Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change Of Variable

  1. Jul 3, 2008 #1
    How to calculate, non-graphically, the new limits of integration when change of variables r done to a double integral?
     
  2. jcsd
  3. Jul 3, 2008 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Perhaps if you posted a specific example, we could help you out?
     
  4. Jul 3, 2008 #3
    Plz wait 4 a day., its midnight in INDIA n ma brain had almost slept. I 'ill b vhappy if u could solv ma prob...... Thanks in advance!!!
     
  5. Jul 3, 2008 #4
    Wow...what kind of english do they learn in India?

    To answer the question, it always helps to draw a graph but if you post a specific example we'll show you how to use some algebra to find the new limits.
     
  6. Jul 4, 2008 #5
    "inte[x=a to b] [y=c to d] f(x,y) dxdy = inte[r=sqrt(a^2+c^2) to r =sqrt(b^2+d^2)] [ß=arctan(c/a) to arctan(d/b)] f(rcosß,rsinß) rdrdß "
    is this expression currect ???
     
  7. Jul 4, 2008 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Edit: I see that I was mistaken.
     
    Last edited: Jul 4, 2008
  8. Jul 4, 2008 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]\int_{x=a}^b\int_{y= c^d} f(x,y)dxdy[/tex]
    [tex]\int_{r=\sqrt{a^2+ c^2}}^{\sqrt{b^2+ d^2}}\int_{\theta= arctan(c/a)}^{arctan(d/b)} f(rcos(\theta}, rsin(\theta)) r dr d\theta[/tex]

    No those are not the same. The first is over a rectangle in the plane while the second is over a portion of an annulus- that is, over the region between two circles bounded by two angles. Those will always be what you get if your limits of integration in rectangular and polar coordinates are all constants.

    In general, the limits of integration for the "inside" integral will depend on the other variable. If you really had to put that first integral into polar coordinates (not a good idea because of the lack of symmetry) you would probably do best to break it into four areas: draw lines from the origin to the four vertices and do a separate integral on each of those areas in order to avoid the "discontinuity" at the corners. For example, the first integral might be with [itex]\theta[/itex] between arctan(c/b) and arctan(c/a) to get the region between those two lower vertices (I am assuming that d-c> b- a. Otherwise you will "hit" the vertex (b,d) before (a,c).) Now on each line between those, r must go from the lower line, y= c, to the vertical line x= b. In polar coordinates, that is [itex]r sin(\theta)= c[/itex] to [itex]r cos(\theta)= b[/itex]. That means that r varies from [itex]c/sin(\theta)[/itex] to [itex]b/cos(\theta)[/itex].
    The first of the four integrals would be
    [tex]\int_{\theta= arctan(d/b)}^{arctan(c/a)}\int_{r= c/sin(\theta)}^{b/cos(\theta)}f(r cos(\theta),r sin(\theta)) r dr d\theta[/tex]

    How you change a double integral from one set of coordinates to another depends very strongly on the specific geometry of the situation.
     
  9. Jul 4, 2008 #8
    Thanx !!! , both u geniees ......
     
  10. Jul 4, 2008 #9
    Plz show me how u arrived at the new limits '' "
     
  11. Jul 4, 2008 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I DID:
    Those values are the angle of the straight line from the origin to each of the two points (b,c), (a,c).

     
  12. Jul 5, 2008 #11
    If r varies frm c/sinß to b/cosß ,
    what are the values of ß in both the cases ???
     
  13. Jul 5, 2008 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It is a variable. I said before that taking constants for all limits of integration will give only very limited areas- in the case of polar coordinates it would be a part of an annulus. For more general areas- and in polar coordinates a rectangle is a "general area", only the limits of the "outer integral" must be constants. The limits of integration of the "inner integrals" may be functions of the variiables that still remain.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Change Of Variable
  1. Change of variables. (Replies: 3)

  2. Change of variable (Replies: 1)

  3. Change of variable (Replies: 5)

  4. Change of variable (Replies: 7)

Loading...