Sometimes it is more convenient to evaluate a double integral in polar coordinates. To do that, we can perform the substitutions

A caveat is that the area element in polar coordinates, and is instead given by

The geometric justification for that is shown in the following figure:

The area of the small sector is approximately that of a rectangle with width and arc length , hence the area .

After finding the area element in the polar coordinate, we can evaluate the integral as an iterated integral with and as variables.

Example Integrate where

We can set this up as an iterated integral of and , but the result is cumbersome

\end{align}

\begin{align} & \frac{2}{3} \int_{x=0}^1 (1 - x^2)^{3/2} , dx \ &= \frac{2}{3} \int_{\theta=0}^{\pi/2} (1 - \sin^2 \theta)^{3/2} \cos(\theta) ,d\theta \ &= \frac{2}{3} \int_{\theta=0}^{\pi/2} (\cos(\theta))^3 \cos(\theta) ,d\theta \ &= \frac{2}{3} \int_{\theta=0}^{\pi/2} \cos^4 \theta ,d\theta \ &= \frac{2}{3} \int_{\theta=0}^{\pi/2} \left( \frac{1 + \cos{2\theta}}{2} \right)^2 , d\theta \ &= \frac{2}{3} \int_{\theta=0}^{\pi/2} \left( \frac{1}{4} + \frac{1}{2} \cos{2\theta} + \frac{1}{4}\cos^2 2\theta \right) , d\theta \ &= \frac{2}{3} \int_{\theta=0}^{\pi/2} \left( \frac{1}{4} + \frac{1}{2} \cos{2\theta} + \frac{1}{4}\left(\frac{1 + \cos{2\theta}}{2} \right) \right) , d\theta \ &= \dots \ &= \frac{\pi}{8} \end{align}

If we use polar coordinates, the program become much simpler: