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Quick question about finding area for polar coordinates

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the area of the shaded region.
    r=sqrt(θ)



    2. Relevant equations

    A = integral from a to b 1/2r^2dθ

    3. The attempt at a solution

    I know how to solve the question, I just don't know what to use for a and b. I tried 0 and 2pi but I am getting the wrong answer. Here is a picture:

    http://www.webassign.net/mapleplots/a/8/0c017c152b2f6bb8c6df656e3a1bc2.gif

    Thank you for the help. I just need to know what parameters to use for a and b :).
     
  2. jcsd
  3. Apr 30, 2012 #2

    Curious3141

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    Which quadrant does the shaded area reside in?

    Which values of [itex]\theta[/itex] define that quadrant?
     
  4. Apr 30, 2012 #3

    SammyS

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    If you are to find the area of the shaded region, the values of θ bounding the region look quite obvious.

    attachment.php?attachmentid=46811&stc=1&d=1335767843.gif

    You will also need to know r as a function of θ. It appears that r is directly proportional to θ. The constant of proportionality should be given in your problem.
     

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  5. Apr 30, 2012 #4
    can you tell the answer because it does not seem very hard
     
  6. Apr 30, 2012 #5

    Curious3141

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    Actually, his original problem statement gives [itex]r = \sqrt{\theta}[/itex], which makes the integration even easier.
     
  7. Apr 30, 2012 #6

    SammyS

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    OOPS!

    I missed that. r=sqrt(θ) probably would have been more recognizable with a radical or fractional exponent.

    Yes, that makes things even easier!

    I think we've given all the clues we should until we hear back from OP .
     
  8. Apr 30, 2012 #7
    Haha, wow sorry about that guys. It was late last night and I was trying to study so I was over thinking everything. Got the answer now, thanks :P.
     
  9. Apr 30, 2012 #8

    sharks

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    Yeah, i think the answer should be: [tex]\frac{1}{2}\left[(\ln (2\pi)-\ln (\frac{3\pi}{2})\right][/tex]
     
  10. Apr 30, 2012 #9
    Yeah, that was it, lol. Thanks.
     
  11. Apr 30, 2012 #10

    ehild

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    Are you sure?

    ehild
     
  12. Apr 30, 2012 #11
    Regardless, I got the answer, lol. It was form 3pi/2 to 2pi. Sorry about asking such a dumb question guys, hah.
     
  13. Apr 30, 2012 #12

    sharks

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    Well, the integration gives:
    [tex]A=\frac{1}{2}\left[(\ln \theta)\right] ^{\theta=2\pi}_{\theta=\frac{3\pi}{2}}=\frac{1}{2}\left[(\ln (2\pi)-\ln (\frac{3\pi}{2})\right][/tex]
    So, to answer your question, i am sure. :smile: Thanks for asking.
     
    Last edited: Apr 30, 2012
  14. Apr 30, 2012 #13
    Where does the [itex]\ln \theta[/itex] come from?
     
  15. Apr 30, 2012 #14

    sharks

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    [tex]A=\frac{1}{2}\int \frac{1}{\theta}\,.d\theta [/tex]
     
  16. Apr 30, 2012 #15
    Are you sure? Please explain. I fail to see where you get [itex]\frac{1}{\theta}[/itex] from.
     
  17. Apr 30, 2012 #16

    sharks

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    Show your work and i'll tell you where you went wrong. :smile:
     
  18. Apr 30, 2012 #17
    Isn't the area equal to
    [tex]\frac{1}{2}\int_\frac{3\pi}{2}^{2\pi}r^2\, d\theta[/tex]
     
  19. Apr 30, 2012 #18

    sharks

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    You have the integrand wrong. Check the first post for the correct integrand.
     
  20. Apr 30, 2012 #19

    SammyS

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    Yes, and [itex]\displaystyle r=\sqrt{\theta}[/itex] so your guess is as good as mine concerning where [itex]\displaystyle \frac{1}{\theta}[/itex] comes from !
     
  21. Apr 30, 2012 #20
    I think we can safely assume that 1/2r^2 has been misinterpreted as 1/(2r^2).
     
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