📐 Concept diagram

07-02 — Triple Integrals in Cylindrical and Spherical Coordinates

Phase: 7 — Calculus IV: Vector Calculus Subject: 07-02 Prerequisites: 07-01 — Triple Integrals, polar coordinates (Phase 6) Next subject: 07-03 — Change of Variables and Jacobians

Learning Objectives

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

1. Convert triple integrals from rectangular to cylindrical coordinates, applying the Jacobian factor r
2. Convert triple integrals from rectangular to spherical coordinates, applying the Jacobian factor ρ² sin φ
3. Identify when cylindrical or spherical coordinates are advantageous based on the symmetry of the region and integrand
4. Describe common 3D regions (cylinders, cones, spheres, paraboloids) in cylindrical and spherical coordinates
5. Apply these coordinate systems to compute volume, mass, center of mass, and moments of inertia

Core Content

1. Cylindrical Coordinates

⚠️ CRITICAL FOUNDATION: Cylindrical (dV = r dz dr dθ) and spherical (dV = ρ² sin φ dρ dφ dθ) coordinates are essential for problems with axial or radial symmetry. NEVER forget the Jacobian factors r and ρ² sin φ — omitting them gives wrong answers.

Definition: A point (x, y, z) in rectangular coordinates is expressed in cylindrical coordinates (r, θ, z) as:

$x = r cos θ
y = r sin θ
z = z
$

Where r ≥ 0, θ ∈ [0, 2π), and z is unrestricted.

Interpretation: Cylindrical coordinates extend polar coordinates into the third dimension by adding the rectangular z-coordinate. "r" is the distance from the z-axis (not from the origin!).

Conversion formulas:

r = √(x² + y²)    (distance from z-axis)
θ = arctan(y/x)   (with quadrant adjustment)
z = z

The Jacobian for cylindrical coordinates:

When changing variables, the volume element changes: dV = r dr dθ dz.

This comes from the Jacobian determinant:

$          |∂x/∂r  ∂x/∂θ  ∂x/∂z|   |cos θ  −r sin θ  0|
J = det   |∂y/∂r  ∂y/∂θ  ∂y/∂z| = |sin θ   r cos θ  0| = r
          |∂z/∂r  ∂z/∂θ  ∂z/∂z|   |  0       0      1|
$

So: ∭_E f(x, y, z) dV = ∭_E f(r cos θ, r sin θ, z) · r dz dr dθ

The factor r is crucial — forgetting it is the #1 error with cylindrical coordinates.

Why r? In polar/cylindrical coordinates, a small "rectangular" patch of size dr × dθ is actually a small sector with area ≈ r dr dθ (length in angular direction is r dθ, not dθ).

2. Common Regions in Cylindrical Coordinates

Cylinder: x² + y² ≤ a², c ≤ z ≤ d

$r ∈ [0, a], θ ∈ [0, 2π], z ∈ [c, d]
$

Paraboloid: z = x² + y² (opening upward)

z = r²
r ∈ [0, √z], or for volume: z ∈ [r², H], r ∈ [0, √H]

Cone: z = √(x² + y²)

$z = r
$

Solid between two surfaces (e.g., under paraboloid z = 4 − x² − y² above disk x² + y² ≤ 1):

$r ∈ [0, 1], θ ∈ [0, 2π], z ∈ [0, 4 − r²]
$

3. Worked Examples — Cylindrical

Example 1: Find the volume of the solid bounded by the paraboloid z = x² + y² and the plane z = 4.

The paraboloid meets the plane when x² + y² = 4, i.e., r = 2.
Region: r ∈ [0, 2], θ ∈ [0, 2π], z ∈ [r², 4].

V = ∫₀^{2π} ∫₀² ∫_{r²}⁴ r dz dr dθ

Inner (z): ∫_{r²}⁴ r dz = r(4 − r²)
Middle (r): ∫₀² (4r − r³) dr = [2r² − r⁴/4]₀² = 8 − 4 = 4
Outer (θ): ∫₀^{2π} 4 dθ = 8π

Volume = 8π.

Check in rectangular coordinates (tedious!): V = ∫₋₂² ∫{-√(4−x²)}^{√(4−x²)} ∫{x²+y²}⁴ dz dy dx The θ-integral alone gives 2π; in rectangular you'd need ∫₋₂² √(4−x²) dx which gives π/2 times something. Much harder.

