Math Core

Try to integrate over a disk in rectangular coordinates and you get limits like $-\sqrt{4 - x^2} \le y \le \sqrt{4 - x^2}$ and integrands full of square roots. Disks, rings, wedges and petal shapes are described far more naturally with $r$ and $\theta$ . Switching to polar coordinates often turns an ugly double integral into two easy single ones.

Polar rectangles

Recall $x = r\cos\theta$ , $y = r\sin\theta$ and $x^2 + y^2 = r^2$ . The polar analogue of a rectangle is a polar rectangle

R = \{(r, \theta) : a \le r \le b,\ \alpha \le \theta \le \beta\},

a piece of a ring cut out by two rays. A disk of radius $2$ is $0 \le r \le 2$ , $0 \le \theta \le 2\pi$ . The ring between radii $1$ and $2$ is $1 \le r \le 2$ , $0 \le \theta \le 2\pi$ .

The annulus 1 ≤ r ≤ 2. In rectangular coordinates it needs several pieces; in polar it is a single rectangle in r and θ.Open in grapher →

Why the extra factor of r

Slice a polar rectangle with a grid of circles (spacing $\Delta r$ ) and rays (spacing $\Delta \theta$ ). A small cell at radius $r$ is almost a rectangle: one side has length $\Delta r$ , and the other is an arc of length $r\,\Delta\theta$ . So its area is about

\Delta A \approx r\,\Delta r\,\Delta\theta.

Cells far from the origin are bigger than cells near it, even though they have the same $\Delta r$ and $\Delta\theta$ . The factor $r$ accounts for exactly that stretching.

Double integrals in polar coordinates

If $f$ is continuous on a polar region $D = \{(r, \theta) : \alpha \le \theta \le \beta,\ h_1(\theta) \le r \le h_2(\theta)\}$ with $0 \le \beta - \alpha \le 2\pi$ , then

\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.

In short: replace $x$ and $y$ by $r\cos\theta$ and $r\sin\theta$ , and replace $dA$ by $r\,dr\,d\theta$ .

Common mistake

The most common error is forgetting the $r$ in $dA = r\,dr\,d\theta$ . A quick check: the area of a disk of radius $a$ is $\int_0^{2\pi}\int_0^a r\,dr\,d\theta = \pi a^2$ . Without the $r$ you would get $2\pi a$ , which is not even an area.

Integrals over disks and rings

Worked example: A radially symmetric integrand

Evaluate $\iint_D (x^2 + y^2)\,dA$ , where $D$ is the disk $x^2 + y^2 \le 4$ .

Solution. In polar, $x^2 + y^2 = r^2$ and $D$ is $0 \le r \le 2$ , $0 \le \theta \le 2\pi$ :

\int_0^{2\pi} \int_0^2 r^2 \cdot r\,dr\,d\theta = \int_0^{2\pi} d\theta \cdot \int_0^2 r^3\,dr = 2\pi \cdot 4 = 8\pi.

Worked example: Volume under a paraboloid

Find the volume of the solid under the paraboloid $z = 4 - x^2 - y^2$ and above the $xy$ -plane.

Solution. The paraboloid meets the plane $z = 0$ on the circle $x^2 + y^2 = 4$ , so the base is the disk $r \le 2$ and the height is $4 - r^2$ :

V = \int_0^{2\pi} \int_0^2 (4 - r^2)\,r\,dr\,d\theta = 2\pi\left[2r^2 - \frac{r^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi.

One famous use of polar coordinates is the Gaussian integral. Over the disk of radius $a$ ,

\iint_{r \le a} e^{-(x^2 + y^2)}\,dA = \int_0^{2\pi}\int_0^a e^{-r^2}\,r\,dr\,d\theta = \pi\left(1 - e^{-a^2}\right).

Letting $a \to \infty$ gives $\pi$ over the whole plane. That whole-plane integral also equals $\left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right)^2$ , so $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ : a single-variable fact that is very hard to reach any other way.

Regions bounded by polar curves

When the outer boundary is a polar curve $r = h(\theta)$ , the inner limits become $0 \le r \le h(\theta)$ . Integrating $1$ over such a region recovers the polar area formula from Calculus II:

A = \int_{\alpha}^{\beta} \int_0^{h(\theta)} r\,dr\,d\theta = \int_{\alpha}^{\beta} \frac{1}{2}h(\theta)^2\,d\theta.

The region inside the cardioid r = 1 + cos(θ). For each angle θ, r runs from 0 out to the curve.Open in grapher →

Worked example: Area inside a cardioid

Find the area of the region inside $r = 1 + \cos\theta$ .

Solution. The curve is traced once for $0 \le \theta \le 2\pi$ :

\begin{aligned} A &= \int_0^{2\pi} \int_0^{1 + \cos\theta} r\,dr\,d\theta = \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \cos^2\theta\right)d\theta \\ &= \frac{1}{2}\left(2\pi + 0 + \pi\right) = \frac{3\pi}{2}. \end{aligned}

Here $\int_0^{2\pi}\cos^2\theta\,d\theta = \pi$ , using $\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)$ .

Converting an iterated integral to polar

Given a rectangular iterated integral, first sketch the region from its limits, then describe that region in polar coordinates.

Worked example: From rectangular to polar

Evaluate $\displaystyle\int_0^{2} \int_0^{\sqrt{4 - x^2}} \frac{1}{1 + x^2 + y^2}\,dy\,dx$ .

Solution. The limits describe $0 \le x \le 2$ and $0 \le y \le \sqrt{4 - x^2}$ : the quarter disk of radius $2$ in the first quadrant. In polar that is $0 \le r \le 2$ , $0 \le \theta \le \dfrac{\pi}{2}$ :

\int_0^{\pi/2} \int_0^2 \frac{r}{1 + r^2}\,dr\,d\theta = \frac{\pi}{2} \cdot \frac{1}{2}\ln(1 + r^2)\Big|_0^2 = \frac{\pi \ln 5}{4}.

Tip

Switch to polar when the region is a disk, ring, sector or polar curve, or when the integrand involves $x^2 + y^2$ . Both conditions together are a strong signal.

Practice

Practice 1

Evaluate $\iint_D \sqrt{x^2 + y^2}\,dA$ , where $D$ is the disk of radius $3$ centered at the origin.