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Polar coordinates

  1. Dec 3, 2011 #1
    i just need to know something about the integral with polar coordinates , to know the interval of the teta angle of any domain the proffesor said that we put the pen on the x-axis and move it , i for one moment was not focusing and the doctor had to go to another class can any one explain ??? thank you
  2. jcsd
  3. Dec 3, 2011 #2


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    Well, I have no clue what is meant here, or even what question you are asking. Perhaps he/she was saying that the angle, for a point (x,y) in Cartesian coordinates, is the angle the line through (0, 0) and (x, y) makes with the x-axis. That is, you measure the angle from the x-axis to that line.
  4. Dec 3, 2011 #3
    i mean if you have a graph and you want to change your coordinates from cartesian to polar so that the double intergral is easier
    okay here is an example a semi circle of center O(2,0) if i want to find teta in polar coordinates it will be 90 degree why
  5. Dec 6, 2011 #4
    Any help is much appreciated

    View attachment 41642

    A= 1/2 ∫ [(4 sin⁡θ )^2-(2)^2 ]dθ

    4 sin⁡θ=2 θ=π/6 ,5π/6
  6. Dec 6, 2011 #5


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    What, exactly, are you asking? How they got those limits of integration?

    You have two circles, one with center at (0, 0) and radius 2 and so equation [itex]x^2+ y^2= 4[/itex], the other with center at (0, 2) and radius 2 and so equation [itex]x^2+ (y- 2)^2= x^2+ y^2- 4y+ 4= 4[/itex]. Since those are both equal to 4, they are equal to each other: [itex]x^2+ y^2= x^2+ y^2- 4y+ 4[/itex] which reduces to 4y= 4 and y= 1. Then we have [itex]x^2+ y^2= x^2+ 1= 4[/itex] so [itex]x^2= 3[/itex] and [itex]x= \pm\sqrt{3}[/itex].
    The two points of intersection are [itex](\sqrt{3}, 1)[/itex], [itex](-\sqrt{3}, 1)[/itex].

    In terms of polar coordinates, [itex]\theta= tan^{-1}(y/x)[/itex] which gives [itex]\theta= tan^{-1}(1/\sqrt{3})= \pi/6[/itex] and [itex]\theta= tan^{-1}(-1/\sqrt{3})= 5\pi/6[/itex].
  7. Dec 6, 2011 #6
    Thanks for replying back soo fast I appreciate it i was trying to find the shade parts are in attachment I also think I have it figured out but wouldnt mind a second pair of eyes to look over it. See attachments


    Thanks again
  8. Dec 6, 2011 #7
    ok nice exercise now i got it
  9. Dec 7, 2011 #8


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    How did the "[itex]-8 cos(2\theta)+ 4[/itex]" on the fifth line become "[itex]-4 cos(2\theta)+ 4[/itex]" on the sixth line?

    On the very next line, after integrating you have [itex]-4 sin(2\theta)+ 4\theta[/itex] which is correct because the integral of [itex]cos(2\theta)[/itex] is [itex](sin(2\theta))(1/2)[/itex] but there is no reason to divide by 2 before the integration.
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