Guides / Integrals

Triple integrals: limits from the inside out, r dz dr dθ in cylindrical, and ρ² sin φ in spherical

A triple integral adds up a quantity over a solid: its volume when the integrand is 1, its mass when the integrand is a density, a centroid when it is divided by the volume. It is three single integrals nested inside each other, and nearly every mistake happens before any integrating starts: in limits that should depend on the outer variables, in a volume element that lost its r or its sin φ, or in an angle given the wrong range.

Short answer

Limits: work from the inside out. The innermost limits may depend on both outer variables, the middle limits on the outer one only, and the outer limits are numbers. Cylindrical (x = r cos θ, y = r sin θ, z): dV = r dz dr dθ, for solids with a round cross-section. Spherical (x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ): dV = ρ² sin φ dρ dφ dθ, with 0 ≤ φ ≤ π measured down from the positive z-axis and 0 ≤ θ ≤ 2π. Centroid: z̄ = (∭ z dV)/V.

Why it works

An iterated integral slices the solid. Fix x and y, and the inner integral runs along a vertical segment from the bottom surface to the top one, so its limits are functions of x and y. The middle integral then sweeps that segment across a strip of the shadow region, and the outer integral sweeps the strips across the whole shadow. Each level can depend only on the variables still outside it, which is why the outer limits must be numbers.

The extra factors are the volumes of small boxes. In cylindrical coordinates a step dr, dθ, dz carves out a wedge whose sides are dr, r dθ and dz: the arc length of a step in angle grows with the radius, so dV = r dz dr dθ. In spherical coordinates the sides are dρ, ρ dφ and ρ sin φ dθ, because the circle traced by θ at angle φ from the z-axis has radius ρ sin φ. Multiplying gives ρ² sin φ. Since φ is measured from the z-axis, it only needs to go from 0 to π to reach the south pole; θ supplies the full turn.

Example 1: inner limits that depend on the outer variables

E is the solid above the triangle 0 ≤ y ≤ x ≤ 1 in the xy-plane and below the plane z = x + y. Find its volume and ∭_E x dV.

∫₀¹ ∫₀^x ∫₀^x+y x dz dy dxz runs from the floor to the plane, y across the triangle at a fixed x, and x over numbers.
∫₀^x+y x dz = x(x + y)x is a constant for the inner integral.
∫₀^x (x² + xy) dy = 3x³/2x² · x + x · x²/2.
∫₀¹ 3x³/2 dx = 3/8The outer integral is an ordinary one-variable integral.
V = 1/2, ∭ x dV = 3/8The same limits with integrand 1 give ∫₀¹ 3x²/2 dx = 1/2. Dividing, the centroid has x̄ = 3/4: the solid is taller toward x = 1, so its weight sits on that side.

Example 2: cylindrical coordinates under a slanted plane

Find the volume of the solid inside the cylinder x² + y² = 1, above z = 0 and below the plane z = 2 − y.

0 ≤ z ≤ 2 − r sin θWrite the top surface in cylindrical coordinates. On the unit disk y ≤ 1, so the plane stays above the floor.
∫₀^2π ∫₀¹ (2 − r sin θ) r dr dθThe inner z-integral of 1 is the height. Keep the r from dV.
∫₀¹ (2r − r² sin θ) drMultiply out. θ is a constant for the r-integral.
= 1 − (sin θ)/3r² − (r³/3) sin θ at r = 1.
∫₀^2π (1 − (sin θ)/3) dθ = 2πsin θ integrates to 0 over a full turn. The answer is the disk's area π times the average height 2: the plane tilts up on one side exactly as much as it tilts down on the other.

