1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Converting double integral to polar coordinates

  1. Aug 4, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\int(rsin2\vartheta)drd\vartheta[/tex]
    sorry i dont see how to put the bounds in but they are 0<[tex]\vartheta[/tex]<[tex]\pi[/tex]/2 and 0<r<2acos[tex]\vartheta[/tex]

    2. Relevant equations
    I know that r=sin[tex]\vartheta[/tex]


    3. The attempt at a solution

    Im really not sure where to start my text is terrible. I really dont know how to treat it with the double angle.

    thanks for any help

    sorry meant to put converting polar to rectangular.
     
    Last edited: Aug 4, 2009
  2. jcsd
  3. Aug 5, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Geekchick! :smile:

    (type \int_0^{\pi /2}\int_0^{2acos\theta} :wink:)

    Hint: use sin2θ = 2sinθcosθ :wink:
     
  4. Aug 5, 2009 #3

    jmb

    User Avatar

    I don't think you mean that... (!)

    Well there are basically three things you need to do:
    • Sketch the area of integration and try and work out how you would describe it in rectangular rather than polar coordinates.
    • Convert between integrating infinitesimal units of polar area to infinitesimal units of rectangular area (you can do this either geometrically or mathematically).
    • Convert the integrand into rectangular coordinates --- tiny-tim has already given you a hint on this.

    Apologies if you already knew this! Post your attempts to these issues if tiny-tim's hint isn't enough and somebody will give you additional hints...
     
  5. Aug 5, 2009 #4
    Can I work backwards from the definition [tex]\intf(x,y)dxdy[/tex] = [tex]\intf(rcos\vartheta,rsin\vartheta)rdrd\vartheta[/tex] and then I can say rdrd[tex]\vartheta[/tex]=dA and so all I have is [tex]\int\intsinydxdy[/tex]? And I'm terrible at bounds so where do I start there?
     
  6. Aug 5, 2009 #5

    Mark44

    Staff: Mentor

    What exactly is the problem? Is it to evaluate this integral?
    [tex]\int_{\theta = 0}^{\pi /2} \int_{r = 0}^{2a cos \theta} cos(2 \theta) r dr d\theta[/tex]
    If that's the problem, then all you need to do is evaluate the integral.

    If you're trying to convert this to an iterated integral in rectangular (Cartesian) coordinates, which is not at all obvious from the information you've given, then yes dA = dx dy = r dr d[itex]\theta[/itex], and x = r cos [itex]\theta[/itex], y = r sin [itex]\theta[/itex].

    The region over which you're integrating is a half-circle of radius 2a, with center at (a, 0) in Cartesian coordinates. Keep in mind, though, that your integral is not the area of this region.
     
  7. Aug 5, 2009 #6
    sin instead of cos but yes. my instructions are to convert to Cartesian and then evaluate. But I'm not sure what the integral will be and with what bounds once I convert it...evaluating should be easy.
     
  8. Aug 5, 2009 #7

    Mark44

    Staff: Mentor

    Re: Converting double integral to Cartesian coordinates

    This half-circle determines the limits of integration, whether as a polar iterated integral or a Cartesian iterated integral. Can you find its equation in terms of Cartesian coordinates?
    Also, can you convert sin(2[itex]\theta[/itex]) to Cartesian form?

    These are what you need to do to get the Cartesian integral.
     
  9. Aug 6, 2009 #8

    jmb

    User Avatar

    While this is right, I would really really suggest you don't just treat this as a definition but try to understand where it has come from. For a start this will mean it's not something you need to memorise but can just work it out if you ever forget, but more importantly it will help you to understand geometrically what is happening in integrals like this.

    What I'm basically asking is, do you understand why it should be [itex]r\;{\rm d}r\;{\rm d}\theta[/itex] and not just [itex]{\rm d}r\;{\rm d}\theta[/itex]?

    Another poster has already told you what the bounds mean geometrically, but again I would strongly encourage you to try and sketch it out and make sure you can see this for yourself (of course you should always verify it mathematically after making an initial sketch).

    Once you have a sketch, you need to find a way to parametrise the area of integration in Cartesian coordinates. To understand this it may be worthwhile looking through some simpler examples first. A good way to start to get a feel for the way you parametrise areas in integral bounds is to do some exercises in just changing the order of integration with Cartesian double integrals. Once you are happy with this then move on to simple Cartesian<->Polar transformations and get a good feel for that too. I've attached a few examples. Why don't you work through them and then see if this has taught you enough to solve the bounds problem here. If not we can provide additional hints...
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Converting double integral to polar coordinates
Loading...