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Need help with Polar Coordinates, esp. for integration

  1. Sep 22, 2009 #1
    Any web resources regarding changing the variable of integration from cartesian to polar coordinates that goes beyond the basic :

    x = r cos theta
    y = r sin theta
    r = sq rt (x^2 + y^2)


    I totally don't get how to find the limits of integration using polar coordinates and my undergrad textbooks appears to assume we already know it.

    E.g. r = 2 sin 2 theta , first quadrant. How do we find the limits of integration for r?

    Totally clueless. Tried googling but keep getting just the 3 definitions above. What is the general technique or algorithm?
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 22, 2009 #2

    tiny-tim

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    Hi Legendre! :smile:

    (have a theta: θ and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    The limits when you use r and θ depend on what they are when you use x and y.

    If it's for all x and y, then you use 0 ≤ r < ∞, and 0 ≤ θ ≤ 2π (or -π ≤ θ ≤ π if you prefer).

    If it's for the first quadrant, then you use 0 ≤ r < ∞, and 0 ≤ θ ≤ π/2.

    Whatever it is in x and y, you have to convert …

    do you have an example that you want to try? :smile:
     
  4. Sep 22, 2009 #3
    Hi Legendre
    Since you work is in the first quardrant, then θ takes the values from 0 to π/2
    and r takes the values from 0 _which corresponds to θ=0 i.e r= 2 sin (2 *0)=0 _
    to r= 2 sin (2 θ) since r is a variable and the integral is of the form:
    π/2 2sin(2θ) π/2
    ∫ ∫ dr dθ = ∫ 2sin(2θ) dθ
    0 0 0
    π/2
    = 2(-(1/2)cos(2θ)] = -[cos π -cos 0]=-(-1-1)=2 units of area
    0
    Best Regards
    Riad Zaidan
     
  5. Sep 23, 2009 #4
    r = 2 sin (2θ) in the first quadrant.

    How do we know the lower limit of integration for r is 0?
    How do I know the upper limit of integration for r is 2 sin (2θ)?

    Is this done by sketching the area? I have no idea how to begin sketching r = 2 sin (2θ)!


    # Another question : How do you sketch first quadrant, limits of integration r = 1/cos θ and r = 0?


    Note : I am a lot more eager to find out how we arrive at the answer than the answer itself.

    Thanks for all the help! :)
     
  6. Sep 23, 2009 #5

    tiny-tim

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    (I take it you mean the interior of that curve?)

    ok … since you're restricted to the first quadrant, θ is restricted to [0,π/2].

    And r = 2 sin (2θ) is obviously single-valued for any particular θ (and r = 0 at θ = 0 and π/2) …

    so r is restricted to [0, 2sin(2θ)] for each θ

    So the limits are ∫0π/202sin(2θ) :smile:

    (do you see that you have to pick the coordinates off one at a time? … in this case, do the θ limits first, then for a fixed θ, do the r limits)
    sin(2θ) obviously goes from 0 up to 1 and down to 0, so without sketching it you know it starts at the origin, goes away, and comes back (once).
    1/cosθ goes from 1 upwards to ∞ … if you want to sketch it, just do a freehand curve through (in (x,y) coordinates) (1,0) and (0,∞) :wink:
     
  7. Sep 23, 2009 #6
    Hi tiny-tim
    thanks for the detailes
    Riad Zaidan
     
  8. Sep 24, 2009 #7
    Thanks a lot everyone!

    One more question :

    Regarding the graph of r = 2 sin (2θ), I know how to sketch it if its y = 2 sin 2x but in polar coordinates, we're plotting r on the "x-axis" and θ on the "y-axis" (right??).

    So doesn't this mean it is something like sketching x = 2 sin (2y) on the regular vertical y-axis and horizontal x-axis?

    Can I simply treat r as the vertical axis and θ as the horizontal axis? Then r = 2 sin (2θ) is very easy to plot.
     
  9. Sep 24, 2009 #8
    Pardon me but I don't quite get what you are saying.

    Keeping it in the first quadrant. i.e. 0 <= θ <= π/2.

    Are you saying that if r = f(θ) is single-valued for any particular θ, then r is restricted to the interval [f(0), f(θ)]?

    What do I do with a general r = f(θ) if I know 0 <= θ <= π/2?
     
    Last edited: Sep 24, 2009
  10. Sep 24, 2009 #9

    tiny-tim

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    No no no no noooooo!

    Lines of constant r are circles, and lines of constant θ are straight lines through the origin ("spokes").

    You need to go back to your book and to study this.

    Alternatively, google "graph polar coordinates" and look at sites such as http://www.ies.co.jp/math/java/calc/sg_kyok/sg_kyok.html" [Broken]
     
    Last edited by a moderator: May 4, 2017
  11. Sep 24, 2009 #10
    arghhhh...i hate polar coordinates lol.

    yeah, of course we can't interchange r and θ. r = 2 sin (2θ) is a circle in (r,θ) but is a transformed sin curve in (θ,r). lol what was i thinking.

    yes i need to go back to my book and study this. any books to recommend? i looked through Binmore & Davies (Calculus), Ostaszewski (Advanced Mathematical Methods), Sydsaeter (Further Mathematics for Economic Analysis) as well as two set of lecture notes on line but they all just touch on the basics briefly. any good basic calculus book that has a lot of materials on converting and integration in polar coordinates beyond the basics?


    # regarding the site : very useful, thanks a lot!
     
    Last edited by a moderator: May 4, 2017
  12. Sep 25, 2009 #11

    tiny-tim

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    ?? Polar coordinates are fun when you get used to them! :biggrin:
    Sorry, I don't know.

    You could try putting out an appeal in the Math & Science Learning Materials forum at https://www.physicsforums.com/forumdisplay.php?f=151".

    But maybe it's best just to browse the internet until you find something you like. :smile:
     
    Last edited by a moderator: Apr 24, 2017
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