Example 3: the centroid of a cone

The solid cone between z = √(x² + y²) and the plane z = 2 has its tip at the origin. Find ∭ z dV and the height z̄ of its centroid.

r ≤ z ≤ 2, 0 ≤ r ≤ 2The cone is z = r; it meets the plane where r = 2.
∫_r² z · r dz = r(4 − r²)/2The factor r from dV rides along.
∫₀² (2r − r³/2) dr = 4 − 2 = 2r² − r⁴/8 at r = 2.
∭ z dV = 2π · 2 = 4πNothing depends on θ, so the θ-integral multiplies by 2π.
V = 8π/3, z̄ = 4π/(8π/3) = 3/2The volume is (1/3)π · 2² · 2, the cone formula, and the same integral with integrand 1 agrees. The centroid is three quarters of the way from the tip to the base, as for every cone.

Example 4: spherical coordinates for a hemisphere

Find the height z̄ of the centroid of the solid upper half of the unit ball, x² + y² + z² ≤ 1 with z ≥ 0.

0 ≤ φ ≤ π/2ρ runs from 0 to 1 and θ from 0 to 2π. The upper half is φ up to the equator, π/2, not π.
z dV = ρ³ cos φ sin φ dρ dφ dθz = ρ cos φ, times the volume element ρ² sin φ dρ dφ dθ.
∫₀¹ ρ³ dρ = 1/4The limits are all numbers and the integrand is a product, so the triple integral splits into three single ones.
∫₀^π/2 cos φ sin φ dφ = 1/2It is (sin² φ)/2 from 0 to π/2.
∭ z dV = 2π/8 = π/4The θ-integral gives 2π, times 1/4 and 1/2.
z̄ = (π/4)/(2π/3) = 3/8The volume is half of 4π/3. For a hemisphere of radius a the centroid is at 3a/8.

Common mistakes, and the exact line they happen on

1. A number where the middle limit needs a variable

The limits of Example 1.

0 ≤ z ≤ x + yRight so far.
∫₀¹ ∫₀¹ x(x + y) dy dx = 7/12This is the mistake. Running y from 0 to 1 integrates over the whole unit square, not the triangle under the line y = x. Draw the shadow region: at a fixed x, y only goes up to x. The right value is 3/8.

2. Dropping the r in cylindrical coordinates

The volume of Example 2.

0 ≤ z ≤ 2 − r sin θRight so far.
∫∫ (2 − r sin θ) dr dθ = 4πThis is the mistake. Without the r, every ring of the disk counts the same however large it is, so the small rings near the axis are overweighted. A quick test: the base is a disk of area π and the average height is 2, so the volume must be 2π.

3. Leaving out sin φ in spherical coordinates

The integral of z in Example 4.

z = ρ cos φRight so far.
∭ ρ³ cos φ dρ dφ dθ = π/2This is the mistake. The volume element is ρ² sin φ, not ρ²; without sin φ the thin regions near the poles count as much as the wide band at the equator. The same error makes the volume of the hemisphere π²/3 instead of 2π/3. The right value is π/4.

Practice

Evaluate ∭ (x + z) dV over the box 0 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 ≤ z ≤ 3.
Show answer
∭ x dV = 2 · 1 · 3 = 6 and ∭ z dV = 2 · 1 · 9/2 = 9, so the total is 15.
Find the volume of the solid inside the cylinder r = 2, above z = 0 and below z = x² + y².
Show answer
∫₀^2π ∫₀² r² · r dr dθ = 2π · 4 = 8π.
Find the volume of the spherical shell 1 ≤ ρ ≤ 2.
Show answer
2π · 2 · (8 − 1)/3 = 28π/3: the φ-integral of sin φ from 0 to π is 2, and the ρ-integral of ρ² is 7/3.
Find the volume of the part of the unit ball with 0 ≤ φ ≤ π/3.
Show answer
2π · (1/3) · (1 − cos(π/3)) = 2π · (1/3) · (1/2) = π/3.
The cylinder r ≤ 1, 0 ≤ z ≤ 2 has density z. Find its mass.
Show answer
∫₀^2π ∫₀¹ ∫₀² z · r dz dr dθ = 2π · (1/2) · 2 = 2π.