Integration in 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 for the polar coordinate map, 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.