[SOLVED] Area computations in polar coordinates

• Mar 19th 2008, 08:49 PM
hasanbalkan
[SOLVED] Area computations in polar coordinates
Find the area of the region that lies interior to all three circles:
$\displaystyle r = 1$
$\displaystyle r = 2cos \theta$
$\displaystyle r = 2sin \theta$

I graphed it, but I have problems figuring out how to solve it. Any help with some explanations will be appreciated!
• Mar 20th 2008, 10:50 AM
roy_zhang
Please see the attached graph. We have 3 circles: $\displaystyle r=1$ (in red), $\displaystyle r=2\cos\theta$ (in blue) and $\displaystyle r=2\sin\theta$ (in green). I labeled 2 intersection points of the common interior area of the three circles. They are $\displaystyle (\frac{\sqrt{3}}{2},\frac{1}{2})$ and $\displaystyle (\frac{1}{2},\frac{\sqrt{3}}{2})$ or in polar coordinates $\displaystyle (1,\frac{\pi}{6})$ and $\displaystyle (1,\frac{\pi}{3})$

Look closely to the common interior region among the three circles, we know its area can be obtained by adding three areas together: 1) the area obtained by integrating the green circle ($\displaystyle r=2\sin\theta$) from $\displaystyle 0$ to $\displaystyle \frac{\pi}{6}$, 2) the area obtained by integrating the red circle ($\displaystyle r=1$) from $\displaystyle \frac{\pi}{6}$ to $\displaystyle \frac{\pi}{3}$, 3) the area obtained by integrating the blue circle ($\displaystyle r=2\cos\theta$) from $\displaystyle \frac{\pi}{3}$ to $\displaystyle \frac{\pi}{2}$.

Mathematically, we have:
$\displaystyle A=\frac{1}{2}\int\limits_0^\frac{\pi}{6} (2\sin\theta)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{6}^\frac{\pi}{3} (1)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{3}^\frac{\pi}{2} (2\cos\theta)^2 d\theta$
$\displaystyle =\frac{5}{12}\pi-\frac{\sqrt{3}}{2}$

Note that the first and third area are identical, so we only need to calculate one of those.

Roy
• Mar 21st 2008, 11:47 AM
hasanbalkan
Thank you a lot! I was trying to split it into two parts rather than to three parts. Just one correction there in the integrals for people interested in the same problem:

$\displaystyle A=\frac{1}{2}\int\limits_0^\frac{\pi}{6} (2\sin\theta)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{6}^\frac{\pi}{3} (1)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{3}^\frac{\pi}{2} (2\cos\theta)^2 d\theta$
• Mar 21st 2008, 11:53 AM
roy_zhang
Quote:

Originally Posted by hasanbalkan
Thank you a lot! I was trying to split it into two parts rather than to three parts. Just one correction there in the integrals for people interested in the same problem:

$\displaystyle A=\frac{1}{2}\int\limits_0^\frac{\pi}{6} (2\sin\theta)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{6}^\frac{\pi}{3} (1)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{3}^\frac{\pi}{2} (2\cos\theta)^2 d\theta$

Hi hasanbalkan,

Thank you for your correction! I corrected it.

Roy