Polar Coordinates in Double Integrals

Polar cells widen away from the origin

Use $x=r\cos\theta$ , $y=r\sin\theta$ with $r\ge0$ . A narrow cell has radial thickness $dr$ and approximate arc length $r\,d\theta$ , so its area is $dA=r\,dr\,d\theta$ . This extra factor is geometric; substituting only into the integrand misses it.

Choose angular and radial bounds describing the intended region once, apart from boundary overlaps of area zero. A full disk uses $0\le\theta\le2\pi$ , $0\le r\le R$ . An annulus replaces zero by its inner radius; a sector restricts the angles.

Visual guide

VISUAL GUIDEThe angular width grows with radius

Rays at fixed angles

A polar patch has radial thickness dr and tangential width approximately r·dθ. The two shaded patches use the same angular span and radial thickness, but the outer patch has more area. This explains the factor r in dA = r dr dθ.

Worked example: a radial moment

Over the disk $x^2+y^2\le4$ ,

\iint_R(x^2+y^2)dA=\int_0^{2\pi}\int_0^2r^2r\,dr\,d\theta=8\pi.

The integrand contributes $r^2$ and the area element contributes another $r$ . Since the disk area is $4\pi$ , the average squared distance from the center is $2$ .

Worked example: a shifted circular boundary

The disk inside $x^2+y^2\le2x$ becomes $r^2\le2r\cos\theta$ . For $r>0$ , this gives $0\le r\le2\cos\theta$ , requiring $-\pi/2\le\theta\le\pi/2$ . Its area is

\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}r\,dr\,d\theta=\pi.

The region is a unit disk centered at $(1,0)$ , so ordinary geometry checks the result. Dividing by $r$ during the derivation should not make you discard the origin, which belongs to the boundary and has zero area anyway.

Polar coordinates simplify circular symmetry, but not every region benefits. A rectangle often gives awkward radial bounds. Select coordinates to simplify both the region and the integrand rather than following a formula automatically.

Practice

Find the area of the annulus $1\le x^2+y^2\le9$ .
Integrate $1$ over the first-quadrant unit disk.
What is missing from $\int_0^{2\pi}\int_0^R f(r\cos\theta,r\sin\theta)drd\theta$ as a planar area integral?

Show worked solutions

$\int_0^{2\pi}\int_1^3r drd\theta=8\pi$ .
Bounds $0\le r\le1$ , $0\le\theta\le\pi/2$ give $\pi/4$ .
The Jacobian factor $r$ in the area element.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.