📐 Concept diagram

07-01 — Triple Integrals

Phase: 7 — Calculus IV: Vector Calculus Subject: 07-01 Prerequisites: Double integrals (Phase 6), 06-09 — Double Integrals Next subject: 07-02 — Triple Integrals in Cylindrical and Spherical Coordinates

Learning Objectives

By the end of this subject, you will be able to:

1. Define the triple integral as a limit of Riemann sums and interpret it as volume (when f=1) or mass (when f is density)
2. Set up and evaluate iterated triple integrals over rectangular boxes using Fubini's theorem
3. Describe and classify 3D regions as Type I, Type II, or Type III and set up the corresponding iterated integrals
4. Change the order of integration in a triple integral when one order is computationally intractable
5. Apply triple integrals to compute volume, mass, center of mass, and moments of inertia

Core Content

1. Definition — Riemann Sums for Triple Integrals

⚠️ CRITICAL FOUNDATION: Triple integrals ∭_E f(x,y,z) dV extend double integrals to 3D. The challenge is correctly describing the solid E as Type I, II, or III and setting up nested bounds. When f=1, the integral gives volume; when f is density, it gives mass.

The triple integral of f(x, y, z) over a solid region E is:

$∭_E f(x, y, z) dV = lim    Σ Σ Σ f(xᵢ*, yⱼ*, zₖ*) ΔV
                     Δ→0   i  j  k
$

Where ΔV = Δx Δy Δz is the volume of a small sub-box, and (xᵢ, yⱼ, zₖ*) is a sample point.

Physical interpretations:

• If f(x, y, z) = 1, then ∭_E 1 dV = Volume(E)
• If f(x, y, z) = ρ(x, y, z) (density), then ∭_E ρ dV = Mass(E)
• Average value: f_avg = (1/Volume) ∭_E f dV

2. Fubini's Theorem — Iterated Triple Integrals

Fubini's Theorem for triple integrals: If f is continuous on the box B = [a, b] × [c, d] × [r, s], then:

$              b  d  s
∭_B f dV = ∫  ∫  ∫  f(x, y, z) dz dy dx
             a  c  r
$

And the order can be rearranged in any of the 3! = 6 possible orders — they all give the same result.

How to read iterated triple integrals:

$    b  d  s
∫  ∫  ∫  f(x, y, z) dz dy dx
   a  c  r

means: first integrate w.r.t. z (treating x,y constant),
       then integrate the result w.r.t. y (treating x constant),
       then integrate that result w.r.t. x.
$

Example 1 (box): Evaluate ∭_B xyz dV over B = [0, 1] × [0, 2] × [0, 3].

    1  2  3
∫  ∫  ∫  xyz dz dy dx
   0  0  0

Innermost (z, treating x,y constant):
∫₀³ xyz dz = xy[z²/2]₀³ = (9/2)xy

Middle (y, treating x constant):
∫₀² (9/2)xy dy = (9/2)x[y²/2]₀² = (9/2)x·2 = 9x

Outer:
∫₀¹ 9x dx = 9[x²/2]₀¹ = 9/2

Since this is a box, we can factor it: (∫₀¹ x dx)(∫₀² y dy)(∫₀³ z dz) = (1/2)·(2)·(9/2) = 9/2. ✓

3. Triple Integrals over General Regions

A solid region E in 3D is classified by how it's bounded.

Type I (z-simple): z is bounded between two surfaces that are functions of x and y:

$E = {(x, y, z) : (x, y) ∈ D,  u₁(x, y) ≤ z ≤ u₂(x, y)}

∭_E f dV = ∬_D [∫_{z=u₁}^{u₂} f(x, y, z) dz] dA
         = ∬_D (inner z-integral) dA
$

Then D itself is evaluated as a double integral (Type I or II in xy-plane).

Type II (y-simple): y bounded between surfaces as functions of x and z.

Type III (x-simple): x bounded between surfaces as functions of y and z.

General approach to setting up a triple integral:

1. Project the solid E onto the xy-plane (or xz-plane, or yz-plane) to get the shadow region D
2. Determine z-limits: find the bottom surface z = u₁(x, y) and top surface z = u₂(x, y)
3. Set up the double integral over D with the z-integral innermost
4. Describe D as a Type I or Type II region for the outer double integral

Example 2 (Type I solid): Set up ∭_E z dV where E is the solid under the plane z = 4 − x − y and above the triangle in the xy-plane with vertices (0,0), (1,0), (0,1).

Step 1: Project onto xy-plane. The triangle D: x ≥ 0, y ≥ 0, x + y ≤ 1.
Step 2: z ranges from z = 0 (xy-plane) to z = 4 − x − y.
Step 3: D is Type I: x ∈ [0, 1], y ∈ [0, 1 − x].

