Areas in Polar Coordinates

**mollymcf2009** · April 28th 2009, 08:34 AM

Find the area of the region inside the first curve and outside the 2nd curve:

1st: $r = 6cos(\theta)$

2nd: $r = 3$

Ok, so I have a circle w/ radius 3 with center (0,0) and a circle with radius 3 with center (3,0), and I found that they intersect at $\frac{\pi}{3}$ & $\frac{5\pi}{3}$ and the area I need to find is the right side of $r=6cos\theta$

*Hope I am okay up to here*

So, for my area:

$A = \frac{1}{2} \int^{\frac{\pi}{3}}_{\frac{5\pi}{3}} 36 cos^2\theta d\theta - \frac{1}{2} \int^{\frac{\pi}{3}}_{\frac{5\pi}{3}} 3^2 d\theta$

Did I set this up correctly?
Thanks!!

**Pinkk** · April 28th 2009, 08:55 AM

If $\frac{5\pi}{3}$ is going to be your lower limit, then $\frac{\pi}{3} + 2\pi = \frac{7\pi}{3}$ should be your upper limit. The way you have it written would actually be the area NOT including the region you want to find the area of (since what you have set up is integrating from $\frac{\pi}{3}$ to $\frac{5\pi}{3}$ in a counter-clockwise manner).

**mollymcf2009** · April 28th 2009, 09:07 AM

Originally Posted by Pinkk

If $\frac{5\pi}{3}$ is going to be your lower limit, then $\frac{\pi}{3} + 2\pi = \frac{7\pi}{3}$ should be your upper limit. The way you have it written would actually be the area NOT including the region you want to find the area of (since what you have set up is integrating from $\frac{\pi}{3}$ to $\frac{5\pi}{3}$ in a counter-clockwise manner).

Oh, ok I see! So, my equation for finding the area is correct, but the limits I have are for the area I DON'T want. Right?

**Pinkk** · April 28th 2009, 09:11 AM

Exactly. You can have your limits of integration be $[-\frac{\pi}{3},\frac{\pi}{3}]$ or $[\frac{5\pi}{3},\frac{7\pi}{3}]$ .

**mollymcf2009** · April 28th 2009, 09:17 AM

Originally Posted by Pinkk

If $\frac{5\pi}{3}$ is going to be your lower limit, then $\frac{\pi}{3} + 2\pi = \frac{7\pi}{3}$ should be your upper limit. The way you have it written would actually be the area NOT including the region you want to find the area of (since what you have set up is integrating from $\frac{\pi}{3}$ to $\frac{5\pi}{3}$ in a counter-clockwise manner).

Ok, so what I did was just integrated from $0$ to $\frac{\pi}{3}$ and then multiplied everything by 2. Would that not give me the same result? I am getting $3\pi - 18$ for my area, which is incorrect. It is entirely possible that I have algebra errors (I'm the queen of them). I just want to make sure that that would be my problem and that otherwise I am evaluating this correctly.

**Pinkk** · April 28th 2009, 09:35 AM

I got $3\pi +\frac{\sqrt{3}}{2}$ . I'll go step-by-step after I come back from class if you still can't arrive at that answer. Good luck!

**mollymcf2009** · April 28th 2009, 10:17 AM

Originally Posted by Pinkk

I got $3\pi +\frac{\sqrt{3}}{2}$ . I'll go step-by-step after I come back from class if you still can't arrive at that answer. Good luck!

That isn't it either. This is a problem in webassign, so I know immediately if it's the correct answer. If you don't mind, I would really appreciate you working out the steps. Thanks!!!!!

**Calculus26** · April 28th 2009, 11:23 AM

See attachment

**Pinkk** · April 28th 2009, 11:44 AM

D'oh, left out a factor of nine for the trig term:

This is what I got:

$\displaystyle\int^\frac{\pi}{3}_0 36cos^{2}\theta - 9 \,\,d\theta$

$\displaystyle\int^\frac{\pi}{3}_0 9 + 18cos2\theta \,\,d\theta$

$9(\frac{\pi}{3}) + 9sin\frac{2\pi}{3}= 3\pi + \frac{9\sqrt{3}}{2}$

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