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

$In cylindrical: ρ = r. The cylinder: r ∈ [0, 1], θ ∈ [0, 2π], z ∈ [0, 2].

m = ∭ r · r dz dr dθ = ∫₀^{2π} ∫₀¹ ∫₀² r² dz dr dθ
  = 2π · 2 · ∫₀¹ r² dr = 4π · [r³/3]₀¹ = 4π/3.
$

Example 3: Find the moment of inertia I_z for the solid between z = r² and z = 4 (the paraboloid from Example 1) with density 1.

$I_z = ∭ (x² + y²) dV = ∭ r² · r dz dr dθ

Region: r ∈ [0, 2], θ ∈ [0, 2π], z ∈ [r², 4].

I_z = ∫₀^{2π} ∫₀² ∫_{r²}⁴ r³ dz dr dθ
Inner: r³(4 − r²) = 4r³ − r⁵
Middle: ∫₀² (4r³ − r⁵) dr = [r⁴ − r⁶/6]₀² = 16 − 64/6 = 16 − 32/3 = (48−32)/3 = 16/3
Outer: 2π · 16/3 = 32π/3.
$

4. Spherical Coordinates

Definition: A point (x, y, z) is expressed in spherical coordinates (ρ, φ, θ) as:

$x = ρ sin φ cos θ
y = ρ sin φ sin θ
z = ρ cos φ
$

Where: - ρ ≥ 0 is the distance from the origin (not the z-axis!) - φ ∈ [0, π] is the angle from the positive z-axis (the colatitude or polar angle) - θ ∈ [0, 2π) is the same azimuthal angle as in cylindrical coordinates

NOTE: In physics, θ and φ are often swapped (θ for polar, φ for azimuthal). In this curriculum, we follow the mathematical convention: ρ = radial distance, φ = polar angle from z-axis, θ = azimuthal angle in xy-plane.

Conversion formulas:

ρ = √(x² + y² + z²)       (distance from origin)
φ = arccos(z/ρ)           (angle from z-axis)
θ = arctan(y/x)           (same as cylindrical)

The Jacobian for spherical coordinates:

$          |∂x/∂ρ  ∂x/∂φ  ∂x/∂θ|
J = det   |∂y/∂ρ  ∂y/∂φ  ∂y/∂θ|
          |∂z/∂ρ  ∂z/∂φ  ∂z/∂θ|

    |sin φ cos θ  ρ cos φ cos θ  −ρ sin φ sin θ|
= det|sin φ sin θ  ρ cos φ sin θ   ρ sin φ cos θ| = ρ² sin φ
    |cos φ       −ρ sin φ          0           |
$

So: ∭_E f(x, y, z) dV = ∭_E f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) · ρ² sin φ dρ dφ dθ

The Jacobian ρ² sin φ is crucial. For φ ∈ [0, π], sin φ ≥ 0, so ρ² sin φ ≥ 0.

Geometric intuition: A small volume element in spherical coordinates is a "curved brick" of dimensions: - radial: dρ - polar: ρ dφ (arc length in the φ-direction at radius ρ) - azimuthal: ρ sin φ dθ (arc length in the θ-direction at radius ρ, accounting for the latitude)

Volume = dρ × ρ dφ × ρ sin φ dθ = ρ² sin φ dρ dφ dθ.

5. Common Regions in Spherical Coordinates

Sphere of radius R: x² + y² + z² ≤ R²

$ρ ∈ [0, R], φ ∈ [0, π], θ ∈ [0, 2π]
$

Spherical shell (hollow ball): R₁ ≤ ρ ≤ R₂

$ρ ∈ [R₁, R₂], φ ∈ [0, π], θ ∈ [0, 2π]
$

Cone: z = √(x² + y²) → ρ cos φ = ρ sin φ → tan φ = 1 → φ = π/4. For the solid cone with half-angle φ₀: φ ∈ [0, φ₀].

Ice cream cone (sphere above cone): ρ ≤ R, 0 ≤ φ ≤ φ₀.

Paraboloid in spherical: z = x² + y² → ρ cos φ = ρ² sin² φ → ρ = cos φ / sin² φ (messy!). Paraboloids are usually better in cylindrical.

6. Worked Examples — Spherical

Example 4: Find the volume of a sphere of radius R.

$Sphere: x² + y² + z² ≤ R². In spherical: ρ ∈ [0, R], φ ∈ [0, π], θ ∈ [0, 2π].

V = ∭ ρ² sin φ dρ dφ dθ

= ∫₀^{2π} dθ · ∫₀^π sin φ dφ · ∫₀^R ρ² dρ

= 2π · [−cos φ]₀^π · [ρ³/3]₀^R

= 2π · (−(−1) − (−1)) · (R³/3)

= 2π · 2 · (R³/3) = (4/3)πR³.
$

This is the familiar formula V = (4/3)πR³, derived cleanly from spherical coordinates.