    1  1−x  4−x−y
∫  ∫   ∫       z dz dy dx
   0  0    0

Inner (z): ∫₀^{4−x−y} z dz = [z²/2]₀^{4−x−y} = (4−x−y)²/2

Middle (y): ∫₀^{1−x} (4−x−y)²/2 dy
Let u = 4−x−y, du = −dy.
When y = 0: u = 4−x. When y = 1−x: u = 4−x−(1−x) = 3.
= (1/2)∫_{4−x}³ u² (−du) = (1/2)∫₃^{4−x} u² du = (1/2)[u³/3]₃^{4−x}
= (1/6)[(4−x)³ − 27]

Outer (x): (1/6)∫₀¹ [(4−x)³ − 27] dx
∫₀¹ (4−x)³ dx: let v = 4−x, dv = −dx, limits 4→3
= ∫₄³ v³ (−dv) = ∫₃⁴ v³ dv = [v⁴/4]₃⁴ = (256−81)/4 = 175/4

∫₀¹ 27 dx = 27

Total: (1/6)[175/4 − 27] = (1/6)[175/4 − 108/4] = (1/6)(67/4) = 67/24

4. Changing the Order of Integration

Just as with double integrals, sometimes the given order is impossible to integrate in closed form, but a different order works. With triple integrals, there are 6 possible orders (permutations of dx, dy, dz).

Strategy for reversing order: 1. Identify the region E from the given bounds 2. Draw/sketch E (or its projections) 3. Describe E in the new order, finding new bounds

Example 3: Change the order of ∫₀¹ ∫₀^z ∫₀^y f(x, y, z) dx dy dz to end with dz.

Original: z ∈ [0, 1], y ∈ [0, z], x ∈ [0, y].

Region E: x ≤ y ≤ z, 0 ≤ z ≤ 1.
In the xyz-coordinate system, this is a tetrahedron with vertices (0,0,0), (0,0,1), (0,1,1), (1,1,1).

Project onto xy-plane: the shadow D is the triangle x ∈ [0, 1], y ∈ [x, 1].
For each (x, y), z ranges from z = y (the plane z = y, since y ≤ z) to z = 1.

New integral: ∫₀¹ ∫_x¹ ∫_y¹ f(x, y, z) dz dy dx.

Example 4 (integral that requires reordering): Evaluate ∫₀¹ ∫_z¹ ∫₀^y e^(x²) dx dy dz.

Inner ∫₀^y e^(x²) dx cannot be expressed in elementary functions.

But reversing the order helps. Original region: z ∈ [0, 1], y ∈ [z, 1], x ∈ [0, y].

The plane projection approach:
- x ranges from 0 to 1
- For fixed x: y ranges from x to 1 (since x ≤ y ≤ 1), and z ranges from 0 to y

Wait, let's be careful. The region: 0 ≤ x ≤ y, z ≤ y ≤ 1, 0 ≤ z ≤ 1.
So z goes from 0 to x? Let's think...

Original: x ∈ [0, y], y ∈ [z, 1], z ∈ [0, 1].

In the x-direction first (outer = dz):
Start with x: 0 ≤ x ≤ ?, then y: x ≤ y ≤ ?, then z: 0 ≤ z ≤ ?.

The region in xyz: 0 ≤ x ≤ y, and 0 ≤ z ≤ y ≤ 1.

For fixed x: y goes from x to 1. For fixed x and y: z goes from 0 to y.
So: ∫₀¹ ∫_x¹ ∫₀^y e^(x²) dz dy dx.

Inner: ∫₀^y e^(x²) dz = e^(x²)·y
Middle: ∫_x¹ y e^(x²) dy = e^(x²)[y²/2]_x¹ = e^(x²)(1 − x²)/2
Outer: ∫₀¹ (1 − x²)/2 · e^(x²) dx

Since e^(x²) has no elementary antiderivative, try a different integration order — put x innermost.

Alternative: x ∈ [0, y], y ∈ [z, 1], z ∈ [0, 1].
Keep x innermost: ∫₀¹ ∫_z¹ ∫₀^y e^(x²) dx dy dz — same as original!

