§ 01Why Not Just Use Cartesian Coordinates?
Triple integrals in Cartesian form look like ∭ f(x,y,z) dx dy dz. For a box, that is fine. For a sphere, a cone, or a cylinder, the limits of integration become a nightmare of square roots and nested conditions.
Consider computing the volume of the unit sphere x² + y² + z² = 1 in Cartesian coordinates. The outer integral runs from z = −1 to z = 1; the middle from y = −√(1−z²) to y = √(1−z²); the inner from x = −√(1−y²−z²) to x = √(1−y²−z²). Each bound depends on the others, and the integrand f = 1 still requires evaluating three nested square-root expressions.
In spherical coordinates, the same volume is simply ∫₀²π ∫₀π ∫₀¹ ρ²sin φ dρ dφ dθ, constant limits, clean integrand, trivial to evaluate.
§ 02Cylindrical Coordinates (r, θ, z)
Cylindrical coordinates are the most natural extension of 2D polar coordinates into three dimensions. You simply keep the z-axis as in Cartesian coordinates, but describe the position in the x-y plane using a radial distance r and angle θ.
Conversion Formulas: Cylindrical ↔ Cartesian
y = r sin θ
z = z
These are identical to 2D polar, with z passing through unchanged.
θ = arctan(y/x)
z = z
Use the correct quadrant for θ. When x = 0 and y > 0, θ = π/2; when y = 0 and x > 0, θ = 0.
This is the single most useful identity when switching to cylindrical coordinates. Whenever you see x² + y² in a Cartesian integral, replace it immediately with r².
Standard Range of Variables
For a full sweep in cylindrical coordinates: r ≥ 0, 0 ≤ θ ≤ 2π, and z is unrestricted. The variable r must always be non-negative: if a problem yields negative r, something has gone wrong in the setup.
The crucial factor is the extra r: this is the Jacobian of the transformation. Never forget it. The order of dz, dr, dθ can be rearranged to suit the limits, but r must always appear.
§ 03Spherical Coordinates (ρ, θ, φ)
Spherical coordinates describe a point by its distance from the origin ρ (rho), the azimuthal angle θ (the same θ as in cylindrical: rotation in the xy-plane), and the polar angle φ (phi) measured downward from the positive z-axis.
Conversion Formulas: Spherical ↔ Cartesian
y = ρ sinφ sinθ
z = ρ cosφ
Note: the xy-radius (distance from z-axis) is ρ sinφ, not ρ itself. Apply the usual 2D polar formulas to that radius to get x and y.
θ = arctan(y/x)
φ = arccos(z/ρ)
ρ is the full 3D distance from the origin (always ≥ 0). θ comes from the projection onto the xy-plane. φ is measured from the positive z-axis.
This is the spherical equivalent of r² = x² + y². Any time you see the sum of three squares, spherical coordinates will simplify things dramatically.
The Jacobian factor is ρ² sinφ, two factors now, not just one. Neither ρ² nor sinφ can ever be omitted. Note also that sinφ ≥ 0 for φ ∈ [0, π], which is why no absolute value is needed.
Standard Limits of Integration
For a full sphere of radius R, the standard limits are: 0 ≤ ρ ≤ R, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π. For an upper hemisphere only, restrict φ ∈ [0, π/2].
§ 04Jacobians: Where dV Comes From
When you change variables in a triple integral, the volume element dV transforms by a factor called the Jacobian determinant. For a transformation (x, y, z) → (u, v, w), the Jacobian is the determinant of the matrix of partial derivatives ∂(x,y,z)/∂(u,v,w).
Jacobian for Cylindrical Coordinates
With x = r cosθ, y = r sinθ, z = z, the Jacobian matrix is:
Expanding: det = 1 · (cosθ · r cosθ − (−r sinθ) · sinθ) = r cos²θ + r sin²θ = r. Hence dV = r dr dθ dz.
Jacobian for Spherical Coordinates
The full 3×3 Jacobian for spherical coordinates has determinant ρ² sinφ. The derivation is lengthier but the result is fundamental:
Since φ ∈ [0, π], we have sinφ ≥ 0 throughout the domain, so |J| = J. The volume element dV = ρ² sinφ dρ dφ dθ is always positive, as required for an area element.
Forgetting the Jacobian factor is the single most common error in coordinate-change problems.
§ 05Which System to Use and When
| Feature / Geometry | Use Cylindrical | Use Spherical | Use Cartesian |
|---|---|---|---|
| Defining surface | Cylinder x²+y²=r², cone z=√(x²+y²) | Sphere x²+y²+z²=R² | Plane, box, tetrahedron |
| Volume element | r dr dθ dz | ρ² sinφ dρ dφ dθ | dx dy dz |
| Key identity | x²+y² = r² | x²+y²+z² = ρ² | No simplification |
| Typical limits | 0≤r≤R, 0≤θ≤2π, a≤z≤b | 0≤ρ≤R, 0≤θ≤2π, 0≤φ≤π | Nested, often complex |
| Best for | Cylinders, cones, radial in xy-plane | Spheres, hemispheres, cones from origin | Rectangular regions, planes |
| Number of Jacobian factors | 1 (just r) | 2 (ρ² and sinφ) | 1 (factor is 1) |
The Decision Flowchart
§ 06Worked Examples — Cylindrical Coordinates
The region is a solid cylinder of radius 2 and height 5. In cylindrical coordinates: 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 5.
