[SOLVED] Area computations in polar coordinates

**hasanbalkan** · March 19th 2008, 08:49 PM

Find the area of the region that lies interior to all three circles:
$r = 1$
$r = 2cos \theta$
$r = 2sin \theta$

I graphed it, but I have problems figuring out how to solve it. Any help with some explanations will be appreciated!

**roy_zhang** · March 20th 2008, 10:50 AM

Please see the attached graph. We have 3 circles: $r=1$ (in red), $r=2\cos\theta$ (in blue) and $r=2\sin\theta$ (in green). I labeled 2 intersection points of the common interior area of the three circles. They are $(\frac{\sqrt{3}}{2},\frac{1}{2})$ and $(\frac{1}{2},\frac{\sqrt{3}}{2})$ or in polar coordinates $(1,\frac{\pi}{6})$ and $(1,\frac{\pi}{3})$

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 ( $r=2\sin\theta$ ) from $0$ to $\frac{\pi}{6}$ , 2) the area obtained by integrating the red circle ( $r=1$ ) from $\frac{\pi}{6}$ to $\frac{\pi}{3}$ , 3) the area obtained by integrating the blue circle ( $r=2\cos\theta$ ) from $\frac{\pi}{3}$ to $\frac{\pi}{2}$ .

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

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

Roy

**hasanbalkan** · March 21st 2008, 11:47 AM

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:

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

**roy_zhang** · March 21st 2008, 11:53 AM

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:

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

Hi hasanbalkan,

Thank you for your correction! I corrected it.

Roy

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