Let me try dz innermost:
Project onto xy-plane: shadow is x ∈ [0, 1], y ∈ [x, 1].
z: from 0 to y (since z ≤ y).
∫₀¹ ∫_x¹ ∫₀^y e^(x²) dz dy dx = ∫₀¹ ∫_x¹ y e^(x²) dy dx
= ∫₀¹ e^(x²) [y²/2]_x¹ dx = ∫₀¹ e^(x²)(1 − x²)/2 dx

Still tough... but actually with dx innermost and a different outer:
Let me try x outermost, then z, then y:
x: 0 ≤ x ≤ 1, then for fixed x, z and y need to satisfy z ≤ y and y ∈ [x, 1] with z ∈ [0, y].
This becomes: z ∈ [0, 1], y ∈ [max(x,z), 1].

This is getting complicated — not all integrals become easy by reordering. The key point is that reordering can sometimes help, and the method is to carefully describe the region.

For a clean example that DOES work: ∫₀¹ ∫₀^{1−z} ∫₀^{1−y−z} f(x, y, z) dx dy dz. This is a tetrahedron. The same tetrahedron in a different order would have different bounds.

### 5. Volume, Mass, and Moments

**Volume:**

V = ∭_E 1 dV

**Mass** (with density ρ(x, y, z)):

m = ∭_E ρ(x, y, z) dV

$
**Center of mass** (x̄, ȳ, z̄):
$

x̄ = (1/m) ∭_E x ρ dV ȳ = (1/m) ∭_E y ρ dV z̄ = (1/m) ∭_E z ρ dV

$
**Moments of inertia** about coordinate axes:
$

I_x = ∭_E (y² + z²) ρ dV (about x-axis) I_y = ∭_E (x² + z²) ρ dV (about y-axis) I_z = ∭_E (x² + y²) ρ dV (about z-axis)

### Common Misconceptions

1. **"Triple integrals are always harder than double integrals."** Not true — over boxes they factor nicely. Over symmetric regions, cylindrical or spherical coordinates (07-02) make them easy.

2. **"The integrand must be non-negative."** No — f can be negative. The integral then represents net accumulation (e.g., net charge when f is charge density).

3. **"Setting up the bounds is like double integrals but with one more variable."** Partly true, but 3D regions are harder to visualize. Always sketch the projection first.

4. **"All six orders of integration are equally easy."** No — one order may yield elementary antiderivatives while others produce non-elementary integrals. The order choice matters critically.

---

---
## Key Terms

- **Center of mass**
- **Describe D**
- **Mass**
- **Mass(E)**
- **Moments of inertia**
- **Project**
- **Set up**
- **Volume(E)**

## Worked Examples

### Example 1: Volume of a Wedge

**Problem:** Find the volume of the solid wedge bounded by z = 0, z = y, x = 0, x = 1, y = 0, y = 2.

**Solution:**

Step 1: Identify the region. - A rectangular base in the xy-plane: x ∈ [0, 1], y ∈ [0, 2] - z ranges from 0 (bottom) to y (top)

Step 2: Set up the triple integral. V = ∭_E 1 dV = ∫₀¹ ∫₀² ∫₀^y dz dy dx

Step 3: Evaluate. Innermost: ∫₀^y 1 dz = y Middle: ∫₀² y dy = [y²/2]₀² = 2 Outer: ∫₀¹ 2 dx = 2

Volume = 2 cubic units.

**Check:** The cross-sectional area at fixed x is a triangle with base 2 and height 2, area = 2. Times length 1 = 2. ✓

### Example 2: Mass with Variable Density

**Problem:** A solid cube [0, 1]³ has density ρ(x, y, z) = x + y + z. Find its mass.

**Solution:**

m = ∭ ρ dV = ∫₀¹ ∫₀¹ ∫₀¹ (x + y + z) dz dy dx

Innermost (z): ∫₀¹ (x + y + z) dz = [(x+y)z + z²/2]₀¹ = x + y + 1/2

Middle (y): ∫₀¹ (x + y + 1/2) dy = [xy + y²/2 + y/2]₀¹ = x + 1/2 + 1/2 = x + 1

Outer (x): ∫₀¹ (x + 1) dx = [x²/2 + x]₀¹ = 1/2 + 1 = 3/2

Mass = 3/2.

**Symmetry check:** Average density = average of (x+y+z) over [0,1]³ = 1/2 + 1/2 + 1/2 = 3/2. Volume = 1. Mass = 3/2. ✓

### Example 3: Center of Mass of a Tetrahedron

**Problem:** Find the center of mass of the tetrahedron with vertices (0,0,0), (1,0,0), (0,1,0), (0,0,1) with uniform density ρ = 1.

**Solution:**

The tetrahedron lies in x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1.

Volume: V = 1/6 (known formula for tetrahedron volume).

Now, x̄ = (1/V) ∭ x dV. By symmetry, x̄ = ȳ = z̄.

