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Double integral help

  1. Oct 29, 2008 #1
    can someone please give me some help on this integral. it should be solvable analytically.

    [tex]\int^{1}_{-1}\int^{2\pi}_{0}\frac{1}{2+x+\sqrt{1-x^2}cos(y)+\sqrt{1-x^2}sin(y)}dxdy[/tex]
     
  2. jcsd
  3. Oct 30, 2008 #2

    Mark44

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    This one might be easier to work in polar coordinates:
    x = r*cos([tex]\theta[/tex])
    y = r*sin([tex]\theta[/tex])
    dxdy = rdrd[tex]\theta[/tex]

    You'd want to change the limits of integration, also, to represent the rectangle R={(x,y): 0 <= x <= 2 [tex]\pi[/tex], -1 <= y <= 1} in its polar form.

    I haven't worked this problem, but that's what I would try.
     
  4. Oct 30, 2008 #3
    i don't think the integral is in cartesian coordinates. x ranges from (-1,1) and y from (0,2pi). i'm not sure what this coordinate system is called, but it's different. so ur suggestion wouldn't work.
     
  5. Oct 30, 2008 #4

    Mark44

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    It's ***definitely*** in Cartesian coordinates, as evidenced by dx and dy. The fact that y ranges from 0 to 2pi has nothing to do with whether this is or isn't in Cartesian coordinates.
     
  6. Oct 30, 2008 #5
    sorry, i did a variable change because i tried to save time by just using x and y instead of greek symbols mu and gamma.

    i hope that clears it up that they are not cartesian coordinates.
     
  7. Oct 30, 2008 #6

    Mark44

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    No, it does not.

    You can start with a double integral over a closed and bounded region R and get this integral:
    [tex]\int \int [/tex] f(x, y) dA
    The Cartesian iterated integral has the form [tex]\int \int [/tex] f(x, y) dx dy.
    The polar iterated integral has the form [tex]\int \int [/tex] f(r, [tex]\theta[/tex]) r dr d[tex]\theta[/tex].

    It doesn't matter if you use mu and gamma instead of x and y, it's still Cartesian.
     
  8. Oct 30, 2008 #7

    gabbagabbahey

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    Do you really have to solve this by hand, analytically? If you are allowed to use Mathematica, I would definitely consider it....anyways, I haven't solved this by hand but if you choose to do so, you will have to do the y-integral first (unless you can find a useful change of variables)....Mathematica's solution for the y-integral involves Cos(y/2) and Sin(y/2) terms, so using the double angle formula might help you.

    I may be able to come up with something more useful if you show me the original problem; there may be a way to avoid this particular integral altogether.
     
  9. Oct 30, 2008 #8

    gabbagabbahey

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    There is no way to determine the nature of the Co-ordinate system from the given integral...for example, what if your [itex]f(r,\theta)=g(r,\theta)/r[/itex]? then the integral looks like [itex]\int g(r,\theta)dr d \theta[/itex] and a simple renaming of r to x and theta to y could make it look like the given integral.

    That being said, I would guess based on the form of the integrand that if this integral represents some sort of physical application; it is given in curvilinear coordinates. In either case, it is unimportant since the integrand is a scalar for this problem.
     
  10. Oct 30, 2008 #9

    Mark44

    Staff: Mentor

    I stand corrected...
     
  11. Oct 31, 2008 #10
    the original integral was [tex]\int_{4\pi}\frac{1}{1+A.\Omega}d\Omega[/tex] where A is a vector and [tex]\Omega[/tex] is the solid angle.
     
  12. Oct 31, 2008 #11

    gabbagabbahey

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    What is the vector [itex]\vec{A}[/itex]? And what surface is [tex]\Omega[/tex] the solid A angle of (sphere, cone...etc)?
     
  13. Oct 31, 2008 #12
    to make it simple, vector A = (1,2,3) in the cartesian coordinates and it is the solid angle over a sphere. i'm pretty sure there's a way to simplify this. i've tried multiplying top and bottom by 1+A.Omega and then separating them into two integrals. One is an even function and one is an odd function. So the odd function drops to zero and I am left with:

    [tex]2.\int_{A.\Omega}\frac{1}{1-(A.\Omega)^2}d\Omega[/tex]

    i'm not sure if i made it worst!
     
    Last edited: Oct 31, 2008
  14. Oct 31, 2008 #13

    gabbagabbahey

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    Usually, the solid angle for a sphere is just the scalar quantity [itex]\Omega=4 \pi[/itex] and [itex] d \Omega= sin(\theta) d \theta d\phi[/itex] in the usual spherical coordinates... Perhaps you'd better explain to me what expressions you are using, and how you are taking he dot product of a vector with a scalar????
     
  15. Oct 31, 2008 #14
    oh ok. [tex]\Omega = (\mu,\sqrt{1-\mu^2}cos(\gamma),\sqrt{1-\mu^2}sin(\gamma))[/tex] in the Cartesian coordinate.
     
  16. Oct 31, 2008 #15

    gabbagabbahey

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    hmmm...I've never seen the solid angle expressed as a vector before.. what are [itex]\mu[/itex] and [itex]\gamma[/itex] and what expression are you using for [itex]d \Omega[/itex]?
     
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