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Math Help - [SOLVED] polar coordinates

  1. #1
    Bop
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    [SOLVED] polar coordinates

    Hello, I don't know how I could represent the are that there is between these two functions (from 0 to 1) of the attached image to be able to integrate it so:

    cartesian coordinates it would be:

    A={(x,y): y\ge0; x^2+y^2\le1; x^2+y^2-2x\le0 }

    so:

    \int^{1/2}_0 \int^{\sqrt{2x-x^2}}_0 + \int^{1}_{1/2} \int^{\sqrt{1-x^2}}_0

    How could I write it in polar coordinates?



    Thank you very much!
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  2. #2
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    Quote Originally Posted by Bop View Post
    \int^{1/2}_0 \int^{\sqrt{2x-x^2}}_0 + \int^{1}_{1/2} \int^{\sqrt{1-x^2}}_0

    How could I write it in polar coordinates?
    Correctly, how else?

    1) Comprehend what it is you are doing. You do seem to have a grip on the two unit circles, one centered at the Origin and the other centered at (1,0).

    2) You appear to have the right area defined, but that is the hard way. ALWAYS exploit symmetries. The second integral looks easier, so why not just double that one and leave the other alone?

    3) You need equations of the two circles in polar coordinates.

    The one centered at the origin in trivial: r = 1
    The other is a little trickier, but realizing that r > 0, we have r = 2\cos(\theta)

    4) We'll need a point of intersection. You know already that this is at x = 1/2. Since it's a Standard Unit Circle, only a little thought is required to realize that this intersection is at y = \sqrt{3} / 2 This is a very well known location on the Standard Unit Circle.

    5) You will have to recognize that the transformation to polar is not direct. You will need two integrals, but they will cover different areas. Whereas, in Cartesian coordinates, each integral covered exactly half the area, in Polar it is quite lop-sided.

    6) We're all done but the setting up. Let's see what you get.
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  3. #3
    Bop
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    Hey! Thanks for answering.

    I have:

     \int^{1/2}_0 \int^{\sqrt{2x-x^2}}_0 f(x,y) dy dx + \int^{1}_{1/2} \int^{\sqrt{1-x^2}}_0 f(x,y) dy dx = \int^{\pi/3}_0 \int^1_0  f(r,\theta) r dr d\theta + \int^{\pi/2}_{\pi/3} \int^{2cos\theta}_1 f(r,\theta) r dr d\theta


    Right?


    Thank you again.
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  4. #4
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    Very good, but I'd still like to see you delete one of the Cartesian Coordinate pieces and simply double the one remaining.

    What else have you?
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  5. #5
    Bop
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    Hello,

    I don't see how I could simplify the integrals, I think I can't take advantage of symmetry because the function we integrate f(x,y) it is not necessarily constant so it is not the same integral from cero to 1/2 that integral from 1/2 to 1, maybe I have misunderstand you.


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  6. #6
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    Fair enough. I was assuming a function with symmetry identical to the region.

    Extra credit for arguing and winning your point!
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  7. #7
    Bop
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    Ok, thank you for helping!
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