Set up as Type I (z-simple): z: 0 to 1−x−y y: 0 to 1−x x: 0 to 1

∭ x dV = ∫₀¹ ∫₀^{1−x} ∫₀^{1−x−y} x dz dy dx

Innermost: ∫₀^{1−x−y} x dz = x(1−x−y) Middle: ∫₀^{1−x} x(1−x−y) dy = x[(1−x)y − y²/2]₀^{1−x} = x[(1−x)(1−x) − (1−x)²/2] = x(1−x)²/2

Outer: ∫₀¹ (x/2)(1−x)² dx = (1/2)∫₀¹ x(1−2x+x²) dx = (1/2)∫₀¹ (x − 2x² + x³) dx = (1/2)[x²/2 − 2x³/3 + x⁴/4]₀¹ = (1/2)(1/2 − 2/3 + 1/4) = (1/2)(6/12 − 8/12 + 3/12) = (1/2)(1/12) = 1/24

x̄ = (1/24) / (1/6) = 1/4.

By symmetry: (x̄, ȳ, z̄) = (1/4, 1/4, 1/4).

### Example 4: Reversing Order

**Problem:** Rewrite ∫₀² ∫₀^{√(4−x²)} ∫₀^{√(4−x²−y²)} f(x, y, z) dz dy dx with dx innermost.

**Solution:**

The region is the solid in the first octant inside the sphere x² + y² + z² = 4. Original: x ∈ [0, 2], y ∈ [0, √(4−x²)], z ∈ [0, √(4−x²−y²)].

For dx innermost (order: dz dy dx → ... dx ...): The full sphere in the first octant has y² + z² ≤ 4 − x².

For dy dz dx order (outer x, then z, then y): After projecting onto xz-plane, y ranges from 0 to √(4−x²−z²). The shadow in xz-plane: x² + z² ≤ 4, x ≥ 0, z ≥ 0. So: x ∈ [0, 2], z ∈ [0, √(4−x²)], y ∈ [0, √(4−x²−z²)].

But we want dx innermost. Let's try z outer, y middle, x innermost: The sphere: x² + y² + z² ≤ 4. x = √(4−y²−z²) on the surface. z ∈ [0, 2], y ∈ [0, √(4−z²)], x ∈ [0, √(4−y²−z²)].

So: ∫₀² ∫₀^{√(4−z²)} ∫₀^{√(4−y²−z²)} f(x, y, z) dx dy dz. ```

Quiz

Q1: A triple integral ∭_E f(x, y, z) dV with f = 1 computes:

A) The surface area of E B) The volume of E C) The mass of E D) The center of mass of E

Correct: B)

• If you chose B: When f = 1, ∭_E 1 dV = Volume(E), just as ∬_R 1 dA = Area(R). Correct!
• If you chose A: Surface area requires a surface integral, not a triple integral.
• If you chose C: Mass requires f to be the density function ρ(x, y, z).
• If you chose D: Center of mass involves additional moments, not just the triple integral.

Q2: Fubini's theorem for triple integrals over a box [a,b]×[c,d]×[p,q] states that the integral can be written as:

A) ∫_a^b ∫_c^d f(x,y,z) dy dx only B) Any of the 6 possible orders of iterated integration, all giving the same result C) Only dx dy dz in that specific order D) The integral cannot be split into iterated integrals

Correct: B)

• If you chose B: Fubini's theorem guarantees all 3! = 6 orders of integration give the same result for continuous f on a box. Correct!
• If you chose A: That's a double integral, not triple — and you can choose any order.
• If you chose C: The order doesn't matter — all 6 permutations work.
• If you chose D: Triple integrals over boxes can always be expressed as iterated integrals.

Q3: For a Type I solid region E, which variable has limits that are functions?

A) All three variables have constant limits B) z-limits are functions of (x, y); x and y have constant limits C) y-limits are functions of (x, z) D) x-limits are functions of (y, z)

Correct: B)

• If you chose B: Type I (z-simple): E = {(x,y,z) : (x,y) ∈ D, g₁(x,y) ≤ z ≤ g₂(x,y)}. z is between two surfaces. Correct!
• If you chose A: That describes a rectangular box, not a general Type I solid.
• If you chose C: That's (roughly) Type II (y-simple).
• If you chose D: That's (roughly) Type III (x-simple).

Q4: When changing the order of integration in a triple integral, you must:

A) Simply swap the limits B) Redescribe the solid region E in the new order, adjusting all limit expressions accordingly C) Multiply by −1 D) Keep the same limits but permute the differentials

Correct: B)

• If you chose B: Changing order requires completely redescribing the solid — the limits are region-dependent, not just swapped. Correct!
• If you chose A: Swapping limits without redescribing gives the wrong region.
• If you chose C: No sign change occurs for order changes (unlike orientation reversal).
• If you chose D: The limit expressions must change to reflect the region in the new order.

