Disks, rings and wedges are awkward in rectangular coordinates: the limits involve $\sqrt{r^2 - x^2}$. In polar coordinates they are rectangles: $a \le r \le b$, $\alpha \le \theta \le \beta$. You met polar area in Calculus 2; now the integrand can be any function.
Key idea
Why it works
When should you switch to polar? Look at both the region and the integrand. Disks, rings, sectors and regions bounded by curves like $r = 2\cos\theta$ have simple limits in polar form, and integrands built from $x^2 + y^2$ become functions of $r$ alone. The circle $x^2 + y^2 = 2x$ is $r^2 = 2r\cos\theta$, that is, $r = 2\cos\theta$ with $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$. Its area is $$\int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} r\,dr\,d\theta = \int_{-\pi/2}^{\pi/2} 2\cos^2\theta\,d\theta = \pi,$$ as it should be for a circle of radius ${}1$.
Worked examples
Worked example: Over a disk
- $x^2 + y^2 = r^2$, and the disk is ${}0 \le r \le 2$, ${}0 \le \theta \le 2\pi$.
- $$\int_0^{2\pi}\int_0^2 r^2 \cdot r\,dr\,d\theta = \int_0^{2\pi} \left[\frac{r^4}{4}\right]_0^2 d\theta = \int_0^{2\pi} 4\,d\theta = 8\pi.$$
Answer: ${}8\pi$
Worked example: A Gaussian
- $\int_0^{2\pi}\int_0^R e^{-r^2}\,r\,dr\,d\theta$: the $r$ from $dA$ makes the substitution $u = r^2$ work.
- $\int_0^R e^{-r^2}r\,dr = \frac{1}{2}\left(1 - e^{-R^2}\right)$, so the integral is $\pi\left(1 - e^{-R^2}\right)$.
- As $R \to \infty$ this tends to $\pi$, the famous fact behind $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$.
Answer: $\pi\left(1 - e^{-R^2}\right)$
Worked example: Half a ring
- ${}1 \le r \le 2$, ${}0 \le \theta \le \pi$, and $y = r\sin\theta$.
- $$\int_0^{\pi}\int_1^2 r\sin\theta \cdot r\,dr\,d\theta = \left(\int_1^2 r^2\,dr\right)\left(\int_0^{\pi}\sin\theta\,d\theta\right) = \frac{7}{3} \cdot 2 = \frac{14}{3}.$$
Answer: $\frac{14}{3}$
Common mistake