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Cartesian/Polar Integration

  1. Nov 17, 2013 #1
    I'm confused with limits of integrating and which to integrate first,

    I've been getting by so far with just knowing the following,
    [itex] \int_0^1\int_0^y f(x,y) dxdy = \int_0^1\int_0^{rsin\theta} f(rcos\theta,rsin\theta) rdrd\theta [/itex]

    But what happens when we switch it around?
    [itex] \int_0^1\int_0^{x} f(x) dydx = ...? [/itex]

    Will it still be the same? That is,
    [itex] = \int_0^1\int_0^{rcos\theta} f(rcos\theta,rsin\theta) rdrd\theta [/itex]

    Thank you for your time!

    edit: I suppose it would be, since the assignment of the particular characters x and y are arbitrary. I'm still confused since I'm not showing it explicitly. I need to review Jacobians I suppose, very short on time this weekend though.
    Last edited: Nov 17, 2013
  2. jcsd
  3. Nov 17, 2013 #2


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    hi tolove! :smile:
    the limits and the jacobian are two separate matters

    finding the correct limits is simply a question of drawing the region, and making sure the new limits give the same region

    your first limits were for 0 ≤ x ≤ y ≤ 1

    your second limits are for 0 ≤ y ≤ x ≤ 1 …

    so how should you change them to get the same region? :wink:
  4. Nov 17, 2013 #3
    What if it's too complicated to draw by hand, or I'm unable to recognize the function?
  5. Nov 17, 2013 #4


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    not gonna happen …

    it'll always be f(y) ≤ x ≤ g(y) and a ≤ y ≤ b :wink:
  6. Nov 17, 2013 #5
    I'm unable to recognize the function rather often. For example,

    [itex] \int_0^1 \int_0^{\sqrt{(1-x^2)}} e^{-x^2-y^2} dydx [/itex]
  7. Nov 17, 2013 #6


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    ok, thats 0 ≤ y ≤ √(1 - x2) ≤ 1

    ie 0 ≤ y2 ≤ 1 - x2 ≤ 1,

    which becomes … ? :smile:
  8. Nov 17, 2013 #7
    Alright, I got it! Thank you very much! There wasn't any graphing involved in this, though? It's all shown algebraically with those neat little inequalities. I suppose there are cases where traps are present, however.

    r = 0..1, theta=0..Pi/2

    Edit: Ahh! When you say graph, you mean the limits, not the actual function being integrated?
    Last edited: Nov 17, 2013
  9. Nov 17, 2013 #8


    Staff: Mentor

    Yes, the limits of integration. They define the region over which integration is to take place. It's very important to have a good understanding of what this region looks like, especially if you're changing from Cartesian to polar or vice-versa.
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