Double Integrals in Polar Coordinates, multivariable and vector calculus lesson

A round pizza is awkward to cut into a grid of little squares, yet easy to cut into slices and rings. Disks, rings and wedges are just as awkward in $x$ and $y$ , where every bound turns into a square root. In polar coordinates the same regions have constant bounds. The price is one extra factor, and getting that factor right is most of the lesson.

Key terms

Polar coordinates and their range

Definition Polar coordinates $(r, \theta)$ describe the point

x = r\cos\theta, \qquad y = r\sin\theta,

where $r \ge 0$ is the distance to the origin and $\theta$ is an angle in radians measured counterclockwise from the positive $x$ axis. Then $r^2 = x^2 + y^2$ . To give each point other than the origin exactly one pair, restrict $\theta$ to a half open interval of length $2\pi$ , such as $0 \le \theta < 2\pi$ or $-\pi < \theta \le \pi$ . The origin has $r = 0$ and any $\theta$ .

In plain words The radius says how far out, the angle says which direction. Keeping $r$ nonnegative and the angle inside one full turn stops the same point from being named twice.

Example The point $(-1, -1)$ has $r = \sqrt{2}$ . It lies in the third quadrant, so $\theta = 5\pi/4$ in the range $[0, 2\pi)$ , or $\theta = -3\pi/4$ in the range $(-\pi, \pi]$ . Note that $\arctan(y/x) = \arctan 1 = \pi/4$ points to the wrong quadrant.

Sector area scaling

Definition A polar rectangle is a set

\begin{aligned} S = \{ (r, \theta) \mid{} & r_1 \le r \le r_2, \\ & \theta_1 \le \theta \le \theta_2 \} \end{aligned}

with $0 \le r_1 \le r_2$ and $0 \le \theta_2 - \theta_1 \le 2\pi$ . Since a sector of radius $r$ and angle $\Delta\theta$ has area $\tfrac{1}{2} r^2 \Delta\theta$ , the area of $S$ is

\begin{aligned} \Delta A &= \tfrac{1}{2}(r_2^2 - r_1^2)\,\Delta\theta \\[0.4em] &= r^* \, \Delta r \, \Delta\theta, \end{aligned}

where $\Delta r = r_2 - r_1$ , $\Delta\theta = \theta_2 - \theta_1$ and $r^* = \tfrac{1}{2}(r_1 + r_2)$ is the average radius.

In plain words A curved patch of a ring is almost a rectangle with sides $\Delta r$ and $r \, \Delta\theta$ . The farther out the patch sits, the longer its arc, so the area grows in proportion to $r$ .

Example For $1 \le r \le 3$ and $0 \le \theta \le \pi/6$ , the sector formula gives

\frac{1}{2}(9 - 1) \cdot \frac{\pi}{6} = \frac{2\pi}{3},

and the average radius formula gives $2 \cdot 2 \cdot \pi/6 = 2\pi/3$ as well. The same angle and radial width at $5 \le r \le 7$ gives an area three times larger, since the average radius is $6$ instead of $2$ .

The polar area element

Definition Let $f$ be continuous on the polar rectangle $R$ given by $0 \le a \le r \le b$ and $\alpha \le \theta \le \beta$ , with $0 \le \beta - \alpha \le 2\pi$ . Then

\iint_R f \, dA = \int_\alpha^\beta \int_a^b f \, r \, dr \, d\theta,

where $f$ on the right is evaluated at $(r\cos\theta, r\sin\theta)$ .

In short, $dA = r \, dr \, d\theta$ . The factor $r$ is the absolute value of the Jacobian of the map $(r, \theta) \mapsto (r\cos\theta, r\sin\theta)$ , which is one-to-one on the region apart from a boundary set of zero area.

In plain words Replace $x$ and $y$ by their polar forms, and never forget the extra $r$ . It is the stretching factor from the sector area above.

Example For the disk $x^2 + y^2 \le 4$ ,

\begin{aligned} &\iint e^{-(x^2 + y^2)} \, dA \\[0.4em] &\quad = \int_0^{2\pi} \int_0^2 e^{-r^2} \, r \, dr \, d\theta \\[0.4em] &\quad = 2\pi \cdot \frac{1 - e^{-4}}{2} = \pi \left( 1 - e^{-4} \right). \end{aligned}

Without the $r$ , the inner integral $\int_0^2 e^{-r^2} \, dr$ has no elementary antiderivative.

Polar regions

Definition A polar region of the general kind is

\begin{aligned} D = \{ (r, \theta) \mid{} & \alpha \le \theta \le \beta, \\ & h_1(\theta) \le r \le h_2(\theta) \}, \end{aligned}

where $0 \le \beta - \alpha \le 2\pi$ and $h_1$ , $h_2$ are continuous on $[\alpha, \beta]$ with $0 \le h_1 \le h_2$ . If $f$ is continuous on $D$ , then

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

with $f$ evaluated at $(r\cos\theta, r\sin\theta)$ .

In plain words For each direction $\theta$ , a ray from the origin enters the region at $r = h_1(\theta)$ and leaves at $r = h_2(\theta)$ . That ray plays the part of the vertical strip from rectangular coordinates.

