Double Integrals in Polar Coordinates

Quick answer

Disks, rings and wedges are awkward in x and y but are rectangles in polar coordinates. Substituting x = r cos θ and y = r sin θ turns a double integral over such a region into an iterated integral in r and θ with constant limits, provided the area element dA becomes r dr dθ. The factor r appears because a small polar rectangle far from the origin is wider than one close to it. Polar integrals give volumes of domes and cones and the famous value √π for the integral of e^(−x²).

What you'll learn

Describe disks, rings and wedges in polar coordinates
Convert a double integral to polar form with dA = r dr dθ
Explain where the factor r comes from
Compute volumes and the Gaussian integral with polar coordinates

Disks are polar rectangles

Integrate $x^2 + y^2$ over the disk $x^2 + y^2 \le 4$ in rectangular coordinates and the limits are $-2 \le x \le 2$ and $-\sqrt{4 - x^2} \le y \le \sqrt{4 - x^2}$ , with square roots everywhere. In polar coordinates the same disk is $0 \le r \le 2$ , $0 \le \theta \le 2\pi$ : a rectangle in the $r\theta$ -plane, with constant limits. Rings and wedges centered at the origin work the same way. A polar rectangle is a region $a \le r \le b$ , $\alpha \le \theta \le \beta$ .

Double integrals in polar coordinates. If $D$ is the polar rectangle $a \le r \le b$ , $\alpha \le \theta \le \beta$ , then

$\displaystyle\iint_D f(x, y)\,dA = \int_\alpha^\beta \int_a^b f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta$

Two changes happen at once: $x$ and $y$ become $r\cos\theta$ and $r\sin\theta$ , and $dA$ becomes $r\,dr\,d\theta$ . The integrand often simplifies too, since $x^2 + y^2 = r^2$ .

Why the extra r

Cut the region with circles $r = \text{constant}$ and rays $\theta = \text{constant}$ , spaced $\Delta r$ and $\Delta\theta$ apart. The pieces are small polar rectangles. One at distance $r$ from the origin has a straight side of length $\Delta r$ and a curved side, an arc of a circle of radius $r$ , of length $r\,\Delta\theta$ . Its area is about $r\,\Delta r\,\Delta\theta$ . The same $\Delta\theta$ sweeps out a wider piece far from the origin than near it, and the factor $r$ records that. A Riemann sum over these pieces is $\sum f\,r\,\Delta r\,\Delta\theta$ , and its limit is the formula above.

A polar rectangle inside a ring

Worked examples

The dome z = 4 − x² − y²

Example 5 — The Gaussian integral

Show that $\displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ .

The function $e^{-x^2}$ has no elementary antiderivative, so no single-variable method works. Call the integral $I$ . Then

I^2 = \int_{-\infty}^{\infty} e^{-x^2}\,dx \int_{-\infty}^{\infty} e^{-y^2}\,dy = \iint_{\mathbb{R}^2} e^{-(x^2 + y^2)}\,dA

The whole plane is $0 \le r < \infty$ , $0 \le \theta \le 2\pi$ , and the factor $r$ makes the inner integral work:

I^2 = \int_0^{2\pi} \int_0^\infty e^{-r^2}\,r\,dr\,d\theta = 2\pi \cdot \frac{1}{2} = \pi

The inner improper integral is $\Big[-\tfrac{1}{2}e^{-r^2}\Big]_0^\infty = \tfrac{1}{2}$ . Since $I > 0$ , $I = \sqrt{\pi} \approx 1.7725$ . This value is the reason $\sqrt{\pi}$ appears in the formula for the normal distribution.

Common mistakes

Practice problems

Use a polar double integral to find the area of the disk of radius $3$ .

Answer

$9\pi$

Full solution

$\int_0^{2\pi} \int_0^3 r\,dr\,d\theta = 2\pi \cdot \tfrac{9}{2} = 9\pi$ .
Find the area of the ring $1 \le x^2 + y^2 \le 4$ .

Answer

$3\pi$

Full solution

$\int_0^{2\pi} \int_1^2 r\,dr\,d\theta = 2\pi \cdot \tfrac{4 - 1}{2} = 3\pi$ , the difference of the areas $4\pi$ and $\pi$ .
Evaluate $\iint_D \sqrt{x^2 + y^2}\,dA$ over the unit disk.

