---
tag: 0T89W1QZ9K
kind: statement
role: theorem
name: Integration in polar coordinates
aliases: [polar-change-of-variables]
msc: [26B15]
uses: [00N5FHP1CV, 0RJZV05MJA, 0NKBMFC18R]
sources:
  - url: https://en.wikipedia.org/wiki/Polar_coordinate_system
    locator: Area
  - url: https://en.wikipedia.org/wiki/Multiple_integral
    locator: Polar coordinates
---

## Statement

Let $0 \le r_1 < r_2$ and $\alpha < \beta \le \alpha + 2\pi$, and let $D = \{(r \cos \theta, r \sin \theta) : r_1 \le r \le r_2, \ \alpha \le \theta \le \beta\}$.
Let $f \colon D \to \mathbb{R}$ be continuous.
Then
$$\iint_D f(x, y) \, dx \, dy = \int_\alpha^\beta \int_{r_1}^{r_2} f(r \cos \theta, r \sin \theta) \, r \, dr \, d\theta.$$

## Remarks

This is the [[change of variables formula|00N5FHP1CV]] for the [[polar coordinate map|0RJZV05MJA]], whose Jacobian determinant is $r$. The boundary rays and the origin have area zero and do not change the integral.
Example: the area of the disc of radius $a$ is $\int_0^{2\pi} \int_0^a r \, dr \, d\theta = \pi a^2$.
For a region $\alpha \le \theta \le \beta$, $0 \le r \le g(\theta)$ with $g$ continuous, the formula gives the area $\tfrac{1}{2} \int_\alpha^\beta g(\theta)^2 \, d\theta$.
The Gaussian integral follows from $\iint_{\mathbb{R}^2} e^{-(x^2 + y^2)} \, dx \, dy = \int_0^{2\pi} \int_0^\infty e^{-r^2} r \, dr \, d\theta = \pi$. Hence $\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$.

## Explanations

### L1 human

In polar coordinates a small piece of the plane is almost a rectangle with sides $dr$ and $r \, d\theta$.
So its area is $r \, dr \, d\theta$, and the factor $r$ appears in the integral.