Difficulty: Advanced | Prerequisites: §15.1–15.5 (double integrals, polar coordinates, surface area), visualising 3D regions from boundary equations.
Triple integrals extend double integrals into three dimensions. Instead of integrating a function over a flat region D, you integrate over a solid region E in space. The main challenges are: describing E with correct bounds on three variables, choosing the best order of integration, and setting up the integral so it is actually computable. This section also introduces mass and centre of mass for three-dimensional objects, using a density function ρ(x, y, z). These calculations combine geometry (what does the solid look like?) with calculus (how do I integrate over it?).
A triple integral ∫∫∫_E f(x, y, z) dV computes the accumulated value of f over a solid E. With f = 1 you get volume; with f = ρ you get mass. The hardest part is describing E with nested inequalities in x, y, z. Changing the order of integration often makes the integral tractable.
Triple integral
∫∫∫_E f(x, y, z) dV, the three-dimensional analogue of the double integral. When f = 1, it computes the volume of E.
Iterated integral
A triple integral written as three nested single integrals, each over one variable. The bounds of the inner integrals typically depend on the outer variables.
Order of integration
The sequence in which you integrate the three variables (e.g. dz dy dx vs dx dz dy). There are six possible orders. Choosing the right one can turn an impossible integral into a routine one.
Tetrahedron
A solid with four triangular faces. In Problem 7, the tetrahedron is bounded by the coordinate planes (x = 0, y = 0, z = 0) and the plane 2x + 3y + z = 6.
Centre of mass (̄x, ̄y, ̄z)
The "balance point" of a solid with density ρ. Computed by dividing the first moments M_{yz}, M_{xz}, M_{xy} by the total mass M.
Hyperboloid (of one sheet)
A surface of the form x² + y² − z² = c. In Problem 10, the equation x² + y² − z² = 1 with z ≤ 0 forms a bowl opening downward.
Maximising a triple integral
In Problem 9, the region E is chosen so that the integrand 9 − 4x² − 4y² − 4z² is never negative. The optimal E is the ellipsoid (here a sphere of radius 3/2) where the integrand equals zero on the boundary.
The tetrahedron is bounded by x = 0, y = 0, z = 0, and 2x + 3y + z = 6, with ρ(x, y, z) = z.
Solve the plane equation for z: z = 6 − 2x − 3y.
Project onto the xy-plane: the triangular region where 2x + 3y ≤ 6, x ≥ 0, y ≥ 0. This gives 0 ≤ x ≤ 3, 0 ≤ y ≤ (6 − 2x)/3.
Mass: M = ∫₀³ ∫₀^{(6−2x)/3} ∫₀^{6−2x−3y} z dz dy dx.
The inner integral gives z²/2 evaluated from 0 to 6 − 2x − 3y, yielding (6 − 2x − 3y)²/2.
For the centre of mass, also compute M_{yz} = ∫∫∫ xz dV, M_{xz} = ∫∫∫ yz dV, and M_{xy} = ∫∫∫ z² dV, then divide each by M.
The original integral is ∫₀³ ∫_{9−x²}⁹ ∫₀^{9−y} f(x, y, z) dz dy dx.
Describing the region E:
x ranges from 0 to 3
For each x, y ranges from 9 − x² to 9
For each (x, y), z ranges from 0 to 9 − y
From the bounds: y = 9 − x² is a downward parabola, y + z = 9 is a plane, and z = 0 is the xy-plane.
To rewrite in other orders, express E as a set of inequalities and reorganise. For dz dx dy: y ranges from 0 to 9, x from 0 to √(9 − y), z from 0 to 9 − y. For dy dz dx: x from 0 to 3, z from 0 to x², y from 9 − x² to 9 − z.
The key skill is visualising the solid from three different projections (onto xy, xz, and yz planes).
The integrand 9 − 4x² − 4y² − 4z² is positive inside the ellipsoid 4x² + 4y² + 4z² ≤ 9, i.e. x² + y² + z² ≤ 9/4.
This is a sphere of radius 3/2. Including any point outside this sphere would subtract from the integral (the integrand is negative there), so the optimal E is the closed ball x² + y² + z² ≤ 9/4.
To evaluate, use spherical coordinates:
∫₀^{2π} ∫₀^π ∫₀^{3/2} (9 − 4ρ²) ρ² sin φ dρ dφ dθ.
The angular integrals give 4π. The radial integral ∫₀^{3/2} (9ρ² − 4ρ⁴) dρ = 9(3/2)³/3 − 4(3/2)⁵/5. Multiply through to get the maximum value.
The region E lies inside x² + y² = 3, above the hyperboloid x² + y² − z² = 1 (so z = −√(x² + y² − 1)), and below z = 0.
In cylindrical coordinates: 0 ≤ θ ≤ 2π, 1 ≤ r ≤ √3 (the hyperboloid requires r² ≥ 1), −√(r² − 1) ≤ z ≤ 0.
Note: r starts at 1, not 0, because the hyperboloid x² + y² − z² = 1 at z = 0 gives x² + y² = 1.
