- From Double to Triple Integrals
- Integration Over a Box, Fubini's Theorem
- Integration Over a General Region
- Setting Up Limits and Switching Order
- Applications: Volume, Mass, Centre of Mass
- Triple Integrals in Cylindrical Coordinates
- Triple Integrals in Spherical Coordinates
- Worked Examples
- Common Mistakes
- 10-Question Quiz
§ 01From Double to Triple Integrals
A double integral ∬ f(x,y) dA sums the values of a function over a 2D region. A triple integral ∭ f(x,y,z) dV does the same over a solid 3D region E.
The idea is the same: slice E into tiny volume elements dV, evaluate f at each point, multiply, and sum. In Cartesian coordinates, dV = dx dy dz. The resulting triple integral is evaluated as three nested single-variable integrals, working from the innermost outward.
When f = 1, the triple integral gives the volume of the region E. When f = ρ(x,y,z) (density), it gives the mass. The integrand determines what physical quantity is computed.
§ 02Integration Over a Box — Fubini's Theorem
The simplest region for a triple integral is a rectangular box: B = [a,b] × [c,d] × [e,f], meaning a ≤ x ≤ b, c ≤ y ≤ d, e ≤ z ≤ f. All limits are constant numbers.
On a rectangular box with continuous f, the triple integral equals the iterated integral in any of the six possible orderings. You are free to choose whichever order makes the arithmetic easiest.
All Six Integration Orders
For a box B = [a,b]×[c,d]×[e,f], the six orderings all give the same result:
§ 03Integration Over a General Region
For a non-box region E, the limits of the inner integrals depend on the outer variables. There are three standard types, depending on which axis is "stacked".
E lies above the surface z = u₁(x,y) and below z = u₂(x,y). The projection of E onto the xy-plane is the 2D region D. Integrate z first (between the surfaces), then integrate the result over D using a double integral.
Same idea but y is the "inner" variable, bounded by two surfaces y = v₁(x,z) and y = v₂(x,z). The 2D region D is the projection onto the xz-plane.
x is the inner variable. In practice this ordering is less common but occasionally the most natural choice.
§ 04Setting Up Limits and Switching Order
Setting up the correct limits is the hardest skill in triple integration. A reliable 4-step strategy works for most problems.
The 4-Step Strategy for Setting Up Limits
Switching the Order of Integration
Sometimes an integral in one order is impossible to evaluate in closed form, but in another order it becomes elementary. The strategy for switching order is:
§ 05Applications: Volume, Mass, and Centre of Mass
Triple integrals are the natural tool for any quantity that is "spread through" a 3D region.
Setting f = 1 integrates the constant function 1 over all of E. The result is the total volume. This is almost always the simplest possible triple integral.
If ρ is the mass density (mass per unit volume), integrating ρ over the solid gives the total mass. Uniform density ρ = constant gives M = ρ·V.
The x-coordinate of the centre of mass is the weighted average of x over E. Similarly for ȳ and z̄. The centre of mass (x̄, ȳ, z̄) is the balance point of the solid.
The moment of inertia about the z-axis measures resistance to rotation. Similar formulas with (y²+z²) and (x²+z²) give Ix and Iy.
§ 06Triple Integrals in Cylindrical Coordinates
When the region E has cylindrical symmetry (a cylinder, cone, or anything involving x² + y²) cylindrical coordinates (r, θ, z) transform the integral into a much simpler form.
The key factor is the Jacobian r. Every cylindrical-coordinate triple integral must include this factor. Also: x² + y² becomes r², which dramatically simplifies most integrands.
§ 07Triple Integrals in Spherical Coordinates
When E is a sphere, hemisphere, or any region where x² + y² + z² appears, spherical coordinates (ρ, θ, φ) give the simplest form. See the Cylindrical & Spherical page for full derivations.
The Jacobian is ρ² sinφ. Both factors are required. For a full sphere of radius R: 0 ≤ ρ ≤ R, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π.
| Region Type | Best System | Key Reason |
|---|---|---|
| Box / planes | Cartesian | Constant limits; no simplification from other systems |
| Cylinder, cone | Cylindrical | x²+y² = r²; Jacobian r; limits become constant |
| Sphere, hemisphere | Spherical | x²+y²+z² = ρ²; Jacobian ρ² sinφ; limits constant |
| Mixed (cone + sphere) | Spherical | Both surfaces simplify in spherical — cone is φ=const, sphere is ρ=const |
| Paraboloid | Cylindrical | z = r² describes the paraboloid; limits become simple |
§ 08Worked Examples
All limits are constants (a box), so Fubini applies immediately. Choose order dz dy dx.
