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I have a question about which order to put limits of integration for polar curves. Say I wanted to find the region enclosed by the curve $r= 2\cos\theta$ from $5\pi/3$ to $\pi/6$ (not the other way around). How would I do this? thanks!

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Since area is always a nonnegative quantity, and since switching the limits of integration just changes the sign, you can actually choose any orientation you want, and then take the absolute value of your answer. –  Jared May 6 '13 at 3:52
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thank you for your answer! what i'm confused about, however, is that the limits seem to be sweeping different areas depending on the order. take, for example, this graph (linked). If I wanted the area between 5pi/3 to pi/6, wouldn't that be a different area than pi/6 to 5pi/3? and together those areas would add up to the area of the entire circle? wolframalpha.com/share/… –  Samantha May 6 '13 at 3:59
    
Ahh, I see. Yes, as joriki mentions in his answer, it would be best then to use the limit of integration $\frac{13\pi}{6}$ instead of $\frac{\pi}{6}$. –  Jared May 6 '13 at 4:27
    
thank you very much! –  Samantha May 6 '13 at 4:39

1 Answer 1

It seems that you mean the area that lies at angles in $[5\pi/3,2\pi]$ or in $[0,\pi/6]$. To get this area, either perform those two integrations separately, or note that adding $2\pi$ makes no difference for your purposes, so you can integrate over $[5\pi/3,13\pi/6]$ in one go.

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thank you very much! –  Samantha May 6 '13 at 4:38
    
as another question, why do I get different answers if I integrate r=2sin(theta) from 5pi/ to pi vs. 5pi/6 to 3pi/2. Isn't the area contained between these limits the same? Thank you! –  Samantha May 6 '13 at 5:27
    
@Samantha: The first limit is incomplete; please clarify. –  joriki May 6 '13 at 5:30
    
I'm sorry, the first limit is also 5pi/6. I have submitted this question separately. thank you very much for all your help. –  Samantha May 6 '13 at 5:49

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