Double Integrals in Polar Coordinates — Mathematics Knowledge System

Overview

Some regions — disks, annuli, circular sectors, cardioid-shaped regions — are far more naturally described using polar coordinates (r, θ) than Cartesian (x, y). Converting a double integral over such a region into polar coordinates replaces awkward Cartesian bounds (often involving square roots) with clean bounds on r and θ, at the cost of an extra factor of r appearing in the area element: dA = r dr dθ. This factor is not optional bookkeeping — it is the Jacobian of the polar transformation, accounting for the fact that a small 'rectangle' of sides dr and dθ in polar coordinates actually sweeps out a physical area of r dr dθ, growing with distance from the origin.

Intuition

In Cartesian coordinates, a small grid cell of width dx and height dy has area exactly dx·dy everywhere, since Cartesian grid lines are straight and evenly spaced. In polar coordinates, the 'grid' is made of concentric circles (constant r) and rays from the origin (constant θ). A small polar 'cell' between r and r+dr, and between θ and θ+dθ, is approximately a thin rectangle with one side of length dr (radial) and the other side of length r dθ (arc length, since arc length = radius × angle) — NOT dθ. So its area is dr · (r dθ) = r dr dθ. Far from the origin, a fixed angular slice dθ sweeps out much more physical area than the same dθ close to the origin — exactly the extra factor of r.

Formal Definition

Definition

For a region D described in polar coordinates as α ≤ θ ≤ β, h₁(θ) ≤ r ≤ h₂(θ), the double integral converts as:

\iint_D f(x,y)\, dA = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta, r\sin\theta)\, r\, dr\, d\theta

Polar double integral

dA = r\, dr\, d\theta

Polar area element

Notation

Notation	Meaning
$dA = r\, dr\, d\theta$	Area element in polar coordinates, including the Jacobian factor r
$D' = \{(r,\theta) : \alpha\le\theta\le\beta,\ h_1(\theta)\le r \le h_2(\theta)\}$	The region D rewritten with polar bounds

Derivation

The r factor is the Jacobian determinant of the transformation x=r cosθ, y=r sinθ, computed as a 2×2 determinant of partial derivatives:

\frac{\partial(x,y)}{\partial(r,\theta)} = \det\begin{pmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\end{pmatrix} = r\cos^2\theta + r\sin^2\theta = r

Expand the determinant using sin²θ+cos²θ=1

dA = dx\, dy = \left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| dr\, d\theta = r\, dr\, d\theta \quad (r \ge 0)

The change-of-variables theorem multiplies by the absolute value of the Jacobian

Properties

Full-disk area check

\iint_{r\le R} 1\, dA = \int_0^{2\pi}\int_0^R r\, dr\, d\theta = \pi R^2 \text{ (recovers the familiar disk-area formula)}

Best suited to circular symmetry

\text{Polar conversion helps most when } D \text{ is a disk, sector, annulus, or when the integrand contains } x^2+y^2

Theorems

Theorem 1: Polar Change of Variables

\text{If } f \text{ is continuous on a polar region } D \text{, then } \iint_D f\, dA = \int_\alpha^\beta\int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,r\sin\theta)\, r\, dr\, d\theta

Applications

The Gaussian integral ∫e^(−x²−y²)dA over the whole plane is evaluated by converting to polar coordinates, a technique central to probability theory and statistical mechanics.

3D Visualization

Worked Examples

Convert to polar: x²+y² = r², and D becomes 0≤r≤2, 0≤θ≤2π.
$\iint_D (x^2+y^2)\, dA = \int_0^{2\pi}\int_0^2 r^2\cdot r\, dr\, d\theta = \int_0^{2\pi}\int_0^2 r^3\, dr\, d\theta$
Integrate with respect to r: ∫₀² r³ dr = [r⁴/4]₀² = 16/4 = 4.
$\int_0^2 r^3\, dr = 4$
Integrate the constant 4 over θ from 0 to 2π.
$\int_0^{2\pi} 4\, d\theta = 8\pi$

Answer: 8π

Practice Problems

Difficulty 5/10

Evaluate ∬_D 1 dA where D is the disk r ≤ 3 (i.e., find its area using a polar double integral).

Difficulty 6/10

Evaluate ∬_D xy dA where D is the quarter disk in the first quadrant with r ≤ 2 (0≤θ≤π/2).

Difficulty 7/10

A circular pond of radius 4 m has depth given by d(r) = 4 − r (in meters, with r the distance from the center). Find the total volume of water using a polar double integral.

Common Mistakes

Common Mistake

Forgetting the extra factor of r and writing dA = dr dθ.

dA = r dr dθ, NOT dr dθ — the Jacobian of the polar transformation contributes this extra r, and omitting it is the single most common error when switching to polar coordinates.

Common Mistake

Using the wrong bounds for r when the region is bounded by a curve like r = f(θ) instead of a full disk.

Sketch the region first: r often ranges from 0 (or an inner boundary curve) up to an outer boundary r = f(θ) that DEPENDS on θ, not a fixed constant — this is exactly analogous to variable inner bounds in Cartesian iterated integrals.

Common Mistake

Applying polar conversion to a region with no circular symmetry, making the bounds more complicated rather than simpler.

Polar coordinates only simplify integrals over disks, sectors, annuli, or when the integrand naturally involves x²+y²; for a rectangular or triangular region, Cartesian coordinates usually remain easier.

Quiz

Flashcards

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Summary

Converting a double integral to polar coordinates replaces dA with r dr dθ, picking up the Jacobian factor r.
Polar coordinates simplify integrals over circularly symmetric regions (disks, sectors, annuli) and integrands involving x²+y².
Bounds for r are often functions of θ (r from an inner curve to r=f(θ)); sketching the region first avoids setup errors.
A classic application is evaluating the Gaussian integral ∫e^{-x²}dx via polar coordinates over the whole plane.
The most common mistake is forgetting the extra factor of r in the area element.

References

BookStewart, J. Calculus: Early Transcendentals, 8th ed. Ch. 15.4.
WebsiteWikipedia — Multiple integral (Polar coordinates)