Volume = ∫₀^{2π} ∫₁^{√3} ∫_{-√(r²−1)}^0 r dz dr dθ = 2π ∫₁^{√3} r√(r² − 1) dr.
Substitute u = r² − 1 to evaluate.
For the centre of mass with constant density, ̄x = 0 and ̄y = 0 by symmetry. Compute ̄z = M_{xy}/(M) where M_{xy} = ∫∫∫ z · ρ dV and M = ρ · V.
Triple integral (Cartesian)
∫∫∫E f(x, y, z) dV = ∫a^b ∫{g₁(x)}^{g₂(x)} ∫{h₁(x,y)}^{h₂(x,y)} f(x, y, z) dz dy dx
Volume
V = ∫∫∫_E 1 dV
Mass
M = ∫∫∫_E ρ(x, y, z) dV
First moments
M_{yz} = ∫∫∫ xρ dV, M_{xz} = ∫∫∫ yρ dV, M_{xy} = ∫∫∫ zρ dV
Centre of mass
̄x = M_{yz}/M, ̄y = M_{xz}/M, ̄z = M_{xy}/M
Cylindrical coordinates
x = r cos θ, y = r sin θ, z = z, dV = r dz dr dθ
Spherical coordinates
x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ, dV = ρ² sin φ dρ dφ dθ
Triple integrals compute the mass of engine blocks, the gravitational pull of irregularly shaped asteroids, and the charge distribution inside a semiconductor. Centre of mass calculations are essential in aerospace engineering (where does the rocket balance?) and structural engineering (where does the load concentrate?). The Osmia artifact problem (Problem 10) is a toy version of how archaeologists estimate the mass and balance point of artefacts from their geometry.
Students frequently mix up which moment corresponds to which coordinate. M_{xy} (integrating zρ) gives you ̄z, not ̄x. The subscript tells you which plane the moment is about.
When changing the order of integration, students rewrite the bounds algebraically without re-checking them against the geometry. Always re-sketch or verify that the new bounds describe the same solid.
For Problem 9, students sometimes try calculus of variations or Lagrange multipliers. The argument is simpler: the integrand is non-negative inside the sphere and negative outside, so include exactly the non-negative part.
In cylindrical coordinates, students forget that r starts at 1 (not 0) in Problem 10 because the hyperboloid has a waist radius of 1.
⚠️ Changing the order of integration in a triple integral is a classic exam question. You must be able to describe E from any projection.
⚠️ Mass and centre of mass with a given ρ(x, y, z) will almost certainly appear. Know the formulas cold and watch the moment subscripts.
⚠️ "Find the region that maximises the integral" is an unusual problem type. The principle, include only where the integrand is non-negative, applies whenever you see it.
⚠️ Cylindrical coordinates are the natural choice when the region involves cylinders or surfaces of revolution about the z-axis. Recognise the signal.
True or false: ∫∫∫_E 1 dV gives the mass of E.
False. It gives the volume. Mass requires integrating the density ρ.
Fill in the blank: The moment M_{xy} = ∫∫∫ ___ ρ dV, and it is used to find ̄___.
z; ̄z
True or false: To maximise ∫∫∫_E g(x,y,z) dV, you should always choose E to be all of ℝ³.
False. You include only the region where g ≥ 0, because negative contributions reduce the total.
Fill in the blank: In cylindrical coordinates, dV = ___.
r dz dr dθ
Q: For the tetrahedron bounded by x = 0, y = 0, z = 0, and x + y + z = 1, set up the integral for volume in the order dz dy dx.
A: ∫₀¹ ∫₀^{1−x} ∫₀^{1−x−y} dz dy dx. The inner limit comes from solving z = 1 − x − y, the middle from y = 1 − x, and x runs from 0 to 1.
Q: Rewrite ∫₀¹ ∫₀^x ∫₀^y f dz dy dx in the order dy dx dz.
A: The region is 0 ≤ z ≤ y ≤ x ≤ 1. In order dy dx dz: z from 0 to 1, x from z to 1, y from z to x. So ∫₀¹ ∫_z¹ ∫_z^x f dy dx dz.
Q: Why does the region in Problem 10 have r ≥ 1 rather than r ≥ 0?
A: The hyperboloid x² + y² − z² = 1 requires x² + y² ≥ 1 (since z² ≥ 0). At z = 0 the waist of the hyperboloid has radius 1, so the solid has a hole in the middle.
Q: A solid E has constant density ρ = 5 and volume 12. What is its mass? Where is its centre of mass if E is symmetric about the z-axis?
A: Mass = 5 × 12 = 60. By symmetry ̄x = 0, ̄y = 0. ̄z requires computing M_{xy}/M.
Triple integrals in Cartesian coordinates are the foundation for cylindrical (§15.7) and spherical (§15.8) coordinate integrals, which are just triple integrals with a different coordinate system and Jacobian. Centre of mass calculations reappear in physics (rigid-body dynamics) and in §16 when you integrate vector fields over surfaces and volumes. The order-of-integration skill transfers directly to probability theory, where you change the order of integration in joint density functions.
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