2}.
Step 4 — outer (\theta). A constant integrand
over a full turn multiplies by 2\pi:
\int_0^{2\pi} \frac{a^2 h}{2}\, d\theta = 2\pi \cdot \frac{a^2 h}{2} = \pi a^2 h.
The familiar V = \pi a^2 h — base area times height — drops out,
and the extra r was exactly what made it right.
Spherical coordinates: for balls and shells
For spherical symmetry use radius \rho, polar angle
\varphi (down from the z-axis), and
azimuth \theta:
x = \rho\sin\varphi\cos\theta,
y = \rho\sin\varphi\sin\theta,
z = \rho\cos\varphi. Its volume element carries a
\rho^2 \sin\varphi:
dV = \rho^2 \sin\varphi\; d\rho\, d\varphi\, d\theta.
For a continuous f on a solid E, the
triple integral is an iterated integral whose volume element depends on the coordinates:
-
Cartesian: dV = dx\, dy\, dz.
-
Cylindrical (x=r\cos\theta,\ y=r\sin\theta,\ z=z):
dV = r\, dz\, dr\, d\theta.
-
Spherical
(x=\rho\sin\varphi\cos\theta,\ y=\rho\sin\varphi\sin\theta,\ z=\rho\cos\varphi):
dV = \rho^2 \sin\varphi\, d\rho\, d\varphi\, d\theta.
-
Volume: \operatorname{vol}(E) = \iiint_E 1\, dV
in any of them.
The art of a triple integral is rarely the integration — it is the setup, and the
setup is dictated by the solid's symmetry. A rule of thumb:
- Box-shaped solid, or limits that are flat planes → Cartesian.
-
A circular cross-section, an axis of rotational symmetry (a can, a pipe, a cone) →
cylindrical: the r\, d z\, d r\, d\theta
element fits like a glove.
-
A ball, a spherical shell, or anything that depends only on distance from a centre →
spherical: then x^2+y^2+z^2 = \rho^2
collapses the integrand to a function of \rho alone.
For instance, the volume of a ball of radius a is brutal in
Cartesian coordinates but a one-liner in spherical:
V = \int_0^{2\pi}\!\int_0^{\pi}\!\int_0^a \rho^2 \sin\varphi\; d\rho\, d\varphi\, d\theta = 2\pi \cdot 2 \cdot \frac{a^3}{3} = \frac{4}{3}\pi a^3.
(The three factors are \int_0^{2\pi} d\theta = 2\pi,
\int_0^\pi \sin\varphi\, d\varphi = 2, and
\int_0^a \rho^2\, d\rho = a^3/3.) Pick the coordinates the solid
is shaped like, and the hard integral becomes three easy ones.