Example 5: Find the volume of the "ice cream cone" — the solid above the cone z = √(x² + y²) and inside the sphere x² + y² + z² = 4.

$Cone: φ = π/4. Sphere: ρ = 2.
Region: ρ ∈ [0, 2], φ ∈ [0, π/4], θ ∈ [0, 2π].

V = ∫₀^{2π} ∫₀^{π/4} ∫₀² ρ² sin φ dρ dφ dθ

Inner: ∫₀² ρ² dρ = 8/3
Middle: (8/3) ∫₀^{π/4} sin φ dφ = (8/3)[−cos φ]₀^{π/4} = (8/3)(1 − √2/2) = (8/3)(2−√2)/2 = (4/3)(2−√2)
Outer: 2π · (4/3)(2−√2) = (8π/3)(2−√2).

Volume = (8π/3)(2 − √2) ≈ 4.91.
$

Example 6: Find the center of mass of a solid hemisphere x² + y² + z² ≤ R², z ≥ 0, with uniform density.

$Hemisphere: ρ ∈ [0, R], φ ∈ [0, π/2], θ ∈ [0, 2π].

Volume: V = (2/3)πR³ (half the sphere).

By symmetry: x̄ = ȳ = 0 (axisymmetric about z-axis).
z̄ = (1/V) ∭ z dV = (1/V) ∭ ρ cos φ · ρ² sin φ dρ dφ dθ

Numerator: ∭ ρ³ cos φ sin φ dρ dφ dθ
= ∫₀^{2π} dθ ∫₀^R ρ³ dρ ∫₀^{π/2} cos φ sin φ dφ

∫₀^{π/2} cos φ sin φ dφ: Let u = sin φ, du = cos φ dφ, limits 0→1.
= ∫₀¹ u du = 1/2.

∫₀^R ρ³ dρ = R⁴/4.

Numerator = 2π · (R⁴/4) · (1/2) = πR⁴/4.

z̄ = (πR⁴/4) / (2πR³/3) = (πR⁴/4) · (3/(2πR³)) = 3R/8.

So the center of mass is at (0, 0, 3R/8).
$

7. Choosing Between Cylindrical and Spherical

Rule of thumb: - If the region involves x² + y² but not z in a coupled way → cylindrical - If the region involves x² + y² + z² → spherical - Paraboloids: cylindrical (z = r² is simple; in spherical it's messy) - Cones with spheres: spherical if the top is spherical; cylindrical if bounded by z = constant

Key Terms

• Ice cream cone
• Solid between two surfaces

Worked Examples

Example 1: Cylindrical — Volume Between Surfaces

Problem: Find the volume of the solid bounded above by z = 9 − x² − y² and below by z = 0.

Solution:

$The paraboloid z = 9 − r² meets z = 0 at r = 3.
Region: r ∈ [0, 3], θ ∈ [0, 2π], z ∈ [0, 9 − r²].

V = ∫₀^{2π} ∫₀³ ∫₀^{9−r²} r dz dr dθ
  = 2π ∫₀³ r(9 − r²) dr
  = 2π [9r²/2 − r⁴/4]₀³
  = 2π (81/2 − 81/4)
  = 2π (81/4) = 81π/2.
$

Example 2: Spherical — Volume Inside Sphere and Cone

Problem: Find the volume of the solid inside the sphere ρ = 2 and outside the cone φ = π/3.

Solution:

"Outside" the cone means φ ranges from π/3 to π (the lower part).
Wait — the cone φ = π/3 opens upward (narrow cone near z-axis).
"Outside" typically means outside the cone's solid, so φ ∈ [π/3, π].

But let's verify: "inside sphere AND outside cone." The cone φ = π/3 is the surface z = r/√3 (since cos φ = z/ρ = √3/2 at φ=π/3, but let's use tan: r/z = tan(π/3) = √3, so z = r/√3).

Region: ρ ∈ [0, 2], φ ∈ [π/3, π], θ ∈ [0, 2π].

V = ∫₀^{2π} ∫_{π/3}^π ∫₀² ρ² sin φ dρ dφ dθ
  = 2π · (8/3) · ∫_{π/3}^π sin φ dφ
  = (16π/3) · [−cos φ]_{π/3}^π
  = (16π/3) · (1 − (−1/2))
  = (16π/3) · (3/2) = 8π.

Example 3: Conversion — Rectangular to Spherical

Problem: Rewrite ∫₋₁¹ ∫{-√(1−x²)}^{√(1−x²)} ∫{√(x²+y²)}^{√(2−x²−y²)} f(x, y, z) dz dy dx in spherical coordinates.

