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Integration of a Circle in Polar Coordinates

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi, I'm trying to find the area of a circle in polar coordinates.I'm doing it this way because I have to put this into an excel sheet to have a matrix of areas of multiple circles.
    Here is an example of the problem.
    4rtWQ.png
    a= radius of small circle
    (gamma, r0) = polar coordinate system where phi is the angle and r0 is the distance to the center of the circle
    gamma = phi (sorry for the confusion) just another constant
    They are all constants except for theta.

    Using the equation provided by wikimedia
    d52097e7e82669349c8da7c64617c8f5.png
    and the equation
    eq0001MP.gif

    2. Relevant equations
    Integrating this would give a constant C, but this value does not matter, does it?
    Also, are my trig identities correct?
    3. The attempt at a solution
    KcHwN.png
    from interval a to b

    *found a mistake in my work: should be sin^2 not sin
     
    Last edited: Oct 2, 2012
  2. jcsd
  3. Oct 2, 2012 #2
    I asked a math professor at my university about this and we tried to do this converting a circle from cartesian to polar coordinates. The equation of the circle would result in a +- equation in front of the square root through quadratic formula.
    He used the equation:
    eq0008M.gif
    and told me to use the positive answer for r0 and the negative for ri. I thought this applies to two different curves but in this case the circle should be one curve.

    Am I doing this right?

    If I am wrong, what equation do I use if the limits of integration only has one curve (side of the circle)
     
    Last edited: Oct 2, 2012
  4. Oct 2, 2012 #3

    LCKurtz

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    I'm sorry, but we aren't mind readers here. What are ##a,r_0, \theta, \phi##? And is the ##r## in the first equation supposed to be the same ##r## in the second equation?
     
  5. Oct 2, 2012 #4
    Sorry I'll upload a better picture with a description

    Yes the r's are the same. I changed phi to gamma. Not sure why I did.
     
    Last edited: Oct 2, 2012
  6. Oct 2, 2012 #5

    SammyS

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    Last question first:

    If you're asking,
    "Do I need to include the Integration constant from the anti-derivative when evaluating a definite integral?"
    The answer is, No.


    Now that you've updated your figure:

    For a circle of radius, a, centered at (r0, φ ) in polar coordinates, the equation in polar coordinates, (r, θ), is like the one you found on wikipedia:
    [itex]\displaystyle
    r=r_0\cos(\theta-\varphi)\pm\sqrt{a^2-r_0^2\sin^2(\theta-\varphi)\,}[/itex]​

    Added in Edit:

    If a ≤ r0, then the limits of integration for θ should be [itex]\displaystyle \varphi\pm\sin^{-1}\left(\frac{a}{r_0}\right)\ .[/itex]
     
    Last edited: Oct 2, 2012
  7. Oct 2, 2012 #6
    Yes they're the same.
     
  8. Oct 2, 2012 #7

    LCKurtz

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    No they aren't. Your original post left off the ##\pm## sign. You can use your professor's suggestion for ##r_{outer}^2-r_{inner}^2## using the plus or minus sign accordingly (I think).
     
  9. Oct 2, 2012 #8
    On wikipedia it says: "the solution with a minus sign in front of the square root giving the same curve"
    http://en.wikipedia.org/wiki/Polar_coordinate_system

    If they're the same curve then I wouldn't need to use ##r_{outer}^2-r_{inner}^2## would I?

    What if my limits of integration was between these points? Which equations would I use?
    vRW08.png
     
    Last edited: Oct 2, 2012
  10. Oct 2, 2012 #9

    LCKurtz

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    Yes. It's just like when you have a circle ##x^2+y^2=4## and you solve for the upper and lower curves ##y = \pm\sqrt{4-x^2}## for the separate equations of the upper and lower curve if you were going to find its area with a dydx integral. It's all the same circle, but different parts. Your r is apparently double valued function of ##\theta## just like a circle in xy is a double valued function of x. I say "apparently" because I have not myself derived that equation.
     
  11. Oct 2, 2012 #10
    Okay I'll try that way. So the picture of my previous post, would I use the -sqrt since its the lower curve?
     
  12. Oct 2, 2012 #11

    SammyS

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    As that math prof. said,

    Using the plus sign of ±√, gives part (at least a little more than half) of the circle, which he called ro, for rOUTER . The minus sign of ±√, gives part of the circle he called ri for rINNER .

    Find the area from the origin to rOUTER. Subtract from that the area from the origin to rINNER

    Below is rOUTER .

    attachment.php?attachmentid=51455&stc=1&d=1349217383.gif

    Below is rINNER .

    attachment.php?attachmentid=51456&stc=1&d=1349217383.gif

    (From WolframAlpha)
     

    Attached Files:

  13. Oct 2, 2012 #12
    Okay I understand now. This is the general equation I should be using throughout excel.

    After working this out, I came to this equation. How do I integrate this?

    i85gJ.png
     
  14. Oct 2, 2012 #13
    SwPVm.png
    Is this correct? Not sure where to go after this.

    *ignore 2nd to last line*
     
  15. Oct 2, 2012 #14

    SammyS

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    It seems crazy, but the next thing to do is a trig substitution.

    Let u = sin(t) ...
     
  16. Oct 3, 2012 #15
    Thanks, I see how this works.

    One more question. Are my trig identities correct where:
    u = sin(theta-phi)
    du = cos(theta-phi)
     
  17. Oct 3, 2012 #16

    SammyS

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    They're not really identities, and they're almost correct.

    du = cos(theta-phi) d(theta)
     
  18. Oct 4, 2012 #17
    I tried to do this and checked my math with Wolfram Alpha. I'm confused on one of the steps

    BzeDd.png
    *for some reason I input x-1 but it kept putting 1-x*

    My question is how does this work?
    sin(2p) ----> s√(1-s^2 )
     
    Last edited: Oct 4, 2012
  19. Oct 4, 2012 #18
    So I checked my work on Wolfram and I think this is correct. But when I try to plug in values, they don't work.

    I'm not sure what I'm doing wrong.

    Here is my work

    7BSL6.png
    CLU0R.png
     
  20. Oct 4, 2012 #19

    SammyS

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    If you're asking in regards to sin(2sin-1(s)), then the double angle formula for sine.

    cos(sin-1(s))=√(1-s2) .
     
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