Example Take the upper half of the ring $1 \le x^2 + y^2 \le 4$ , so $0 \le \theta \le \pi$ and $1 \le r \le 2$ . Then

\begin{aligned} \iint_D y \, dA &= \int_0^\pi \int_1^2 (r\sin\theta) \, r \, dr \, d\theta \\[0.4em] &= \int_0^\pi \sin\theta \, d\theta \cdot \int_1^2 r^2 \, dr \\[0.4em] &= 2 \cdot \frac{7}{3} = \frac{14}{3}. \end{aligned}

Bounds from the geometry

Definition To set up polar bounds, convert each boundary curve to polar form using $x = r\cos\theta$ , $y = r\sin\theta$ and $x^2 + y^2 = r^2$ , and keep only the branch with $r \ge 0$ . Then find the angles of the rays that touch the region, which give $\alpha$ and $\beta$ , and for each such angle read off where the ray enters and leaves. Intersection angles come from setting the two polar equations equal.

In plain words Stand at the origin, sweep a ray through every direction that hits the region, and note where the ray goes in and comes out.

Example The circle $x^2 + y^2 = 2x$ has center $(1, 0)$ and radius $1$ . In polar form $r^2 = 2r\cos\theta$ , so $r = 2\cos\theta$ , which is nonnegative for $-\pi/2 \le \theta \le \pi/2$ . The volume under the cone $z = \sqrt{x^2 + y^2}$ above this disk is

\begin{aligned} &\int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} r \cdot r \, dr \, d\theta \\[0.4em] &\quad = \int_{-\pi/2}^{\pi/2} \frac{8}{3}\cos^3\theta \, d\theta \\[0.4em] &\quad = \frac{8}{3} \cdot \frac{4}{3} = \frac{32}{9}. \end{aligned}

For a region between two curves, take the part inside $r = 2\cos\theta$ and outside $r = 1$ . The curves meet where $2\cos\theta = 1$ , at $\theta = \pm\pi/3$ , so its area is

\begin{aligned} &\int_{-\pi/3}^{\pi/3} \int_1^{2\cos\theta} r \, dr \, d\theta \\[0.4em] &\quad = \int_{-\pi/3}^{\pi/3} \frac{4\cos^2\theta - 1}{2} \, d\theta \\[0.4em] &\quad = \frac{\pi}{3} + \frac{\sqrt{3}}{2}. \end{aligned}

Avoiding overlap

Definition The polar formula counts every point of $D$ once only if the map $(r, \theta) \mapsto (r\cos\theta, r\sin\theta)$ is one-to-one on the parameter region, apart from a set of zero area there. Typical exceptions are the edge $r = 0$ , which all goes to the origin, and the two edges $\theta = \alpha$ and $\theta = \beta$ when $\beta - \alpha = 2\pi$ , which go to the same ray. So the angle interval must not exceed length $2\pi$ , every bound must satisfy $0 \le h_1(\theta) \le h_2(\theta)$ , and when a curve $r = h(\theta)$ traces itself more than once, the angle interval must cover only one tracing.

In plain words Running the angle too far paints part of the region twice, and the integral counts it twice. Check how much angle the curve needs to be drawn once.

Example The circle $r = 2\sin\theta$ has center $(0, 1)$ and radius $1$ , and it is traced once as $\theta$ runs from $0$ to $\pi$ . Its area is

\begin{aligned} \int_0^\pi \int_0^{2\sin\theta} r \, dr \, d\theta &= \int_0^\pi 2\sin^2\theta \, d\theta \\[0.4em] &= \pi, \end{aligned}

as it should be. Now let $\theta$ run from $0$ to $2\pi$ by mistake. For $\pi < \theta < 2\pi$ the upper bound $2\sin\theta$ is negative, so there is no $r$ with $0 \le r \le 2\sin\theta$ and those directions contain no point of the disk. Yet the inner integral computed blindly is still $2\sin^2\theta > 0$ , and the total comes out as $2\pi$ . If negative $r$ were allowed, those angles would trace the same circle a second time and sweep the disk again, and that second sweep is exactly what the extra $\pi$ counts.

Check yourself

Use polar coordinates to evaluate the integral of $x^2$ over the unit disk $x^2 + y^2 \le 1$ .

Show the answer

With $x^2 = r^2\cos^2\theta$ and $dA = r \, dr \, d\theta$ ,

\begin{aligned} &\int_0^{2\pi} \int_0^1 r^3 \cos^2\theta \, dr \, d\theta \\[0.4em] &\quad = \frac{1}{4} \int_0^{2\pi} \cos^2\theta \, d\theta = \frac{\pi}{4}. \end{aligned}

Convert to polar coordinates and evaluate

\int_0^2 \int_0^{\sqrt{4 - x^2}} xy \, dy \, dx.

Show the answer

The region is the quarter disk of radius $2$ in the first quadrant, so $0 \le \theta \le \pi/2$ and $0 \le r \le 2$ . The integrand times $r$ is $r^3\cos\theta\sin\theta$ , and

\begin{aligned} &\int_0^{\pi/2} \cos\theta\sin\theta \, d\theta \cdot \int_0^2 r^3 \, dr \\[0.4em] &\quad = \frac{1}{2} \cdot 4 = 2. \end{aligned}

Find the area of one petal of the rose $r = \cos 2\theta$ , and say which angle range to use.

Show the answer

The petal on the positive $x$ axis is traced once for $-\pi/4 \le \theta \le \pi/4$ , where $\cos 2\theta \ge 0$ . Its area is

\begin{aligned} &\int_{-\pi/4}^{\pi/4} \int_0^{\cos 2\theta} r \, dr \, d\theta \\[0.4em] &\quad = \int_{-\pi/4}^{\pi/4} \frac{\cos^2 2\theta}{2} \, d\theta = \frac{\pi}{8}. \end{aligned}

Double Integrals in Polar Coordinates

Key terms

Polar coordinates and their range

Sector area scaling

The polar area element

Polar regions

Bounds from the geometry

Avoiding overlap

Check yourself

How this lesson connects