Solution:

The region: z goes from the cone z = √(x²+y²) to the sphere z = √(2−x²−y²).
In xy-plane: x² + y² ≤ 1 (the unit disk).

In spherical:
- Cone: φ = π/4 (since tan φ = r/z = 1)
- Sphere: x² + y² + z² = 2, so ρ² = 2 → ρ = √2
- On the sphere z = √(2−x²−y²), ρ² sin² φ + ρ² cos² φ = 2 → ρ = √2, for all φ, θ.
- The cone meets the sphere: at ρ = √2, φ = π/4, z = √2 cos(π/4) = 1. x²+y² = 2 sin²(π/4) = 1. This matches the xy-project boundary.

ρ ranges from 0 to √2. For each ρ, φ ranges from 0 (z-axis) to the cone φ = π/4... wait.
z goes from the cone UP to the sphere. So the lower bound is the cone: φ = π/4 at the lower surface.
Actually: at given ρ and φ, z = ρ cos φ. For z ≥ √(x²+y²) = ρ sin φ, we need cos φ ≥ sin φ, i.e., φ ≤ π/4.
So φ ∈ [0, π/4].

θ ∈ [0, 2π] (full rotation because xy-projection is the full disk).

In spherical:
∫₀^{2π} ∫₀^{π/4} ∫₀^{√2} f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) · ρ² sin φ dρ dφ dθ.

Quiz

Q1: In cylindrical coordinates (r, θ, z), the volume element dV is:

A) dr dθ dz B) r dr dθ dz C) r² dr dθ dz D) dz dr dθ

Correct: B)

• If you chose B: dV = r dr dθ dz — the factor r is the Jacobian. Same as polar but with dz added. Correct!
• If you chose A: Missing the Jacobian factor r.
• If you chose C: The Jacobian in cylindrical is r (not r²); r² is for spherical.
• If you chose D: Correct factors but also needs the r.

Q2: In spherical coordinates (ρ, φ, θ), the volume element dV is:

A) dρ dφ dθ B) ρ dρ dφ dθ C) ρ² sin φ dρ dφ dθ D) ρ² dρ dφ dθ

Correct: C)

• If you chose C: dV = ρ² sin φ dρ dφ dθ. The ρ² sin φ factor is the Jacobian determinant. Correct!
• If you chose A: Missing the entire Jacobian — this gives wrong answers.
• If you chose B: Missing both the extra ρ and the sin φ.
• If you chose D: Missing the sin φ factor (accounts for latitude shrinking near poles).

Q3: A solid sphere of radius R centered at the origin is described in spherical coordinates as:

A) ρ: 0 to R, φ: 0 to π/2, θ: 0 to π B) ρ: 0 to R, φ: 0 to π, θ: 0 to 2π C) ρ: −R to R, φ: 0 to π, θ: 0 to 2π D) ρ: 0 to R, φ: 0 to 2π, θ: 0 to π

Correct: B)

• If you chose B: Full sphere: ρ ∈ [0, R], φ ∈ [0, π] (pole to pole), θ ∈ [0, 2π] (full rotation). Correct!
• If you chose A: φ only covers the northern hemisphere.
• If you chose C: ρ must be nonnegative.
• If you chose D: φ ∈ [0, 2π] would double-cover the sphere.

Q4: A circular cylinder x² + y² ≤ 4, 0 ≤ z ≤ 5 is best integrated in:

A) Rectangular coordinates B) Cylindrical coordinates C) Spherical coordinates D) Any coordinates — they're all equally efficient

Correct: B)

• If you chose B: The cross-section is circular (x² + y² ≤ 4) and z is independent — cylindrical coordinates are perfect. Correct!
• If you chose A: The circular cross-section makes rectangular limits messy (√(4−x²)).
• If you chose C: Spherical would work but is unnecessarily complex for a cylinder.
• If you chose D: Cylindrical is clearly most efficient here.

Q5: A cone z = √(x² + y²) for 0 ≤ z ≤ 3 is best described in:

A) Rectangular coordinates only B) Cylindrical: 0 ≤ r ≤ z, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 3 C) Spherical: φ = π/4 (constant), 0 ≤ ρ ≤ 3√2, 0 ≤ θ ≤ 2π D) Both B and C are good choices

Correct: D)

• If you chose D: The cone z = r in cylindrical is 0 ≤ r ≤ z; in spherical, the cone is φ = π/4. Both work well. Correct!
• If you chose A: Rectangular would give messy √(x²+y²) limits.
• If you chose B: This is correct but incomplete — spherical also works nicely.
• If you chose C: This is correct but incomplete — cylindrical also works nicely.

