# Thread: Areas in Polar Coordinates

1. ## Areas in Polar Coordinates

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

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

2nd:$\displaystyle r = 3$

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 $\displaystyle \frac{\pi}{3}$ & $\displaystyle \frac{5\pi}{3}$ and the area I need to find is the right side of $\displaystyle r=6cos\theta$

*Hope I am okay up to here*

So, for my area:

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

Did I set this up correctly?
Thanks!!

2. If $\displaystyle \frac{5\pi}{3}$ is going to be your lower limit, then $\displaystyle \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}$ 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 $\displaystyle \frac{\pi}{3}$ to $\displaystyle \frac{5\pi}{3}$ in a counter-clockwise manner).

3. Originally Posted by Pinkk
If $\displaystyle \frac{5\pi}{3}$ is going to be your lower limit, then $\displaystyle \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}$ 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 $\displaystyle \frac{\pi}{3}$ to $\displaystyle \frac{5\pi}{3}$ 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?

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

5. Originally Posted by Pinkk
If $\displaystyle \frac{5\pi}{3}$ is going to be your lower limit, then $\displaystyle \frac{\pi}{3} + 2\pi = \frac{7\pi}{3}$ 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 $\displaystyle \frac{\pi}{3}$ to $\displaystyle \frac{5\pi}{3}$ in a counter-clockwise manner).
Ok, so what I did was just integrated from $\displaystyle 0$ to $\displaystyle \frac{\pi}{3}$ and then multiplied everything by 2. Would that not give me the same result? I am getting $\displaystyle 3\pi - 18$ 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.

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

7. Originally Posted by Pinkk
I got $\displaystyle 3\pi +\frac{\sqrt{3}}{2}$. 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!!!!!

8. See attachment

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

This is what I got:

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

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

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