Change of Variables in Multiple Integrals: Polar, Cylindrical and Spherical Coordinates

A change of variables in a multiple integral replaces the area or volume element by |J| du dv or |J| du dv dw, where J is the Jacobian determinant of the transformation; the standard cases are polar (dA = r dr dθ), cylindrical (dV = r dr dθ dz) and spherical (dV = ρ² sin φ dρ dφ dθ) coordinates. In brief: the area of a disc of radius R in polar coordinates is ∫₀^2π ∫₀^R r dr dθ = πR², and the volume of a ball in spherical coordinates is ∫₀^2π ∫₀^π ∫₀^R ρ² sin φ dρ dφ dθ = 4πR³/3.

In one variable, the substitution rule multiplies by the derivative: dx = g′(t) dt. In several variables, the derivative becomes a matrix and the multiplying factor is its determinant. This guide explains the Jacobian, then polar, cylindrical and spherical coordinates, with worked examples, the Gaussian integral and a table of coordinate systems.

What is the Jacobian and the general rule?

Let (x, y) = (x(u, v), y(u, v)) map a region S in the uv-plane one-to-one onto D. The Jacobian is the determinant J = ∂(x, y)/∂(u, v) = x_u y_v − x_v y_u. Then ∬_D f(x, y) dx dy = ∬_S f(x(u, v), y(u, v)) |J| du dv. For three variables, J is a 3 × 3 determinant.

The absolute value of the determinant measures how much the map stretches areas locally. It is the multi-dimensional analogue of |g′(t)|.

How do polar coordinates work in double integrals?

With x = r cos θ and y = r sin θ: J = cos θ · r cos θ − (−r sin θ)(sin θ) = r. So dA = r dr dθ. The factor r is essential: a ring of radius r and width dr has the area ≈ 2πr dr, not 2π dr.

Example 1: the area of a disc

∫₀^2π ∫₀^R r dr dθ = 2π · R²/2 = πR².

Example 2: an integral over a disc

∬ (x² + y²) dA over x² + y² ≤ 4: ∫₀^2π ∫₀² r² · r dr dθ = 2π · 4 = 8π.

Example 3: the volume under a paraboloid

The volume under z = 9 − x² − y² above z = 0: the disc has radius 3. V = ∫₀^2π ∫₀³ (9 − r²) r dr dθ = 2π [9r²/2 − r⁴/4]₀³ = 2π (40.5 − 20.25) = 40.5π ≈ 127.2.

Example 4: the Gaussian integral

Let I = ∫_−∞^∞ e^−x2 dx. Then I² = ∬_ℝ² e^−(x2+y2) dA = ∫₀^2π ∫₀^∞ e^−r2 r dr dθ = 2π · 1/2 = π. Hence I = √π, a famous result that is not obtainable by elementary antiderivatives. It is an improper integral.

Example 5: a region that is not centered

The disc x² + y² ≤ 2x has the polar boundary r = 2 cos θ for −π/2 ≤ θ ≤ π/2. Its area: ∫_−π/2^π/2 (1/2)(2 cos θ)² dθ = 2 ∫ cos² θ dθ = 2 · π/2 = π ✓ (a disc of radius 1).

How do cylindrical coordinates work?

Cylindrical coordinates (r, θ, z) are polar coordinates in the xy-plane plus the height z: x = r cos θ, y = r sin θ, z = z. The Jacobian is again r, so dV = r dr dθ dz. They suit solids with an axis of symmetry: cylinders, cones, paraboloids.

Example 6: a cylinder

Radius 2, height 5: V = ∫₀^2π ∫₀² ∫₀⁵ r dz dr dθ = 2π · 2 · 5 = 20π ✓ (πr²h).

Example 7: a solid between a paraboloid and a plane

The solid above z = r² and below z = 4. V = ∫₀^2π ∫₀² (4 − r²) r dr dθ = 2π (8 − 4) = 8π.

Example 8: a cone

The solid between z = r and z = 3 (height 3, radius 3): V = ∫₀^2π ∫₀³ (3 − r) r dr dθ = 2π [1.5r² − r³/3]₀³ = 2π (13.5 − 9) = 9π ✓ (π r²h/3 = 9π).

How do spherical coordinates work?

Spherical coordinates (ρ, φ, θ): ρ is the distance from the origin, φ ∈ [0, π] the angle from the positive z-axis and θ ∈ [0, 2π) the azimuth. Then x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ, and the Jacobian is ρ² sin φ, so dV = ρ² sin φ dρ dφ dθ. Note that conventions differ between textbooks (physicists often swap the roles of θ and φ).

Example 9: the volume of a ball

V = ∫₀^2π dθ ∫₀^π sin φ dφ ∫₀^R ρ² dρ = 2π · 2 · R³/3 = 4πR³/3.

Example 10: an integral over a ball

∭ (x² + y² + z²) dV over the unit ball = ∫ ρ² · ρ² sin φ dρ dφ dθ = 2π · 2 · 1/5 = 4π/5.

Example 11: an ice-cream cone

The region inside the sphere ρ = 2 and above the cone φ = π/4 (0 ≤ φ ≤ π/4): V = 2π ∫₀^π/4 sin φ dφ ∫₀² ρ² dρ = 2π (1 − √2/2)(8/3) = (16π/3)(1 − √2/2) ≈ 4.91.

How do you use a custom substitution?

Choose new variables that make the region a rectangle or simplify the integrand.

Example 12: a parallelogram

Compute ∬ (x + y) dA over the parallelogram bounded by x + y = 1, x + y = 3, x − y = 0, x − y = 2. Let u = x + y, v = x − y, so x = (u + v)/2, y = (u − v)/2 and J = (1/2)(−1/2) − (1/2)(1/2) = −1/2, |J| = 1/2. The region is 1 ≤ u ≤ 3, 0 ≤ v ≤ 2. The integral: ∫₁³ ∫₀² u · (1/2) dv du = (1/2)(2)(4) = 4.

Example 13: the area of an ellipse

For x²/a² + y²/b² ≤ 1, let x = a u, y = b v with J = ab. The region becomes the unit disc, so the area is ab · π = πab.

Table of coordinate systems

Where do these coordinates come from?

Euler used a change of variables in double integrals in the 1760s, and the determinant form for general transformations was developed in the following decades by Lagrange (for three variables, in 1773) and later systematized by Carl Gustav Jacob Jacobi, who studied functional determinants in the 1840s and after whom the Jacobian is named. Polar coordinates were used by several mathematicians in the seventeenth century, with Newton and Jakob Bernoulli often credited with early systematic use.

Which mistakes are common?

Forgetting the factor r or ρ² sin φ

Writing dr dθ instead of r dr dθ is the most common error. The element must contain the Jacobian.

Negative Jacobians

Use the absolute value |J|. A negative determinant only means that orientation is reversed.

Wrong angle limits

For a disc not centered at the origin, the limits of r may depend on θ, and the range of θ may be smaller than 2π.

Mixing spherical conventions

Check which angle is measured from the z-axis. Using φ as the azimuth with the wrong factor gives wrong results.