Q5: The triple integral ∭_E ρ(x, y, z) dV where ρ is density gives:

A) Volume B) Surface area C) Mass D) Average density

Correct: C)

• If you chose C: Integrating density over a solid gives its total mass. Correct!
• If you chose A: That requires f = 1.
• If you chose B: Surface area is a surface integral.
• If you chose D: Average density would be (1/Volume) ∭_E ρ dV.

Q6: To verify your limits for a triple integral, you can:

A) Check that all limits are constants B) Integrate f = 1 and confirm the result equals the known volume of E C) Take the derivative of the result D) Set the integral equal to zero

Correct: B)

• If you chose B: Computing ∭_E 1 dV should give the geometric volume — this is an excellent sanity check. Correct!
• If you chose A: Only boxes have all constant limits.
• If you chose C: Differentiation doesn't verify triple integral limits.
• If you chose D: The integral is generally nonzero.

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Evaluate ∭_E (x + 2y + 3z) dV over E = [0, 1] × [0, 2] × [0, 3].

Problem 2: Find the volume of the solid bounded by the coordinate planes and the plane x + y/2 + z/3 = 1.

Problem 3: A solid occupies the region under z = x² + y² above the square [0, 1] × [0, 1]. Set up and evaluate the triple integral for its volume.

Problem 4: Rewrite ∫₀¹ ∫₀^{1−z} ∫₀^{1−y−z} f(x, y, z) dx dy dz with dy innermost.

Problem 5: Find the mass of the solid cylinder x² + y² ≤ 1, 0 ≤ z ≤ 2, with density ρ(x, y, z) = z.

Problem 6: Find the moment of inertia I_z (about z-axis) for the solid box [0, a] × [0, b] × [0, c] with uniform density ρ = 1.

Problem 7: Evaluate ∫₀¹ ∫₀^{1−x} ∫₀^{1−x−y} 6z dz dy dx.

Summary

1. The triple integral ∭_E f dV generalizes double integrals to 3D: it represents volume (when f=1), mass (when f is density), or other accumulated quantities over a solid region
2. Fubini's theorem allows triple integrals to be evaluated as iterated integrals over boxes in any of 6 possible orders, and the choice of order matters critically for computational tractability
3. General 3D regions are described by projecting onto the coordinate planes and finding bounds for the innermost variable as functions of the other two — the bounds on the "inner" integral may depend on the "outer" variables, but the outermost bounds must be constants
4. Changing the order of integration can transform an intractable integral into an elementary one, but requires careful geometric reasoning to determine the new bounds
5. Triple integrals compute physical quantities: volume ∭ 1 dV, mass ∭ ρ dV, center of mass (1/m)∭ (x,y,z)ρ dV, and moments of inertia ∭ (r²)ρ dV

Pitfalls

• Setting the outermost bounds as variables. In an iterated triple integral, the outermost limits must be CONSTANTS (numbers), the middle limits can depend on the outermost variable only, and the innermost limits can depend on both outer variables. A common error is to put functions of y or z in the outermost x-limits, producing an integral where the final result still contains variables.
• Skipping the projection step. The most reliable way to set up a triple integral is to project the solid onto a coordinate plane first, get the shadow region D, then determine the innermost variable's bounds. Trying to determine all six bounds simultaneously without a projection almost always leads to errors.
• Getting the integration order wrong when reordering. In a triple integral, there are 6 possible orders. Describing the same solid in a new order requires carefully identifying how each variable is bounded by the others. A mistake in any one bound produces a completely different integral that evaluates to a different number.
• Confusing moment of inertia formulas. I_z = ∭ (x² + y²)ρ dV uses distance squared to the z-axis (x² + y²), NOT z². I_x uses (y² + z²) and I_y uses (x² + z²). Mixing these up — especially using the wrong squared-distance formula for a given axis — is a common error on physics-oriented problems.
• Treating all regions as Type I. Not every solid is naturally z-simple. Some are better described as y-simple (Type II) or x-simple (Type III), and forcing a Type I setup on the wrong solid can result in needing to split the region or dealing with messy bounds. Always consider which variable has the simplest bounding surfaces.

Next Steps

Move on to 07-02 — Triple Integrals in Cylindrical and Spherical Coordinates to learn how coordinate transformations simplify triple integrals over regions with cylindrical or spherical symmetry, including the Jacobian factors r (cylindrical) and ρ² sin φ (spherical).