∫02 ∫01 ∫03 xyz dz dy dxThe integrand xyz = x · y · z factors completely. Apply Fubini's product rule:
= [∫02 x dx] · [∫01 y dy] · [∫03 z dz]The tetrahedron is bounded by x ≥ 0, y ≥ 0, z ≥ 0, and the plane x + y + z = 1 (equivalently z ≤ 1 − x − y).
Integrate z first from 0 to 1−x−y. Then the projection D onto the xy-plane is the triangle x ≥ 0, y ≥ 0, x + y ≤ 1.
V = ∫01 ∫01−x ∫01−x−y dz dy dxConfirmed by the formula V = (1/6)|det[edges]| = 1/6 for this standard tetrahedron ✓
The paraboloid z = 4 − x² − y² meets z = 0 when x² + y² = 4 (circle of radius 2). So E: 0 ≤ z ≤ 4−x²−y², (x,y) in the disk x²+y² ≤ 4.
The remaining double integral is over the disk x²+y² ≤ 4. Switch to polar: x²+y² = r², dA = r dr dθ, r ∈ [0,2], θ ∈ [0,2π].
∬ (4−r²)²/2 · r dr dθ = ∫02π ∫02 r(4−r²)²/2 dr dθLet u = 4−r², du = −2r dr:
∫02 r(4−r²)²/2 dr = (1/2)∫40 u²(−du/2) = (1/4)∫04 u² du = (1/4)(64/3) = 16/3x² + y² = 9 → r = 3. The plane z = x + 4 becomes z = r cosθ + 4. Limits: 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ r cosθ + 4.
The ∫cosθ term vanishes over a full revolution, the tilted plane contributes zero net volume above the symmetric cylinder.
Cone: z = r. Region: r ≤ z ≤ 2, which requires r ≤ 2. So 0 ≤ r ≤ 2, r ≤ z ≤ 2, 0 ≤ θ ≤ 2π. Integrand x²+y² = r².
∫02π ∫02 ∫r2 r² · r dz dr dθUnit ball: 0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π. Integrand f = 1.
V = ∫02π ∫0π ∫01 ρ² sinφ dρ dφ dθThe integrand ρ² sinφ separates completely:
= [∫01 ρ² dρ] · [∫0π sinφ dφ] · [∫02π dθ] = (1/3) · 2 · 2π = 4π/3The classical formula for the volume of a sphere of radius 1, derived in three lines.
x²+y²+z² = ρ², so (x²+y²+z²)^(3/2) = ρ³. Integrand becomes e^(ρ³). This is impossible in Cartesian.
∫02π ∫0π ∫02 e^(ρ³) · ρ² sinφ dρ dφ dθLet u = ρ³, du = 3ρ² dρ → ρ² dρ = du/3:
∫02 ρ²e^(ρ³)dρ = (1/3)[e^(ρ³)]02 = (e⁸ − 1)/3The inner integral ∫y1 e^(x²) dx has no closed form (e^(x²) has no elementary antiderivative). We must switch order.
From the limits: 0 ≤ z ≤ 1 (z-variable is free, constant limits), 0 ≤ y ≤ 1, y ≤ x ≤ 1. The region in the xy-plane: 0 ≤ y ≤ x, 0 ≤ x ≤ 1 (triangle). So E = {(x,y,z) : 0 ≤ z ≤ 1, 0 ≤ y ≤ x, 0 ≤ x ≤ 1}.
New order: dz dy dx. Since z is independent, keep 0 ≤ z ≤ 1. For the xy-triangle, switch to: 0 ≤ x ≤ 1, 0 ≤ y ≤ x.
= ∫01 ∫0x ∫01 e^(x²) dz dy dxe^(x²) is treated as a constant when integrating in y.
Upper hemisphere: 0 ≤ ρ ≤ 3, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2. Density δ = z = ρ cosφ.
M = ∫02π∫0π/2∫03 ρ cosφ · ρ² sinφ dρ dφ dθThe pyramid has square base 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and apex at z = 1. At height z, the cross-section is a square of side (1−z). So: 0 ≤ z ≤ 1, 0 ≤ x ≤ 1−z, 0 ≤ y ≤ 1−z.
The pyramid is symmetric about x = (1−z)/2 and y = (1−z)/2. So x̄ = ȳ = 1/4 by the symmetry of a uniform square section.
The centre of mass lies one-quarter of the way from the base to the apex, consistent with the general rule that a solid pyramid's centroid is at 1/4 of the height.
§ 09Common Mistakes
§ 1010-Question Quiz
Test your understanding of iterated triple integrals, limits, applications, and coordinate systems.