Double Integrals in Polar Coordinates: r dr dθ & Solved Examples

The change of variables

Before you start: read double integrals first.

Cartesian to polar

\[x = r\cos\theta, \quad y = r\sin\theta, \quad x^2 + y^2 = r^2, \quad dx\,dy = r\,dr\,d\theta\]

\[\iint_R f(x, y)\,dx\,dy = \iint_{R'} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta\]

Use polar coordinates when the region is a disc, a sector or a ring, or the integrand contains \(x^2 + y^2\).

Region	Polar limits
Disc \(x^2 + y^2 \le a^2\)	\(0 \le r \le a\), \(0 \le \theta \le 2\pi\)
Quarter disc in the first quadrant	\(0 \le r \le a\), \(0 \le \theta \le \tfrac{\pi}{2}\)
Ring \(a^2 \le x^2 + y^2 \le b^2\)	\(a \le r \le b\), \(0 \le \theta \le 2\pi\)
Circle \(x^2 + y^2 = 2ax\)	\(0 \le r \le 2a\cos\theta\), \(-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}\)
Circle \(x^2 + y^2 = 2ay\)	\(0 \le r \le 2a\sin\theta\), \(0 \le \theta \le \pi\)

Where the extra r comes from

In polar coordinates the small piece of area is almost a rectangle with sides \(dr\) and \(r\,d\theta\). That is why \(dx\,dy\) becomes \(r\,dr\,d\theta\), and the extra factor \(r\) must never be forgotten.

Solved examples

Example 1: Over a disc

Easy

Evaluate \(\displaystyle\iint_R (x^2 + y^2)\,dA\) over the disc \(x^2 + y^2 \le 1\).

\(\displaystyle\int_0^{2\pi}\!\!\int_0^1 r^2\cdot r\,dr\,d\theta = 2\pi\cdot\frac{1}{4}\).

Answer\(\dfrac{\pi}{2}\)

Example 2: A quarter disc

Medium

Evaluate \(\displaystyle\int_0^a\!\!\int_0^{\sqrt{a^2 - x^2}}(x^2 + y^2)\,dy\,dx\).

The limits describe the quarter disc of radius \(a\) in the first quadrant: \(0 \le r \le a\), \(0 \le \theta \le \tfrac{\pi}{2}\).
\(\displaystyle\int_0^{\pi/2}\!\!\int_0^a r^3\,dr\,d\theta = \frac{\pi}{2}\cdot\frac{a^4}{4}\).

Answer\(\dfrac{\pi a^4}{8}\)

Example 3: A ring

Medium

Evaluate \(\displaystyle\iint_R \frac{dA}{x^2 + y^2}\) over the ring \(1 \le x^2 + y^2 \le 4\).

\(\displaystyle\int_0^{2\pi}\!\!\int_1^2 \frac{1}{r^2}\,r\,dr\,d\theta = 2\pi\big[\log r\big]_1^2\).

Answer\(2\pi\log 2\)

Example 4: The Gaussian integral

Exam level

Show that \(\displaystyle\int_0^\infty\!\!\int_0^\infty e^{-(x^2 + y^2)}\,dx\,dy = \frac{\pi}{4}\), and hence that \(\displaystyle\int_0^\infty e^{-x^2}dx = \frac{\sqrt\pi}{2}\).

The first quadrant is \(0 \le r < \infty\), \(0 \le \theta \le \tfrac{\pi}{2}\): \(\displaystyle\int_0^{\pi/2}\!\!\int_0^\infty e^{-r^2}r\,dr\,d\theta = \frac{\pi}{2}\cdot\frac{1}{2}\).
But the double integral also equals \(\left(\displaystyle\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy\right) = I^2\). So \(I^2 = \dfrac{\pi}{4}\).

Answer\(I = \dfrac{\sqrt\pi}{2}\). This is the proof behind \(\Gamma\left(\tfrac{1}{2}\right) = \sqrt\pi\) in the Gamma function lesson.

Example 5: A circle not centred at the origin

Exam level

Find the area enclosed by \(r = 2a\cos\theta\) using a double integral.

This is the circle \(x^2 + y^2 = 2ax\), covered as \(\theta\) runs from \(-\tfrac{\pi}{2}\) to \(\tfrac{\pi}{2}\).
\(\displaystyle\int_{-\pi/2}^{\pi/2}\!\!\int_0^{2a\cos\theta} r\,dr\,d\theta = \int_{-\pi/2}^{\pi/2} 2a^2\cos^2\theta\,d\theta = 2a^2\cdot\frac{\pi}{2}\).

Answer\(\pi a^2\), the area of a circle of radius \(a\), as expected.

Vipul Sir's tip

Write \(r\,dr\,d\theta\) as one unit every single time, before you do anything else. Forgetting the \(r\) is the most common way to lose the whole question.

Common mistakes

1. Forgetting the factor \(r\)

\(dx\,dy\) becomes \(r\,dr\,d\theta\), not \(dr\,d\theta\).

2. Wrong range of \(\theta\)

A full disc needs \(0\) to \(2\pi\); a quarter disc in the first quadrant needs \(0\) to \(\tfrac{\pi}{2}\). Read it from the sketch.

3. Putting \(\theta\) in the outer limits wrongly

Usually integrate \(r\) first (its limits may depend on \(\theta\)), then \(\theta\) with constant limits.

Practice questions

Q1\(\displaystyle\iint e^{-(x^2 + y^2)}\,dA\) over the unit disc

\(\displaystyle 2\pi\int_0^1 e^{-r^2}r\,dr = \pi(1 - e^{-1})\).

Q2\(\displaystyle\int_0^\pi\!\!\int_0^{a\sin\theta} r\,dr\,d\theta\)

\(\displaystyle\int_0^\pi\frac{a^2\sin^2\theta}{2}\,d\theta = \frac{\pi a^2}{4}\): the area of the circle of diameter \(a\).

Q3\(\displaystyle\int_0^a\!\!\int_0^{\sqrt{a^2 - y^2}}\sqrt{x^2 + y^2}\,dx\,dy\)

Quarter disc: \(\displaystyle\int_0^{\pi/2}\!\!\int_0^a r^2\,dr\,d\theta = \frac{\pi a^3}{6}\).

Q4\(\displaystyle\iint xy\,dA\) over the quarter of the unit disc in the first quadrant

\(\displaystyle\int_0^{\pi/2}\!\!\int_0^1 r^3\cos\theta\sin\theta\,dr\,d\theta = \frac{1}{4}\cdot\frac{1}{2} = \frac{1}{8}\).

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Frequently asked questions

Why does dx dy become r dr dθ?

A small polar element is nearly a rectangle with sides \(dr\) and \(r\,d\theta\), so its area is \(r\,dr\,d\theta\). (Formally, \(r\) is the Jacobian of the change of variables.)

When should I use polar coordinates?

When the region is circular, or the integrand involves \(x^2 + y^2\).

Where this leads next

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Written by Vipul Sir

Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.

Double integrals in polar form: dx dy = r dr dθ.