Answer

$\tfrac{2\pi}{3}$

Full solution

The integrand is $r$ , so the integral is $\int_0^{2\pi} \int_0^1 r^2\,dr\,d\theta = 2\pi \cdot \tfrac{1}{3}$ .
Evaluate $\iint_D y\,dA$ over the upper half of the disk $x^2 + y^2 \le 4$ .

Answer

$\tfrac{16}{3}$

Full solution

The half disk is $0 \le r \le 2$ , $0 \le \theta \le \pi$ , and $y = r\sin\theta$ . The integral is $\int_0^\pi \sin\theta\,d\theta \cdot \int_0^2 r^2\,dr = 2 \cdot \tfrac{8}{3} = \tfrac{16}{3}$ .
Find the volume of the solid above the cone $z = \sqrt{x^2 + y^2}$ and below the plane $z = 2$ .

Answer

$\tfrac{8\pi}{3}$

Full solution

The cone meets the plane where $r = 2$ . The height of the solid is $2 - r$ , so the volume is $\int_0^{2\pi} \int_0^2 (2 - r)\,r\,dr\,d\theta = 2\pi\left(4 - \tfrac{8}{3}\right) = \tfrac{8\pi}{3}$ . This matches the cone formula $\tfrac{1}{3}\pi r^2 h$ with $r = h = 2$ .
Use polar coordinates to find the volume of a ball of radius $1$ .

Hint

The top half lies under $z = \sqrt{1 - x^2 - y^2}$ .

Answer

$\tfrac{4\pi}{3}$

Full solution

The volume is twice $\int_0^{2\pi} \int_0^1 \sqrt{1 - r^2}\,r\,dr\,d\theta$ . With $u = 1 - r^2$ , the inner integral is $\tfrac{1}{2}\int_0^1 \sqrt{u}\,du = \tfrac{1}{3}$ . So the volume is $2 \cdot 2\pi \cdot \tfrac{1}{3} = \tfrac{4\pi}{3}$ .
Write $\displaystyle\int_0^2 \int_0^{\sqrt{4 - x^2}} (x^2 + y^2)\,dy\,dx$ in polar form and evaluate it.

Answer

$\displaystyle\int_0^{\pi/2} \int_0^2 r^3\,dr\,d\theta = 2\pi$

Full solution

The limits describe the quarter disk of radius $2$ in the first quadrant: $0 \le r \le 2$ , $0 \le \theta \le \tfrac{\pi}{2}$ . The integrand $x^2 + y^2$ is $r^2$ , times $r$ from $dA$ . The value is $\tfrac{\pi}{2} \cdot 4 = 2\pi$ , one quarter of Example 1.
A student computes Example 1 as $\int_0^{2\pi} \int_0^2 r^2\,dr\,d\theta$ and gets $\tfrac{16\pi}{3}$ . Find the error.

Answer

The factor $r$ in $dA = r\,dr\,d\theta$ is missing; the answer is $8\pi$ .

Full solution

The integrand $x^2 + y^2 = r^2$ must be multiplied by $r$ : $\int_0^{2\pi} \int_0^2 r^3\,dr\,d\theta = 2\pi \cdot 4 = 8\pi$ . Without the $r$ , the pieces near the origin count as much as the wider pieces far away.

Frequently asked questions

How do you convert a double integral to polar coordinates?

Replace x by r cos θ and y by r sin θ, replace dA by r dr dθ, and describe the region with limits on r and θ.

Why is there an extra r in dA = r dr dθ?

A small polar rectangle with sides Δr and Δθ has a curved side of length about r Δθ, so its area is about r Δr Δθ. Farther from the origin, the same Δθ sweeps out a wider piece.

When should you use polar coordinates?

When the region is a disk, ring or wedge centered at the origin, or when the integrand contains x² + y², which becomes r².

What are the limits for a full disk of radius a?

r from 0 to a and θ from 0 to 2π.

What is the Gaussian integral?

The integral of e^(−x²) over the whole real line. It equals √π, a value found by squaring it, writing the square as a double integral over the plane, and switching to polar coordinates.

What to learn next

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