V = ∫₀²π ∫₀² ∫₀⁵ r dz dr dθSo V = ∫₀²π ∫₀² 5r dr dθ
Confirmed by πr²h = π(4)(5) = 20π ✓
x² + y² = r² in cylindrical coordinates. The integrand becomes r². Limits: 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π, −2 ≤ z ≤ 2.
∭ r² · r dz dr dθ = ∫₀²π ∫₀³ ∫₋₂² r³ dz dr dθThe region is above the cone and below the plane z = 4. In cylindrical: z = r (since z = √(x²+y²) = r). The cone meets z = 4 at r = 4. For fixed r and θ, z runs from r (the cone) up to 4.
V = ∫₀²π ∫₀⁴ ∫ᵣ⁴ r dz dr dθx² + y² ≤ 4 (circle of radius 2 in xy-plane), and 0 ≤ z ≤ √(4−x²−y²). The upper limit on z is √(4−r²), the upper hemisphere of a sphere of radius 2.
Upper hemisphere of radius 3: 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ √(9−r²). Density δ = r (heavier near the outer edge).
M = ∫₀²π ∫₀³ ∫₀^√(9−r²) r · r dz dr dθ = ∫₀²π ∫₀³ ∫₀^√(9−r²) r² dz dr dθLet u = 9 − r², du = −2r dr, so r dr = −du/2. When r = 0, u = 9; when r = 3, u = 0.
∫₀³ r² √(9−r²) dr = ∫₀³ r · r √(9−r²) drUse integration: ∫₀³ r²√(9−r²) dr = (9/4)(π·9/2) ... via trig sub: let r = 3 sinα, dr = 3 cosα dα, √(9−r²) = 3cosα:
= ∫₀^(π/2) 9sin²α · 3cosα · 3cosα dα = 81 ∫₀^(π/2) sin²α cos²α dα = 81 · π/16 = 81π/16§ 07Worked Examples — Spherical Coordinates
The ball B: ρ ≤ R, with full ranges 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π.
V = ∫₀²π ∫₀π ∫₀ᴿ ρ² sinφ dρ dφ dθThe classical formula, derived in two lines.
x² + y² + z² = ρ², so (x²+y²+z²)^(3/2) = ρ³. The integrand becomes e^(ρ³). This is impossible in Cartesian but trivial in spherical.
∭ e^(ρ³) · ρ² sinφ dρ dφ dθLimits: 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.
The integrand factors completely: ρ² e^(ρ³) depends only on ρ; sinφ depends only on φ; nothing depends on θ.
[∫₀¹ ρ² e^(ρ³) dρ] · [∫₀π sinφ dφ] · [∫₀²π dθ]Let u = ρ³, du = 3ρ² dρ:
∫₀¹ ρ² e^(ρ³) dρ = (1/3)∫₀¹ eᵘ du = (e−1)/3Above the cone means smaller φ (closer to the z-axis). The region: 0 ≤ ρ ≤ 2, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/3.
V = ∫₀²π ∫₀^(π/3) ∫₀² ρ² sinφ dρ dφ dθz = ρ cosφ. Upper hemisphere: 0 ≤ ρ ≤ a, 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π.
∫₀²π ∫₀^(π/2) ∫₀ᵃ (ρ cosφ) · ρ² sinφ dρ dφ dθ = ∫₀²π ∫₀^(π/2) ∫₀ᵃ ρ³ sinφ cosφ dρ dφ dθUse the identity sinφ cosφ = sin(2φ)/2:
∫₀^(π/2) sinφ cosφ dφ = [sin²φ/2]₀^(π/2) = 1/2A spherical shell: 1 ≤ ρ ≤ 2, full sphere (0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π).
I = ∫₀²π ∫₀π ∫₁² (1/ρ⁴) · ρ² sinφ dρ dφ dθ = ∫₀²π ∫₀π ∫₁² ρ⁻² sinφ dρ dφ dθ§ 08Common Surfaces in Each Coordinate System
Knowing how standard surfaces look in each coordinate system saves enormous time in setting up limits.
The Cone φ = π/4 vs. the Cone z = r
These are the same cone. In spherical: φ = π/4 means the angle from the z-axis is 45°, so z/ρ = cosφ = 1/√2 and the xy-radius over ρ = sinφ = 1/√2, giving z = xy-radius, which in cylindrical means z = r. Always confirm your surfaces using both descriptions.
§ 09Common Mistakes
§ 1010-Question Quiz
Test your understanding of cylindrical and spherical coordinates. Each question gives immediate feedback.