Difficulty: Intermediate to Advanced | Prerequisites: Double integrals, partial derivatives, polar coordinates, basic 3D geometry.
Triple integrals extend double integrals into three dimensions: instead of integrating over a flat region, you integrate over a solid. This set of notes covers how to compute mass, volume, and centre of mass for 3D objects using Cartesian, cylindrical, and spherical coordinates. You should be comfortable with double integrals over non-rectangular regions before tackling this material. If cylindrical and spherical coordinates are new to you, review the coordinate-conversion formulas first.
Triple integrals let you sum a quantity through a volume. Choosing the right coordinate system (Cartesian, cylindrical, or spherical) is often the difference between a manageable integral and an impossible one. Centre of mass requires computing three separate coordinate averages, each weighted by density.
Triple integral
The integral ∭_E f(x, y, z) dV, which sums f over a solid region E in three-dimensional space. Think of it as adding up contributions from every tiny cube inside the solid.
Mass of a solid
M = ∭_E ρ(x, y, z) dV, where ρ is the density function (mass per unit volume). When ρ is constant, the mass is simply density times volume.
Centre of mass (x̄, ȳ, z̄)
The "balance point" of a solid. Each coordinate is the density-weighted average of that coordinate over the solid: x̄ = (1/M) ∭ x ρ dV, and similarly for ȳ and z̄.
Cylindrical coordinates (r, θ, z)
Polar coordinates in the xy-plane plus the usual z-axis. The volume element is dV = r dr dθ dz. Best suited for solids with circular cross-sections or symmetry about the z-axis.
Spherical coordinates (ρ, φ, θ)
ρ is the distance from the origin, φ is the angle down from the positive z-axis, θ is the angle in the xy-plane from the positive x-axis. The volume element is dV = ρ² sin φ dρ dφ dθ. Best for spheres, cones, and regions described by distance from a point.
Jacobian (for coordinate changes)
The scaling factor that adjusts the volume element when you change coordinates. For cylindrical coordinates the Jacobian is r; for spherical it is ρ² sin φ. In simple terms, it accounts for how much a small region stretches or compresses under the transformation.
Tetrahedron
A solid bounded by four planes, forming a shape with four triangular faces. Integration over a tetrahedron requires careful attention to which variable's bounds depend on the others.
Setting up bounds for a tetrahedron
Consider a tetrahedron with vertices at the origin and points along the axes, bounded by the plane 3x + y + 2z = 6. The key steps:
Determine the range of the outermost variable. Here z runs from 0 to 3.
For each z, express the bounds on x: 0 ≤ x ≤ (-2/3)z + 2.
For each (x, z), express the bounds on y: 0 ≤ y ≤ 6 - 3x - 2z.
The order dz dx dy, dy dx dz, or any other permutation is valid, but some orders produce simpler bounds than others. Choose the one that avoids square roots or complicated expressions.
Computing mass with density ρ = z
With density proportional to height (ρ = z), the mass integral becomes:
M = ∫₀³ ∫₀^(-2/3 z + 2) ∫₀^(6 - 3x - 2z) z dy dx dz
The inner integral (over y) gives z(6 - 3x - 2z). After integrating over x and then z, the mass works out to 9/2.
Computing centre of mass
Each coordinate of the centre of mass requires its own triple integral:
x̄ = (1/M) ∭ xz dy dx dz
ȳ = (1/M) ∭ yz dy dx dz
z̄ = (1/M) ∭ z² dy dx dz
For this tetrahedron with ρ = z, the centre of mass is (2/6, 6/6, 6/5), which simplifies to approximately (1/3, 1, 6/5). The z̄ coordinate being above the geometric centroid makes sense: the density increases with z, so the "weight" is shifted upward.
The problem
Maximise ∭_E (9 - 4x² - 4y² - 4z²) dV over all possible regions E.
The integrand is positive when 9 - 4x² - 4y² - 4z² > 0, which gives 4(x² + y² + z²) < 9, or equivalently x² + y² + z² < 9/4. In other words, 4ρ² < 9, so ρ < 3/2.
The optimal region is the ball of radius 3/2 centred at the origin: include every point where the integrand is positive, and exclude every point where it is negative.