Q6: When computing the volume of a sphere with ∭_E 1 dV in spherical coordinates, the result is:

A) 2πR³ B) (4/3)πR³ C) πR³ D) (2/3)πR³

Correct: B)

• If you chose B: V = ∫₀^{2π} ∫₀^π ∫₀^R ρ² sin φ dρ dφ dθ = 2π · 2 · (R³/3) = (4/3)πR³. Correct!
• If you chose A: Missing the 2/3 factor from the φ integration.
• If you chose C: This is the volume of a cylinder of radius R and height R, not a sphere.
• If you chose D: This is the volume of a hemisphere.

Practice Problems

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

Problem 1: Use cylindrical coordinates to find the volume of the solid cylinder x² + y² ≤ 4, −1 ≤ z ≤ 3.

Problem 2: Find the mass of the solid bounded by the paraboloid z = x² + y² and the plane z = 9, with density ρ(x, y, z) = 1.

Problem 3: Use spherical coordinates to find the volume inside both the sphere ρ = 2 and the cone φ = π/4.

Problem 4: Convert the integral ∫₀¹ ∫₀^{√(1−x²)} ∫₀^{√(1−x²−y²)} (x² + y² + z²) dz dy dx to spherical coordinates and evaluate.

Problem 5: A solid cone of height H and base radius R has uniform density. Using cylindrical coordinates, find its center of mass (z-coordinate). The cone is given by z ∈ [0, H], r ≤ R(1 − z/H).

Problem 6: Find I_z (moment of inertia about z-axis) for the solid sphere x² + y² + z² ≤ R² with uniform density 1. Use spherical coordinates.

Problem 7: Evaluate ∭_E √(x² + y²) dV where E is the solid bounded by the paraboloid z = 9 − x² − y² and the xy-plane.

Summary

1. Cylindrical coordinates (r, θ, z) extend polar coordinates with the rectangular z; the Jacobian is r, giving dV = r dz dr dθ — ideal for cylinders, paraboloids, and regions with rotational symmetry about the z-axis
2. Spherical coordinates (ρ, φ, θ) use distance from origin ρ and polar angle φ from the z-axis; the Jacobian is ρ² sin φ, giving dV = ρ² sin φ dρ dφ dθ — ideal for spheres, spherical caps, and cones with spherical boundaries
3. The Jacobian factor (r or ρ² sin φ) is NOT optional — it accounts for the non-uniform scaling of coordinate transformations and must always be included
4. Converting bounds from rectangular to curvilinear coordinates requires expressing all surfaces in the new coordinates and using the geometry to determine ranges for each variable
5. The choice between cylindrical and spherical is driven by the integrand's dependence on x²+y² vs. x²+y²+z², and whether z appears independently or only in combination with x and y

Pitfalls

• Forgetting the Jacobian factor. The #1 error in curvilinear coordinates: dV is NOT dz dr dθ or dρ dφ dθ. It is r dz dr dθ (cylindrical) and ρ² sin φ dρ dφ dθ (spherical). Omitting the Jacobian gives results that are wrong by a factor that varies with position — and the error is usually not caught because the integral still evaluates to some number.
• Confusing r (cylindrical) with ρ (spherical). In cylindrical coordinates, r = √(x² + y²) is the distance from the z-axis. In spherical coordinates, ρ = √(x² + y² + z²) is the distance from the origin. These are different quantities! Using r when you mean ρ (or vice versa) leads to completely wrong bounds and integrands.
• Using the wrong range for φ in spherical coordinates. The polar angle φ runs from 0 (positive z-axis, "north pole") to π (negative z-axis, "south pole"). Students sometimes copy the θ range [0, 2π) for φ, or use the geographic latitude range [−π/2, π/2], both of which are incorrect in standard mathematical spherical coordinates.
• Forcing spherical coordinates on paraboloids. A paraboloid z = x² + y² becomes z = r² in cylindrical (simple) but ρ cos φ = ρ² sin² φ → ρ = cos φ / sin² φ in spherical (messy). Just because a problem involves 3D doesn't mean spherical is the answer — choose the coordinate system based on which surfaces become simplest.
• Confusing the θ and φ conventions. In mathematics, φ is the polar angle (from the z-axis) and θ is the azimuthal angle (in the xy-plane). In physics, these are often swapped. When working problems from different sources, always check which convention is being used — blindly applying formulas from a different convention is a guaranteed error.

Next Steps

Move on to 07-03 — Change of Variables and Jacobians to learn the general theory of coordinate transformations, including the general change-of-variables formula for multiple integrals and how to compute and interpret Jacobian determinants for arbitrary transformations.