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
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
After finding the area element in the polar coordinate, we can evaluate the integral as an iterated integral with
Example Integrate
We can set this up as an iterated integral of
Cartesian coordinate approach
\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:
The polar coordinate way