Converting to spherical coordinates
The integral becomes:
∫₀^(2π) ∫₀^π ∫₀^(3/2) (9 - 4ρ²) ρ² sin φ dρ dφ dθ
The integrand expands to 9ρ² sin φ - 4ρ⁴ sin φ. Integrating over ρ from 0 to 3/2:
∫₀^(3/2) 9ρ² dρ = 3ρ³ |₀^(3/2) = 3(27/8) = 81/8
∫₀^(3/2) 4ρ⁴ dρ = (4/5)ρ⁵ |₀^(3/2) = (4/5)(243/32) = 243/40
After the φ integral (∫₀^π sin φ dφ = 2) and the θ integral (∫₀^(2π) dθ = 2π), the result simplifies to 81π/5.
Key insight
To maximise a triple integral, integrate over exactly the region where the integrand is non-negative. This is a standard exam question pattern.
The region
The solid E lies between the hyperboloid z² = x² + y² - 1 (below, where z ≤ 0) and the plane z = 0 (above), with the cylinder x² + y² = 2 as the outer boundary. In cylindrical coordinates, the lower surface is z = -√(r² - 1), the upper surface is z = 0, and r runs from 1 to √2.
Projecting onto the xy-plane gives an annular ring: 1 ≤ r ≤ √2, 0 ≤ θ ≤ 2π.
Volume calculation
V = ∫₀^(2π) ∫₁^(√2) ∫_{-√(r²-1)}^0 r dz dr dθ
The inner integral gives r√(r² - 1). Using the substitution u = r² - 1, du = 2r dr, the integral over r becomes (2/3)(r² - 1)^(3/2) evaluated from 1 to √2, giving 2/3. Multiplying by 2π:
Volume = (2/3)π cm³
Mass with constant density
With osmium at ρ = 22.59 g/cm³, the mass is M = 22.59 × (2/3)π ≈ 47.312 g.
Centre of mass
By symmetry, x̄ = 0 and ȳ = 0, since the density is constant and the region is symmetric about the z-axis (a "pseudo-donut" shape).
For z̄, the integral involves z² weighted by density:
z̄ = (1/M) ∫₀^(2π) ∫₁^(√2) ∫_{-√(r²-1)}^0 22.59 · z · r dz dr dθ
This evaluates to approximately -3/8, giving the centre of mass at (0, 0, -3/8).
Finding the sphere equation
A sphere of radius 1 centred at (0, 0, 1) has the equation x² + y² + (z - 1)² = 1, which expands to x² + y² + z² = 2z. In spherical coordinates, ρ² = 2ρ cos φ, so ρ = 2 cos φ.
This is the upper bound on ρ. The sphere sits above the xy-plane, so φ ranges from 0 to π/2, and θ from 0 to 2π.
Region description in spherical coordinates
E = {(ρ, φ, θ) : 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2, 0 ≤ ρ ≤ 2 cos φ}
Mass with density ρ_density = ρ (distance from origin)
The mass integral uses the density function equal to the spherical coordinate ρ (distance from origin), with the volume element ρ² sin φ:
M = ∫₀^(2π) ∫₀^(π/2) ∫₀^(2cos φ) ρ · ρ² sin φ dρ dφ dθ
After integrating the ρ⁴ term over ρ, then using the identity for cos⁴ φ sin φ (whose antiderivative is -cos⁵ φ / 5), the mass evaluates to 8π/5.
Centre of mass
By symmetry (constant-density-like symmetry about the z-axis), x̄ = 0 and ȳ = 0.
For z̄, use z = ρ cos φ in the integrand:
z̄ = (1/M) ∫₀^(2π) ∫₀^(π/2) ∫₀^(2cos φ) ρ · ρ cos φ · ρ² sin φ dρ dφ dθ
This involves integrating ρ⁴ cos φ sin φ over ρ, then cos⁶ φ sin φ over φ (antiderivative: -cos⁷ φ / 7). The final result is z̄ = 8/7.
Centre of mass: ⟨0, 0, 8/7⟩. The centre of mass sits above the geometric centre of the sphere (which is at z = 1), because the density increases with distance from the origin, weighting the top of the sphere more heavily.
