Double Integrals in Polar Coordinates

In this lesson you’ll learn how to convert double integrals from Cartesian (x, y) to polar (r, θ) coordinates. This makes many integrals over circular regions dramatically simpler.

Some regions are naturally circular: disks, annuli (rings), sectors (pie slices). Describing these in Cartesian coordinates requires square roots and messy limits. In polar coordinates, a disk is just “r from 0 to R, θ from 0 to 2π.” Much cleaner.

In polar coordinates, x = r cos θ and y = r sin θ. The key substitution for the area element is

$$dA = r \, dr \, d\theta$$

That extra r is critical. It comes from the Jacobian of the coordinate transformation and accounts for the fact that polar “rectangles” (wedge-shaped pieces) get wider as r increases. A double integral becomes

$$\iint_R f(x, y) \, dA = \int_{\alpha}^{\beta} \int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta) \cdot r \, dr \, d\theta$$

The single most common reason to switch to polar is when the integrand contains x² + y², because

$$x^2 + y^2 = r^2$$

This turns complicated expressions into simple powers of r.

unit disk (r ≤ 1) area element r dr dθ

The green line shows r (distance from origin), the purple arc shows θ (angle from the x-axis), and the orange wedge is a tiny area element dA = r dr dθ. Notice how the wedge gets wider at larger r, which is why the extra r factor is needed.

Example 1: Area of the unit disk

Compute the area of the disk x² + y² ≤ 1 using polar coordinates.

In polar, the disk is 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. The integrand for area is f = 1

$$\text{Area} = \int_0^{2\pi} \int_0^1 r \, dr \, d\theta = \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_0^1 d\theta = \int_0^{2\pi} \frac{1}{2} \, d\theta = \pi$$

Exactly what we expect: the area of a circle with radius 1 is π.

Example 2: Integrating x² + y² over the unit disk

Evaluate the double integral of f(x, y) = x² + y² over the unit disk.

In polar, x² + y² = r², so

$$\int_0^{2\pi} \int_0^1 r^2 \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^1 r^3 \, dr \, d\theta$$ $$= \int_0^{2\pi} \left[ \frac{r^4}{4} \right]_0^1 d\theta = \int_0^{2\pi} \frac{1}{4} \, d\theta = \frac{\pi}{2}$$

Try setting this up in Cartesian coordinates and you’ll see why polar is so much nicer here.

Drag to rotate · Scroll/Pinch to zoom

The bowl-shaped surface is z = x² + y² = r², rising from 0 at the center to 1 at the edge of the green unit disk. The volume under this paraboloid over the disk is π/2, which is what the polar integral computed.

Example 3: Quarter-circle region

Evaluate the double integral of f(x, y) = 1 over the quarter-disk in the first quadrant (x ≥ 0, y ≥ 0, x² + y² ≤ 4).

In polar: r from 0 to 2, θ from 0 to π/2

$$\int_0^{\pi/2} \int_0^2 r \, dr \, d\theta = \int_0^{\pi/2} \left[ \frac{r^2}{2} \right]_0^2 d\theta = \int_0^{\pi/2} 2 \, d\theta = \pi$$

This is the area of a quarter of a circle with radius 2: (1/4)(π)(4) = π. Checks out.

r: 0 → 2 θ: 0 → π/2 quarter-disk, first quadrant

The shaded region is the quarter-disk in the first quadrant with radius 2. In polar coordinates, r goes from 0 to 2 and θ goes from 0 to π/2. The area is π.

Polar double integrals show up whenever the geometry is circular:

• Game engines use them for circular area-of-effect calculations (explosions, healing auras, detection radii)
• Physics uses them for mass and moment of inertia of circular plates, disks, and cylinders
• Engineering uses them for stress analysis on circular cross-sections and heat flow through pipes
• Computer graphics uses them for radial blur effects and circular sampling patterns