Double Integrals in Polar Coordinates

Main idea

Round regions are awkward in $x$ and $y$ but simple in polar coordinates, where a disk is just $0 \le r \le a$ . The catch is the area element, $dA = r\,dr\,d\theta$ : patches far from the center are bigger, and that extra $r$ accounts for it.

\iint_D f\,dA = \int_\alpha^\beta\!\int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\,{\htmlData{sym=0}{r\,dr\,d\theta}}

What the symbols mean

$r\,dr\,d\theta$
Polar area element
A small polar box is about r dθ wide and dr deep, so its area is r dr dθ.

Explanation

Polar coordinates describe a point by how far it is from the center ( $r$ ) and which direction it's in ( $\theta$ ), instead of how far right and how far up. For anything round, that's the natural language: a pizza is $0 \le r \le 6$ , $0 \le \theta \le 2\pi$ , and a slice of it is the same with a smaller $\theta$ range. The formula $\iint_D f\,dA = \int\!\int f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta$ looks busy, but it's three moves: rewrite $f$ in terms of $r$ and $\theta$ , describe the region with $r$ and $\theta$ limits, and swap $dA$ for $r\,dr\,d\theta$ . Bear with the extra $r$ ; it has a very good reason to be there.

Where the $r$ comes from. Chop a disk with circles and rays instead of a square grid. Take a small polar box between radii $r$ and $r + dr$ and angles $\theta$ and $\theta + d\theta$ . It's $dr$ deep, and its width is an arc of a circle of radius $r$ , which has length $r\,d\theta$ (that's what radians mean). So its area is about $r\,dr\,d\theta$ . Think about a pizza: cut it into equal-angle slices and rings of equal width, and the pieces near the crust are much bigger than the ones near the center. The factor $r$ is what counts those outer pieces fairly.

A common mistake. It's tempting to write $dA = dr\,d\theta$ , because $dA = dx\,dy$ worked before. It doesn't: try it on a disk of radius $a$ and you get $2\pi a$ , which isn't even an area. With the $r$ , $\int_0^{2\pi}\!\int_0^a r\,dr\,d\theta = \pi a^2$ , as it should be.

When to switch: the region is a disk, ring or sector, or the integrand involves $x^2 + y^2$ , which is simply $r^2$ . The polar curves and areas from Calculus 2 are the same ideas; this just adds a height $f$ on top.

A famous payoff. $\left(\int_{-\infty}^\infty e^{-x^2}dx\right)^2 = \iint e^{-(x^2+y^2)}dA = \int_0^{2\pi}\!\int_0^\infty e^{-r^2}r\,dr\,d\theta = \pi$ , so $\int_{-\infty}^\infty e^{-x^2}dx = \sqrt\pi$ : the normal distribution's constant. Notice the trick. $e^{-x^2}$ has no elementary antiderivative, but squaring the integral turns it into a double integral over the whole plane, and in polar coordinates the extra $r$ is exactly what makes $e^{-r^2}r$ easy to integrate.

See it

The bell surface z = e^(−(x² + y²)) with a circle of radius R growing under it. The volume inside the circle is π(1 − e^(−R²)), and notice how fast it fills up: about 2.0 at R = 1 and already 3.14 by R = 2.5, because the bell is nearly flat beyond that. That total, π, is (∫ e^(−x²) dx)², which is where the √π in the normal distribution comes from.

Worked examples

Example 1

Evaluate $\iint_D e^{x^2 + y^2}\,dA$ over the unit disk.

1
In polar coordinates, $x^2 + y^2 = r^2$ , the unit disk is $0 \le r \le 1$ , $0 \le \theta \le 2\pi$ , and $dA = r\,dr\,d\theta$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Use polar coordinates to find the area of a disk of radius 3.

Problem 2+5 XP

Evaluate $\iint_D\left(x^2 + y^2\right)dA$ over the unit disk.

Problem 3+5 XP

Evaluate $\iint_D e^{-(x^2 + y^2)}\,dA$ over the unit disk.

Problem 4+5 XP

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

Problem 5+5 XP

Evaluate $\iint_D\frac{1}{x^2 + y^2}\,dA$ over the ring $1 \le r \le 2$ .

Problem 6+5 XP

A sprinkler waters a circular lawn of radius $10$ m, delivering water to a depth of $D(r) = 5e^{-0.1r}$ mm per hour at distance $r$ m from the center. How many liters does it spread per hour? ( $1$ mm of water over $1$ m² is $1$ L.)

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

A ball of radius $2$ has a cylindrical hole of radius $1$ drilled straight through its center. Find the volume that's left.

Where it's used

Radar coverage, the light falling on a circular sensor, and the normal distribution's $\sqrt{2\pi}$ all come from polar double integrals.

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