Volume element in Cartesian coordinates
dV = dx dy dz
Volume element in cylindrical coordinates
dV = r dr dθ dz
Volume element in spherical coordinates
dV = ρ² sin φ dρ dφ dθ
Coordinate conversions (cylindrical ↔ Cartesian)
x = r cos θ, y = r sin θ, z = z, r² = x² + y²
Coordinate conversions (spherical ↔ Cartesian)
x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ, ρ² = x² + y² + z²
Mass
M = ∭_E ρ(x, y, z) dV
Centre of mass
x̄ = (1/M) ∭ xρ dV, ȳ = (1/M) ∭ yρ dV, z̄ = (1/M) ∭ zρ dV
Useful antiderivatives
∫ cosⁿ φ sin φ dφ = -cosⁿ⁺¹ φ / (n + 1) + C
Centre-of-mass calculations determine where to support a physical object so it balances. In aerospace, the centre of mass of a fuel tank (whose density changes as fuel is consumed) governs aircraft stability. In robotics, knowing the centre of mass of each link in a robot arm is essential for motion planning.
Triple integrals with density functions are used in medical imaging (CT scans reconstruct density distributions) and in geophysics (estimating the mass of geological formations from density maps).
Forgetting the Jacobian when converting to cylindrical or spherical coordinates. The volume element is not just dρ dφ dθ; it is ρ² sin φ dρ dφ dθ in spherical and r dr dθ dz in cylindrical.
Mixing up φ and θ in spherical coordinates. In the convention used in most US calculus courses, φ is the polar angle (from the z-axis) and θ is the azimuthal angle (in the xy-plane). Some physics texts swap these.
Assuming the centre of mass lies at the geometric centroid when the density is non-uniform. A heavier top shifts the centre of mass upward; a heavier bottom shifts it downward.
Setting the wrong upper bound for ρ in spherical coordinates. When the sphere is not centred at the origin, the bound on ρ depends on φ (e.g. ρ = 2 cos φ), not a constant.
⚠️ Converting a sphere equation to spherical coordinates (x² + y² + z² = 2z becomes ρ = 2 cos φ) is a standard exam move. Practise it until it is automatic.
⚠️ Using symmetry to set x̄ = 0 and ȳ = 0 saves significant computation. Always check whether the region and density are symmetric about an axis before integrating.
⚠️ The pattern "maximise ∭_E f dV by integrating where f ≥ 0" appears regularly. Recognise it quickly.
True or false: in spherical coordinates, the volume element is ρ² cos φ dρ dφ dθ. (False. It is ρ² sin φ dρ dφ dθ.)
Fill in the blank: the sphere x² + y² + z² = 2z, in spherical coordinates, becomes ρ = ____. (2 cos φ)
True or false: if a solid has constant density and is symmetric about the z-axis, then x̄ = 0 and ȳ = 0. (True.)
Fill in the blank: to maximise ∭_E f dV, you should integrate over the region where f is ____. (non-negative / positive)
True or false: the mass of a solid with density ρ = 5 and volume V = 3 is 15. (True. M = ρV = 5 × 3 = 15.)
Q: A solid tetrahedron has vertices at (0,0,0), (2,0,0), (0,4,0), and (0,0,3). The density is ρ = 1. What integral gives the mass, in the order dz dy dx?
A: The bounding plane is x/2 + y/4 + z/3 = 1, so z = 3(1 - x/2 - y/4). Bounds: x from 0 to 2, y from 0 to 4(1 - x/2), z from 0 to 3(1 - x/2 - y/4). The mass is ∫₀² ∫₀^(4-2x) ∫₀^(3-3x/2-3y/4) dz dy dx.
Q: Convert the equation x² + y² + z² = 4z to spherical coordinates and state the bounds on ρ, φ, and θ.
A: ρ² = 4ρ cos φ gives ρ = 4 cos φ. Bounds: 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2, 0 ≤ ρ ≤ 4 cos φ.
Q: A solid of revolution about the z-axis has density ρ = 5. Its volume is (2/3)π. What are x̄ and ȳ?
A: Both are 0, by symmetry. The constant density and rotational symmetry about the z-axis guarantee this.
Q: You want to maximise ∭_E (16 - x² - y² - z²) dV. What region E should you choose, and what coordinate system is most natural?
A: Integrate where the integrand is non-negative: x² + y² + z² ≤ 16, i.e. the ball of radius 4. Spherical coordinates are the natural choice for a ball centred at the origin.
Triple integrals are the gateway to surface integrals and flux (Stokes' and Divergence theorems), which appear later in Calculus III. The spherical-coordinate machinery practised here is essential for those theorems when applied to spheres and hemispheres.
Centre-of-mass calculations generalise to moments of inertia (second moments), which measure how mass is distributed relative to an axis. Moments of inertia are central to rotational dynamics in physics.
The optimisation idea ("integrate where the integrand is positive") connects to variational methods used in physics and engineering.
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